Exercises¶
Here are a selection of exercises covering various aspects of the course material. Their aim is to make you think about how to solve a problem using code. These are not assessed, but you are encouraged to try these out, because practice is the best way to learn how to code!
In several cases there are already exist functions, e.g., in NumPy, for performing some of these exercise problems. While generally you should use existing functions from well maintained libraries (they will be very well tested and robust), here (unless asked to use an appropriate library) the aim is for you to think about how you would code up the function yourself.
Python basics¶
Exercise 1¶
Part 1
Open a Python/IPython terminal and declare two variables as floating point numbers. Add the two variables and store the output as a new variable. Print out the new variable value to the screen.
Solution
$ ipython
>>> x = 1.2774392 # variable containing a floating point number
>>> y = -3.4374323 # another variable containing a floating point number
>>> z = x + y # add the two variables and store result in z
>>> print(z) # print the result z to the screen
Part 2
Print out the resulting variable to 3 decimal places.
Solution
Several options exist, e.g.,
>>> print("{0:.3f}".format(z))
or
>>> print(f"{z:.3f}")
or
>>> print("%.3f" % z)
Part 3
Perform the same task, but this time write the code in a text file saved with the .py
extension. Run the code in VS Code and also from the command line.
Exercise 2¶
Question
In a Python/IPython terminal, or in a script, import an appropriate library to calculate the sine of a list of angles that are given in degrees.
Solution
A method using the math
library is:
import math
# make a list of angles (assumed to be in degrees)
angles = [0, 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, 165, 180]
sines = []
# loop over angles
for angle in angles:
rad = math.radians(angle) # convert angle to radians
# rad = angle * math.pi / 180 # an alternative
sines.append(math.sin(rad)) # calculate sine an append to list
A different method using the NumPy
library is:
import numpy as np
rads = np.deg2rad(angles) # convert angles to radians
sines = np.sin(rads)
Exercise 3¶
Part 1
Given a list of numbers, without using the built-in
max()
method, find the maximum value
in the list. Try two different ways of looping through the list values.
Solution
Two different options are:
x = [-2, 3, 6, -5, 7, 8, 12, -9]
maximum = x[0] # use first value as initial comparitor
for v in x[1:]:
if v > maximum:
maximum = v # update comparitor
or (using the range()
and
len()
functions):
maximum = x[0]
for i in range(1, len(x)):
if x[i] > maximum:
maximum = x[i]
Part 2
Given two equal length lists of numbers, create a new list containing the sum of the pairs of values from those lists. Try two different methods of doing this.
Solution
Two different options are:
x = [-2, 3, 6, -5, 7, 8, 12, -9]
y = [0, 12, -5, 2, 8, -8, 11, 2]
z = []
for i in range(len(x)):
z.append(x[i] + y[i])
or (using the zip()
function):
z = []
for vx, vy in zip(x, y):
z.append(vx + vy)
Part 3
Given two equal length lists of numbers, add the numbers from the first list onto the values in the second list (do not create a new list).
Solution
Using the enumerate()
function
you can do:
x = [-2, 3, 6, -5, 7, 8, 12, -9]
y = [0, 12, -5, 2, 8, -8, 11, 2]
for i, v in enumerate(x):
y[i] += v
Exercise 4¶
Part 1
Use list comprehension to generate a list containing the square roots of all integers between 1 and 50.
Solution
import math
sqroots = [math.sqrt(x) for x in range(1, 51)]
Part 2
Now use list comprehension to generate the square roots of only even numbers.
Solution
import math
sqroots = [math.sqrt(x) for x in range(1, 51) if x % 2 == 0]
Exercise 5¶
Question
Given a dictionary containing the following information
personaldata = {
"firstname": "Sammy",
"lastname": "Scissors",
"housenumber": 2,
"streetname": "Long Lane",
"city": "Lancaster",
"postcode": "LA1 8TF"
}
construct a single format string that will allow the address to be output in the form:
Sammy Scissors
2 Long Lane
Lancaster
LA1 8TF
Solution
A possible solution is
addressstring = (
"{firstname} {lastname}\n"
"{housenumber} {streetname}\n"
"{city}\n"
"{postcode}\n"
)
print(addressstring.format(**personaldata))
Exercise 6¶
Part 1
Create an \(3 \times 4\) (3 rows, 4 columns) array using a list of lists, where each value in the array is initialised to be 1.
Solution
The best way is to use list comprehension, e.g.,:
x = [[1 for _ in range(4)] for _ in range(3)]
Part 2
Set the corner values of the array to be zero.
Solution
x[0][0] = 0
x[-1][0] = 0
x[0][-1] = 0
x[-1][-1] = 0
Exercise 7¶
Part 1
Create a 2D \(3 \times 3\) matrix of numbers (using lists). Loop over the rows in the matrix and print out the sum of each row.
Solution
x = [[0.1, 0.5, 1.2], [-2.3, 4.5, 0.3], [5.7, -0.3, 1.4]]
# loop over each row
for row in x:
print(sum(row))
Part 2
Now loop over the columns and print out the products of each column.
Solution
# loop over each column
for i in range(len(x)):
# column has to be explicitly extracted
column = [row[i] for row in x]
product = 1.0
for cv in column:
product *= cv
print(product)
Note that this is easier with NumPy arrays, for which the rows and column can be transposed, e.g.:
import numpy as np
y = np.array(x)
for row in y:
print(np.sum(row))
for col in y.T: # transpose of y
print(np.prod(col))
Exercise 8¶
Part 1
Given the following dictionary:
charge = {
"electron": -1,
"positron": 1,
"proton": 1,
"neutron": 0,
"up": 2/3,
"down": -1/3
}
get a list of the dictionary keys and a list of the dictionary values.
Solution
The lists of keys and values can be extracted with:
keys = list(charge.keys()) # need to explicitly convert iterator to list
values = list(charge.values())
Part 2
Add a new particle and its charge to the dictionary.
Solution
charge["strange"] = -1/3
Part 2
Find the number of positively charged particles in the dictionary and get a list of their names.
Solution
One option is:
npos = 0 # counter for positive particles
positive = [] # list of positive particles
for particle, c in charge.items():
if c > 0:
npos += 1
positive.append(particle)
Another method, using list comprehension, is:
positive = [c for c in charge.values() if c > 0]
npos = len(positive)
Exercise 9¶
Part 1
Create a dictionary with three keys "a"
, "b"
and "c"
, where each key value is an empty list.
Solution
data = {"a": [], "b": [], "c": []}
# alternative
# data = dict(a=[], b=[], c=[])
Part 2
Add another key, "d"
, into the dictionary that is also an empty list.
Solution
data["d"] = []
Part 3
For each string in the list:
alpha = ["aadb", "bbcd", "aaaa", "bccc", "dddd", "cbcb", "daca"]
append the numbers of each letter to the appropriate list in the dictionary.
Solution
# loop over the list
for a in alpha:
# loop over each letter in the dictionary
for letter in data:
# count number of letters in each string
numlet = a.count(letter)
# append to list in dictionary
data[letter].append(numlet)
Exercise 10¶
Question
Suppose you have a set of files containing the results of multiple consecutive experiments/simulations. To distinguish the files each file name is suffixed by an integer with preceding zeros, such the number is always 3 digits long (assuming no more than 1000 files exist), e.g.,:
experimental_results_000.txt
experimental_results_001.txt
...
experimental_results_258.txt
experimental_results_259.txt
Assuming you know how many files you have and the file name format, how might you loop over all the files to read them in?
Solution
A possible solution is:
N = 260 # total number of files
basename = "experimental_results_{0:03d}.txt"
# loop over files and read in results
results = []
for i in range(N):
thisfile = basename.format(i)
# read in the results in some form
with open(thisfile, "r") as fp:
results.append(fp.read())
Python functions¶
Exercise 11¶
Question
In a Python file, write a function that asks the user to input a date in the format
"YYYY-MM-DD" and then prints out the day of the week (see the Python
datetime
library). In another Python
script, or Python/IPython terminal, import the function that you have written and run it.
Solution
Create a Python (called, say, weekday.py
) file containing:
from datetime import datetime
def getweekday():
# ask user for input
datestr = input("Input a date in the format YYYY-MM-DD: ")
# split string into parts
year, month, day = datestr.split("-")
# convert into datetime object
d = datetime(int(year), int(month), int(day))
# get the day of the week
weekday = d.strftime("%A")
# output day of the week
print(f"The date {datestr} was on a {weekday}")
You may want to add a check that the date in is the correct format. This function could then be used with:
$ ipython
>>> from weekday import getweekday
>>> getweekday() # run the function
Exercise 12¶
Part 1
Write a function that takes in a list of numbers as an argument and returns their mean.
Answer
An example of how to do this is:
def mean(values):
s = 0.0 # variable to hold sum of values
for value in values:
s += value
# return the mean
return s / len(values)
You could add some checking that values is indeed a list. You could also use the built-in
Python sum()
function rather
than using the for loop.
Part 2
Write a function that takes in a list of numbers as an argument and returns their standard deviation. Can the function from Part 1 be re-used?
Solution
An example of how to do this is:
def std(values):
# get the mean of the values (re-use the previous function)
mu = mean(values)
# get the variance (re-use mean function again)
var = mean([(x - mu)**2 for x in values])
# return the standard deviation
return var ** 0.5
Part 3
Write a function that takes in list of numbers as an argument and returns the median.
Solution
An example of how to do this is:
def median(values):
# use the built-in sorted function to sort the values in ascending order
sortvals = sorted(values)
# get the halfway index
half = int(len(values) / 2)
# check if values contains an odd or even number of values
if len(values) % 2 == 0:
# an even number, so return average of middle two numbers
return (values[half - 1] + values[half]) / 2
else:
# an odd number, so return middle number
return values[half]
Exercise 13¶
Question
Write a function that:
- takes in a list of strings as an argument,
- finds the unique strings,
- counts the number of occurrences of each of those unique strings in the list
- returns those number counts in a dictionary keyed by the unique string values.
E.g.,
>>> animals = ['cat', 'dog', 'dog', 'dog', 'cat', 'horse']
>>> counts = count_occurrences(animals)
>>> print(counts)
{'cat': 2, 'dog': 3, 'horse': 1}
Solution
A way of doing this is:
def count_occurrences(values):
# get unique strings by converting to a set
unique = set(values)
# create empty dictionary for counts
counts = {}
# loop over unique strings and count occurrences
for word in unique:
count = 0
for w in values:
if w == word:
count += 1
# short method (use count method of a list)
#counts[word] = values.count(word)
return counts
Exercise 14¶
Part 1
Write a function that takes in a list as an argument and returns a new list containing the square of every \(n\)th index (starting at the 0 index), where \(n\) is another argument to the function with a default value of 2.
E.g.,
>>> values = [2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> sq = square_index(values)
>>> print(sq)
[4, 16, 36, 64, 100]
Solution
A way of doing this is:
def square_index(values, step=2):
squ = []
# loop over list in steps of "step"
for i in range(0, len(values), step):
squ.append(values[i] ** 2)
return squ
# short method (using slice notation instead)
#return [x ** 2 for x in values[::step]]
You may want to include checks that values
is a list and that step
is an integer.
Part 1
Alter the function so that it takes in another argument, reverse
, which defaults to False
, but
if True
makes the function return the list in reverse order.
Solution
A way of doing this is:
def square_index(values, step=2, reverse=False):
squ = []
# get indices to return
if reverse:
idxs = range(len(values), 0, -step)
else:
idxs = range(0, len(values), step)
# loop over list in steps of "step"
for i in idxs:
squ.append(values[i] ** 2)
return squ
The slice()
function could also be
used rather than the range()
function.
Exercise 15¶
Part 1
Given a square 2D matrix, e.g.,:
M = [[1.5, 2.1, 3.6, 4.1], [-0.2, 6.1, 7.2, -5.0], [3.4, 10.1, 1.7, 12.9], [-13.0, 1.3, -2.4, 0.8]]
write a function that takes in the matrix as an argument and returns its diagonal elements as a list.
Solution
def diag(M):
"""
Return the diagonal elements of a square 2D matrix.
Parameters
----------
M: matrix
A square 2D matrix
Returns
-------
list:
A list of the diagonal elements of the matrix.
"""
de = []
# loop over length of matrix
for i in range(len(M)):
de.append(M[i][i])
return de
You may want to include tests that the matrix is two-dimensional and square.
Part 2
Write a function to calculate the determinant of the matrix.
Hint: The built-in Python
itertools
module can
calculate permutations. You will also need to calculate the signature (or
parity) of the permutation.
Solution
from itertools import permutations
def sgn(permutation):
"""
Get the signature, or parity of a permutation (based on
https://gist.github.com/lycantropos/217710b0afc40b3031762274275c204a)
of numbers between 0 and N, where N is the length of the permutation
list.
Parameters
----------
permutation: list
A permutation of the numbers from 0 to len(list)
Returns
-------
int:
A 1 for an even permutation, -1 for an odd permutation.
"""
if len(permutation) == 1:
return 1
transitions_count = 0
for idx, element in enumerate(permutation):
for next_element in permutation[idx + 1:]:
if element > next_element:
transitions_count += 1
return 1 if not (transitions_count % 2) else -1
def det(M):
"""
Calculate the determinant of a square 2D matrix.
Parameters
----------
M: matrix
A square 2D matrix
Returns
-------
float:
The determinant value.
"""
# length of matrix
n = len(M)
D = 0.0 # variable to sum up determinant
# get permutations (use Leibniz formula)
for perm in permutations(range(n)):
subD = 1.0
for i in range(n):
subD *= M[i][perm[i]]
# get signature of permutation
D += sgn(perm) * subD
return D
Exercise 16¶
Part 1
Write a function to "bin" a list of numbers, i.e., count how many of the numbers are in each of a set of intervals over the full range (e.g., the bin sizes in a histogram). The function arguments should be the list of numbers, the number of bins (defaulting to 10), and the lower and upper bin edges (if not given by the user these should default to use the smallest and largest number in the input list, respectively).
Try doing this without using NumPy!
Solution
def binned(samples, nbins=10, low=None, high=None):
"""
Count the number of values within a set of bins.
Parameters
----------
samples: list
A list of numbers which will be "binned"
nbins: int
The number of bins into which to split the range of numbers
low: float
The edge of the lowest bin (defaults to the smallest value in `samples`)
high: float
The edge of the highest bin (defaults to the largest values in `samples`)
Returns
-------
tuple
A tuple containing two lists: the bin edges and the number counts in each
bin
"""
# get the bin ranges
if low is None:
low = min(samples)
if high is None:
high = max(samples)
# step size between bins
binstep = (high - low) / nbins
# lists to contain bin edges and number counts
binedges = [low]
bincounts = []
# loop over bins
for i in range(nbins):
# set upper edge of bin
binedges.append(binedges[-1] + binstep)
# count number of samples in bin
bincount = 0
for sample in samples:
if binedges[i] <= sample < binedges[i+1]:
bincount += 1
# add in amy samples that equal max value in the final bin
if i == (nbins - 1) and sample == max:
bincount += 1
bincounts.append(bincount)
return binedges, bincounts
Part 2
Edit the function to take another argument, norm
, which if True
normalises the bin counts
so that the area under the histogram \(A = \sum_i^{N_{\rm bins}} n_i \Delta x\) adds up to 1.
Solution
def binned(samples, nbins=10, low=None, high=None, norm=False):
"""
Count the number of values within a set of bins.
Parameters
----------
samples: list
A list of numbers which will be "binned"
nbins: int
The number of bins into which to split the range of numbers
low: float
The edge of the lowest bin (defaults to the smallest value in `samples`)
high: float
The edge of the highest bin (defaults to the largest values in `samples`)
norm: bool
If True normalise the bin counts (default is False)
Returns
-------
tuple
A tuple containing two lists: the bin edges and the number counts in each
bin
"""
# get the bin ranges
if low is None:
low = min(samples)
if high is None:
high = max(samples)
# step size between bins
binstep = (high - low) / nbins
# lists to contain bin edges and number counts
binedges = [low]
bincounts = []
# total number of samples
nsamples = len(samples)
# loop over bins
for i in range(nbins):
# set upper edge of bin
binedges.append(binedges[-1] + binstep)
# count number of samples in bin
bincount = 0
for sample in samples:
if binedges[i] <= sample < binedges[i+1]:
bincount += 1
# add in amy samples that equal max value in the final bin
if i == (nbins - 1) and sample == max:
bincount += 1
if norm:
# normalise the bin counts
bincount = (bincount / nsamples) / binstep
bincounts.append(bincount)
return binedges, bincounts
Exercise 17¶
Part 1
Write a function that returns a boxcar function of the form:
where the arguments should be a list containing \(x\) values at which to evaluate the function, the limits \(a\) and \(b\) at which the boxcar starts and stops, and the amplitude \(C\). The following defaults (to give a standard rectangular function) should be set: \(a = -1/2\), \(b=1/2\), and \(C = 1\).
Solution
A potential way of doing this is:
def boxcar(x, a=-0.5, b=0.5, C=1):
vals = []
# loop over x-values
for xs in x:
if a <= xs <= b:
# add box
vals.append(C)
else:
# add zeros
vals.append(0.0)
return vals
You may want to include checks that x
is a list, that a
is less than b
, and that C
is positive.
An alternative way using NumPy array operations would be:
import numpy as np
def boxcar(x, a=-0.5, b=0.5, C=1):
# initialise the array filled with zeros and the same length as x
vals = np.zeros_like(x)
# get indices of x-array with the [a, b] range
idx = np.argwhere((x > a) & (x < b))
# fill in vals with C
vals[idx] = C
return vals
Part 2
Write a function to that returns a triangular function of the form:
where \(x_0\) is the midpoint of the triangle and \({\rm d}y/{\rm d}x\) is the gradient of the sides of the triangle. The arguments should be a list containing \(x\) values at which to evaluate the function, the limits \(a\) and \(b\) at which the triangle starts and stops, and the peak triangle amplitude \(C\). The following defaults (to give a standard normalised triangular function) should be set: \(a = -1\), \(b=1\), and \(C = 1\).
Solution
A potential way to do this is:
def tri(x, a=-1, b=1, C=1):
vals = []
# get half-width of the triangle's base
halfwidth = (b - a) / 2
# mid-point of triangle
mid = a + halfwidth
# gradient of triangle sides dy/dx
grad = C / halfwidth
# loop over x-values
for xs in x:
if a < xs < b:
vals.append(C - abs(grad * (xs - mid)))
else:
vals.append(0.0)
return vals
You may want to include checks that x
is a list, that a
is less than b
, and that C
is positive.
An alternative way using NumPy array operations would be:
import numpy as np
def tri(x, a=-1, b=1, C=1):
# initialise the array filled with zeros and the same length as x
vals = np.zeros_like(x)
# get half-width of the triangle's base
halfwidth = (b - a) / 2
# mid-point of triangle
mid = a + halfwidth
# gradient of triangle sides dy/dx
grad = C / halfwidth
# get indices of x-array with the [a, b] range
idx = np.argwhere((x > a) & (x < b))
# fill in indices with a triangle
vals[idx] = C - np.abs(grad * (x[idx] - mid))
return vals
Part 3
Create a plot showing both a boxcar function and a triangle function evaluated over some range of \(x\)-values. The two functions should be plotted in different colours and there should be a legend giving the function names.
Solution
A potential solution is:
from matplotlib import pyplot as plt
import numpy as np
# create a range of x-values
x = np.linspace(-5, 5, 1000)
# get boxcar function (choosing some input values)
b = boxcar(x, a=-4.5, b=-1.5, C=3.5)
# get triangle function (choosing some input values)
t = tri(x, a=-3.1, b=2.5, C=6.0)
# make the plot
plt.plot(x, b, color="r", label="boxcar")
plt.plot(x, t, color="b", linestyle="--", label="triangle")
plt.legend() # add the legend based on the label values
plt.xlim([x[0], x[-1]]) # make x-axis limits stick to the x range
plt.show()
Reading/writing data¶
Exercise 18¶
Part 1
You have a file containing the student grades for 3 different exercises. The file consists of
a header line (denoted by starting with a #
), followed by lines containing four values
separated by commas ("comma separated values", or CSV):
- unique student ID
- grade for exercise 1 (mark out of 20)
- grade for exercise 2 (mark out of 30)
- grade for exercise 3 (mark out of 40)
E.g.,:
# Student ID, Exercise 1 (20), Exercise 2 (30), Exercise 3 (40)
1234, 12, 23, 29
1235, 9, 28, 31
1236, 13, 8, 25
Read in the file and then calculate each student's total mark (as a percentage rounded to the nearest integer), where the three exercises are weighted at 25%, 25% and 50%, respectively. You can used a library such as NumPy to read in the data.
Solution
A possible way without using, e.g., NumPy
resfile = "results.csv" # the results file
maxmarks = [20, 30, 40] # maximum marks for each exercise
weights = [0.25, 0.25, 0.50] # fractional weights for each exercise
# read in results
grades = {}
with open(resfile, "r") as fp:
for line in fp.readlines():
# ignore header lines starting with a #
if line[0] != "#":
# split the line on commas
splitline = line.split(",")
# get student ID (string trailing whitespace)
studentid = splitline[0].strip()
# get grades (converting to integers)
grades[studentid] = [int(grade.strip()) for grade in splitline[1:]]
# calculate final weighted grades
finalgrades = {}
for studentid in grades:
finalgrade = 0.0
for i in range(len(maxmarks)):
finalgrade += weights[i] * (grades[studentid][i] / maxmarks[i])
finalgrades[studentid] = round(finalgrade * 100) # convert to percentage
This can be more compact using NumPy's
loadtxt()
function or
the more complete
genfromtxt()
function, e.g.,:
import numpy as np
resfile = "results.csv" # the results file
maxmarks = [20, 30, 40] # maximum marks for each exercise
weights = [0.25, 0.25, 0.50] # fractional weights for each exercise
results = np.loadtxt(
resfile,
comments="#", # ignore header (could also use skiprows=1)
delimiter=",", # comma separated values
)
# calculate final weighted grades
gradevalues = np.round(
100 * sum(weights[i] * results[:,i+1] / maxmarks[i] for i in range(len(weights)))
)
for i in range(len(results)):
studentid = str(results[i, 0]) # convert ID to string
finalgrades[studentid] = gradevalues[i]
Another option would be to use the
pandas
read_csv()
function.
Part 2
Write the results to a new CSV file with the total grade as a new fifth column.
Solution
There are again multiple ways of doing this. E.g., using the NumPy savetxt()
function
(assuming we have the results
and finalgrades
as above):
import numpy as np
# append final results onto the "results" array
results = np.hstack((results, [grade for grade in finalgrades.items()]))
outputfile = "final_results.csv"
header = "np.Student ID, Exercise 1 (20), Exercise 2 (30), Exercise 3 (40), Final grade (%)"
np.savetxt(outputfile, results, fmt="%d", delimiter=",", header=header)
Exercise 19¶
Part 1
Suppose you have the following class to hold some experimental data consisting of the electric field strength at a set of positions on a uniform grid:
import numpy as np
class ElectricField:
def __init__(self, E, xpos, ypos, label="Efield"):
"""
Store the measured electric field at a set of 2D coordinates.
Parameters
----------
E: array
A 2D array of values of the electric field strength.
xpos: array
A 1D array of the x-positions of the measurements.
ypos: array
A 1D array of the y-positions of the measurements.
label: str
A label/name for the experiment. Default is "Efield"
"""
# store copy of E-field as the E attribute
self.E = np.array(E)
# store x and y positions
self.x = np.array(xpos)
self.y = np.array(ypos)
# check E and grid positions are consistent
if self.E.shape != (len(self.x), len(self.y)):
raise ValueError("Shape of E is not consistent with grid points")
self.label = label
Add a method to this class that saves the class itself as a
pickle file (you can used the NumPy
save()
function with the
allow_pickle=True
option set). It should use the label
attribute to construct the name of
the save file (with appropriate extension added).
Also add a classmethod
to
read in a saved object.
Create an instance of the object and try saving it at reading it back in.
Part 2
Add a new method to the class that will return the electric field interpolated at any point
within the grid. You may want to use the SciPy
interp2d
class. The method should raise an error if trying to interpolate outside the bounds of the \(x\)-\(y\)
grid.
Try reading in a previously saved object and using this new method on that object.
Solution
An example of the class could be:
import numpy as np
from scipy.interpolate import interp2d
class ElectricField:
"""Store the measured electric field at a set of 2D coordinates.
Parameters
----------
E: array
A 2D array of values of the electric field strength.
xpos: array
A 1D array of the x-positions of the measurements.
ypos: array
A 1D array of the y-positions of the measurements.
label: str
A label/name for the experiment. Default is "Efield"
"""
def __init__(self, E, xpos, ypos, label="Efield"):
# store copy of E-field as the E attribute
self.E = np.array(E)
# store x and y positions
self.x = np.array(xpos)
self.y = np.array(ypos)
# check E and grid positions are consistent
if self.E.shape != (len(self.x), len(self.y)):
raise ValueError("Shape of E is not consistent with grid points")
self.label = label
def save(self, fname=None):
"""
Save the class to a NumPy pickle file.
Parameters
----------
fname: str
The output file name for storing the class. If not given the
`label` attribute of the class will be used and ".npy" will be
extension.
"""
if fname is None:
fname = self.label
np.save(fname, self, allow_pickle=True)
@classmethod
def load(cls, fname):
"""
Load a saved file containing an instance of this class. The file will
be a NumPy pickle object with a ".npy" extension.
Parameters
----------
fname: str
The name of the file to load
Returns
-------
ElectricField
An ElectricField object.
"""
E = np.load(fname, allow_pickle=True)
# NumPy load will load the data as a 0-D NumPy array, so extract the
# ElectricField object
E = E.item()
# check it's the correct type
if not isinstance(E, cls):
raise TypeError("Loaded file does not contain an ElectricField")
return E
def field_strength(self, x, y):
"""
Return the electric field strength at any point (interpolated if not
on the original grid).
Parameters
----------
x: float
The x-coordinate position
y: float
The y-coordinate position
Returns
-------
float
The electric field strength at the given point.
"""
# create interpolator
fieldinterp = interp2d(self.x, self.y, self.E, bounds_error=True)
try:
E = fieldinterp(x, y)[0]
except ValueError:
raise ValueError(f"x-y coordinates ({x}, {y}) are outside grid "
"bounds.")
return E
Writing out the class, reading it back in, and using the field_strength()
method could be
done with (assuming the class is defined in a file called efield.py
):
from efield import ElectricField
# get x-y positions
x = [-2, -1, 0, 1, 2]
y = [-2, -1, 0, 1, 2]
# "measure" E-field magnitude on the grid
measured = [
[0.1, 0.2, 0.3, 0.2, 0.1],
[0.15, 0.25, 0.35, 0.25, 0.15],
[0.17, 0.29, 0.40, 0.31, 0.22],
[0.18, 0.31, 0.46, 0.38, 0.30],
[0.18, 0.32, 0.52, 0.51, 0.48]
]
# create class
E = ElectricField(measured, x, y, label="experiment1")
# save field
E.save()
# re-load experiment data
Edata = ElectricField.load("experiment1.npy")
# get field at given point
xp, yp = -1.5, 0.4
ef = Edata.field_strength(xp, yp)
print(f"Field strength at ({xp}, {yp}) is {ef}")
Exercise 20¶
Question
Write a script that plots the two-dimensional data in the file
density.txt
. Use the information contained in the file
data_description.txt
to define the axes of your plot (note
that "clabel" refers to a label for a colour bar).
Solution
A potential solution is:
#!/usr/bin/env python3
import numpy as np
import matplotlib.pyplot as plt
# read density.txt and store data into a numpy two-dimensional array
data_file = 'density.txt'
density = np.loadtxt(data_file)
# read data_description.txt and store information in the proper variables
with open('data_description.txt',encoding='utf-8') as file:
for line in file:
idx = line.index("=")
if "xmin" in line:
xmin = float(line[idx+1:-1])
elif "xmax" in line:
xmax = float(line[idx+1:-1])
elif "ymin" in line:
ymin = float(line[idx+1:-1])
elif "ymax" in line:
ymax = float(line[idx+1:-1])
elif "xlabel" in line:
xlabel = line[idx+1:-1]
elif "ylabel" in line:
ylabel = line[idx+1:-1]
elif "clabel" in line:
clabel = line[idx+1:-1]
# create a pseudocolor plot
fig = plt.figure()
ax = fig.add_subplot(111)
c = ax.imshow(
density,
cmap='viridis',
extent=[xmin, xmax, ymin, ymax],
interpolation='antialiased',
origin='lower'
)
ax.set_xlabel(xlabel, fontsize=20)
ax.set_ylabel(ylabel, fontsize=20)
ax.tick_params(labelsize=20)
cbar = fig.colorbar(c)
cbar.set_label(clabel, fontsize=20)
cbar.ax.tick_params(labelsize=20)
fig.tight_layout()
plt.show()
A different way of reading the data_description.txt
file could be done with:
lims = {}
labels = {}
with open("data_description.txt", "r") as file:
for line in file:
key, value = line.split("=")
# get limit and label values
if value.strip().replace(".", "").isnumeric():
lims[key.strip()] = float(value.strip())
else:
labels[key.strip()] = value.strip()
where the limits and labels are stored in dictionaries. Or, it could be read in using the TOML package.
Python classes¶
Exercise 21¶
Part 1
Create a class to hold data from an astronomical survey. It will store the name of each object and the object's magnitude. Upon initialisation the class should take in a single argument, which is a dictionary keyed to object names with their magnitudes as the values. By default it should be initialised with an empty dictionary if nothing is passed to it. It should raise an error if supplied with something other than a dictionary.
The class should have a method for adding in new objects, and methods for returning lists of the names and magnitudes of all objects in the survey, respectively.
Solution
A possible class is:
class AstroSurvey:
def __init__(self, survey={}):
# create an attribute that will store the survey - it is useful to use a dictionary
self.survey = {}
if not isinstance(survey, dict):
raise TypeError("Input must be a dictionary")
# add copy of all objects into the survey
for key, value in survey.items():
self.add_object(key, value)
def add_object(self, name, magnitude):
"""
Add a new object into the survey.
Parameters
----------
name: str
The name of the object.
magnitude: float
The magnitude of the object.
"""
if not isinstance(name, str):
raise TypeError("The object name must be a string")
if not isinstance(magnitude, (float, int)):
raise TypeError("The object's magnitude must be a number")
# add into survey attribute
self.survey[name] = magnitude
def object_names(self):
"""
Return a list of object names.
"""
return list(self.survey.keys())
def object_magnitudes(self):
"""
Return a list of object magnitudes.
"""
return list(self.survey.values())
This could be used by doing, e.g.,:
# create an empty survey
survey = AstroSurvey()
# add some objects (first few Messier objects)
survey.add_object("M1", 8.4)
survey.add_object("M2", 6.3)
survey.add_object("M3", 6.2)
survey.add_object("M4", 5.9)
You might note that this currently doesn't do much more than copying one dictionary into another, albeit with some additional type checking!
Part 2
Add a methods that return the name of the brightest and dimmest objects. Note that for magnitudes in astronomy the lower the number the brighter the object!
Solution
class AstroSurvey:
def __init__(self, survey={}):
# create an attribute that will store the survey - it is useful to use a dictionary
self.survey = {}
if not isinstance(survey, dict):
raise TypeError("Input must be a dictionary")
# add copy of all objects into the survey
for key, value in survey.items():
self.add_object(key, value)
def add_object(self, name, magnitude):
"""
Add a new object into the survey.
Parameters
----------
name: str
The name of the object.
magnitude: float
The magnitude of the object.
"""
if not isinstance(name, str):
raise TypeError("The object name must be a string")
if not isinstance(magnitude, (float, int)):
raise TypeError("The object's magnitude must be a number")
# add into survey attribute
self.survey[name] = magnitude
def object_names(self):
"""
Return a list of object names.
"""
return list(self.survey.keys())
def object_magnitudes(self):
"""
Return a list of object magnitudes.
"""
return list(self.survey.values())
def brightest(self):
"""
Return the name of the brightest object.
"""
# get the index of the brightest object
mag = min(self.object_magnitudes())
idx = self.object_magnitudes().index(mag)
return self.object_names()[idx]
def dimmest(self):
"""
Return the name of the dimmest object.
"""
# get the index of the dimmest object
mag = max(self.object_magnitudes())
idx = self.object_magnitudes().index(mag)
return self.object_names()[idx]
Using these with the previous example could be done with, e.g.,
brightest = survey.brightest()
print(f"The brightest object in the survey is {brightest}")
dimmest = survey.dimmest()
print(f"The dimmest object in the survey is {dimmest}")
Exercise 22¶
Question
Create a class that represents a black body. The class should contain class attributes that define the Stefan-Boltzmann constant and Wien's displacement constant. It should be initialised with a temperature (in Kelvin) and the body's radius including checks to make sure these are positive numbers.
It should include two methods that:
- return the black body's bolometric luminosity,
- return the wavelength at which the emission peaks (Wien's Law) in nanometres.
Solution
import numpy as np
class BlackBody:
# Wien's displacement constant (m K)
b = 2.897771955e-3
# the Stefan-Boltzmann constant (W m^-2 K^-4)
sigma = 5.670374419e-8
def __init__(self, T, radius):
"""
A black-body.
Parameters
----------
T: float
The temperature of the black body (K)
radius: float
The radius of the black body (m)
"""
# check T is a number and greater than 0
if not isinstance(T, (float, int)):
raise TypeError("Temperature must be a number")
if T <= 0:
raise ValueError("Temperature must be a positive number")
self.T = float(T)
# check radius is a number and greater than 0
if not isinstance(radius, (float, int)):
raise TypeError("Radius must be a number")
if radius <= 0:
raise ValueError("Radius must be a positive number")
self.radius = float(radius)
def bolometric_luminosity(self):
"""
Return the black body's bolometric luminosity using the Stefan-Boltzmann equation.
"""
# objects surface area
surfarea = 4 * np.pi * self.radius ** 2
return self.sigma * surfarea * self.T ** 4
def peak_wavelength(self):
"""
Return the peak wavelength (in nanometres) of the black body's thermal
radiation using Wien's Law.
"""
return (self.b / self.T) * 1e9
Exercise 23¶
Part 1
Create a class to define Square
objects. The class should be initialised using a tuple or list
that contains four pairs (also tuples or lists) of \(x\) and \(y\) coordinates for the corners of
the square, which should then be stored in th class. The class should contain a method to check
that the input points define a valid square (i.e. all sides are the same length and all angles
between sides are 90 degrees), which should be used during initialisation and an error raised
if it fails the check.
Part 2
Add methods to the class that return the area and perimeter of the square.
Part 3
Add a method to the class that takes in a point, given by a tuple containing its \(x\) and \(y\)
coordinates, and returns True
if the point is within the square and False
if not.
Part 4
Create a Square
and then get a new Square
based on the original, but rotated by 30 degrees
about the first square's centre. Plot the two squares on the same figure.
Hint: you may have already written code, or a method, in the class that rotates a square.
Solution
A possible class is:
import numpy as np
class Square:
"""
A class defining a Square object.
Parameters
----------
vertices: array
A 4x2 array defining the x-y coordinates of the four corners of the
square. The corner coordinates must be consecutive corners in
either the clockwise or anticlockwise direction.
"""
def __init__(self, vertices):
# store copy of vertices as numpy array
self.vertices = np.array(vertices)
# check if valid square
if not self.valid_square():
raise ValueError("Input coordinates do not define a valid square")
# get the centre of the square
self.centre = (self.vertices[0] + self.vertices[2]) / 2.0
def valid_square(self):
"""
Check that vertices define a valid square: four vertices are given;
each vertex has two points; all sides are the same length; all
corners are 90 degrees.
Returns
-------
bool
False is not a valid square otherwise True
"""
# check vertices contain four pairs of points
if self.vertices.shape != (4, 2):
return False
# check side lengths
distances = []
for i in range(4):
distance = self.side_length(self.vertices[i],
self.vertices[(i + 1) % 4])
distances.append(distance)
if not np.allclose(distances, distances[0]):
return False
# check angles between sides
angles = []
for i in range(4):
origin = self.vertices[i]
prevv = self.vertices[(i + 4 - 1) % 4]
nextv = self.vertices[(i + 1) % 4]
vec1 = prevv - origin
vec2 = nextv - origin
angles.append(self.vertex_angle(vec1, vec2))
if not np.allclose(angles, np.pi / 2.0):
return False
return True
@staticmethod
def side_length(x1, x2):
"""
Get the distance between two coordinates.
Parameters
----------
x1: tuple
A pair of x-y coorinates for a point
x2: tuple
A pair of x-y coorinates for a point
Return
------
float:
The distance between points
"""
return np.linalg.norm(x1 - x2)
@staticmethod
def vertex_angle(vec1, vec2):
"""
Get the angle between two vectors.
Parameters
----------
vec1:
A vector (two coordinate points) defined from the origin
vec2: tuple
A vector (two coordinate points) defined from the origin
Return
------
float:
The angle between the vectors
"""
# dot product of two vectors
dp = np.dot(vec1, vec2)
# magnitude of vectors
mag1 = np.linalg.norm(vec1)
mag2 = np.linalg.norm(vec2)
angle = np.arccos(dp / (mag1 * mag2))
return angle
def area(self):
"""
Return the area of the square.
"""
return self.side_length(self.vertices[0], self.vertices[1]) ** 2
def perimeter(self):
"""
Return the perimeter of the square.
"""
return self.side_length(self.vertices[0], self.vertices[1]) * 4
def in_square(self, point):
"""
Check if a given point is in the square.
Parameters
----------
point: (list, tuple)
A list consisting of the x, y coordinates of the point to test.
Returns
-------
bool
Give True if the point is in the square and False otherwise.
"""
# rotate the square and the point, so they are aligned with the x-y axes
vec1 = [1, 0] # unit vector on x-axis
vec2 = self.vertices[1] - self.vertices[0] # a side of the square
# angle between one of the squares sides and the x-axis
angle = self.vertex_angle(vec1, vec2)
# rotated square
rotsquare = self.rotate_square(angle)
# rotated test point about the centre
rot = np.array(
[[np.cos(angle), -np.sin(angle)], [np.sin(angle), np.cos(angle)]]
)
rotpoint = np.dot(rot, point - self.centre)
# check point is within the square
bottom = rotsquare.side("bottom")
top = rotsquare.side("top")
if rotpoint[1] < bottom[0][1] or rotpoint[1] > top[0][1]:
# outside y-extent of square
return False
left = rotsquare.side("left")
right = rotsquare.side("right")
if rotpoint[0] < left[0][0] or rotpoint[0] > right[0][0]:
# outside x-extent of square
return False
return True
def rotate_square(self, angle):
"""
Return a new Square object that is rotated by a given angle about the
square's centre.
Parameters
----------
angle: float
An angle in radian to rotate the square by.
Returns
-------
Square
A new Square object
"""
# set rotation matrix
rot = np.array(
[[np.cos(angle), -np.sin(angle)], [np.sin(angle), np.cos(angle)]]
)
# store the rotation matrix
self.rotation_matrix = rot
# get the rotated vertices
rotverts = np.array(
[np.dot(rot, vertex - self.centre) for vertex in self.vertices]
) + self.centre
return Square(rotverts)
def side(self, which="bottom"):
"""
Return the two vertices for the given side. If two sides are equivalent
(e.g., are both as "low" as each other if given "bottom") then the
first two be found is returned.
Parameters
----------
which: str
A string with either "bottom", "top", "left" or "right" for the
side to return.
Returns
-------
tuple
The two vertices defining the requested side.
"""
if which[0].lower() == "l":
# left side
idx = np.argsort(self.vertices[:, 0])[0]
elif which[0].lower() == "r":
# right side
idx = np.argsort(self.vertices[:, 0])[-1]
elif which[0].lower() == "b":
# bottom side
idx = np.argsort(self.vertices[:, 1])[0]
elif which[0].lower() == "t":
# top side
idx = np.argsort(self.vertices[:, 1])[-1]
else:
raise ValueError(f"Side '{which}' is not valid")
idxs = [idx, (idx + 1) % 4]
return self.vertices[idxs]
Given this class two squares can be plotted with:
import numpy as np
from matplotlib import pyplot as plt
# first square
s1 = Square([[4, 3], [4, 5], [6, 5], [6, 3]])
# rotated square
s2 = s1.rotate_square(np.deg2rad(30))
fig, ax = plt.subplots(figsize=(3.5,3.5),dpi=200)
# plot s1 in blue and s2 in red
for s, c in zip([s1, s2], ["b", "r"]):
x = np.hstack((s.vertices[:, 0], s.vertices[0, 0]))
y = np.hstack((s.vertices[:, 1], s.vertices[0, 1]))
ax.plot(x, y, color=c)
# make axes have an equal aspect ratio
ax.set_aspect("equal")
fig.tight_layout()
plt.show()
A Rectangle
patch could be used instead if you work out the bottom left corner and the required rotation
angle.
Debugging¶
Exercise 24¶
Question
Fix this broken code:
import maths
deg hypotenuse(a=1.0, b)
"""
A function to calculate and return the hypotenuse of a right angle
triangle with opposite and adjacent sides with lengths a and b.
"""
hyp = maths.sqrt(a ** 2 + b ** 2
retrn hy
hp = hypotenus(a="3", b=4)
Solution
The problems are highlighted below
import maths # "maths" is not a standard Python library
# several issues:
# - typo in "def"
# - no colon at the end of the line
# - can't have a keyword argument before a positional argument
deg hypotenuse(a=1.0, b)
"""
A function to calculate and return the hypotenuse of a right angle
triangle with opposite and adjacent sides with lengths a and b.
"""
hyp = maths.sqrt(a ** 2 + b ** 2 # closing bracket was missing, math library fixed
retrn hy # typo in "return" and returned variable name
# a couple of issues:
# - typo in called function "hypotenuse"
# - trying to pass a string when a number is required
hp = hypotenus(a="3", b=4)
A fixed version would be:
import math
def hypotenuse(a, b):
"""
A function to calculate and return the hypotenuse of a right angle
triangle with opposite and adjacent sides with lengths a and b.
"""
hyp = math.sqrt(a ** 2 + b ** 2)
return hyp
hp = hypotenuse(a=3, b=4)
Exercise 25¶
Question
Fix this broken code:
x = [1, 5, 1, 5 6, 2, 5, 7]
# sum the cube of each number
y = ""
for i = range(len(x) + 1)
y += x[i] *** 3
Solution
The problems are highlighted below:
x = [1, 5, 1, 5 6, 2, 5, 7] # missing comma between 5 and 6
# sum the cube of each number
y = "" # should initialise as 0 not an empty string
# various issues:
# - using "=" rather than in
# - missing colon at end of for statement
# - final index will be 1 too big for y
for i = range(len(x) + 1)
y += x[i] *** 3 # should use "**" not "***"
A fixed version would be:
x = [1, 5, 1, 5, 6, 2, 5, 7]
# sum the cube of each number
y = 0.0
for i in range(len(x)):
y += x[i] ** 3
Exercise 26¶
Question
Fix this broken code:
import numpy as npy
# create dictionary of numpy arrays
arrs = {
"one": np.array([1, 2, 3])
"two": np.array([4, 5, 6]),
"three": np.array(7, 8, 9)
}
# concatenate the three arrays
full = np.concatenate(arrs["on"], arrs["two"], arr["three])
# get the shape of the array
shape = full.shape()
Solution
The problems are highlighted below:
import numpy as npy # alias is not the same as used below
# create dictionary of numpy arrays
arrs = {
"one": np.array([1, 2, 3]) # missing comma
"two": np.array([4, 5, 6]),
"three": np.array(7, 8, 9) # missing square brackets
}
# concatenate the three arrays
# the errors are:
# - "on" is not a valid key
# - arr is not a known variable
# - missing closing quotation mark on "three"
# - concatenate requires a tuple
full = np.concatenate(arrs["on"], arrs["two"], arr["three])
# get the shape of the array
shape = full.shape() # shape is a property not a method
A fixed version would be:
import numpy as np
# create dictionary of numpy arrays
arrs = {
"one": np.array([1, 2, 3]),
"two": np.array([4, 5, 6]),
"three": np.array([7, 8, 9])
}
# concatenate the three arrays
full = np.concatenate((arrs["one"], arrs["two"], arrs["three"]))
# get the shape of the array
shape = full.shape
General problems¶
Exercise 27¶
Question
Estimate the value of \(\pi\) using a Monte Carlo method (i.e., through drawing random numbers).
Solution
A potential solution, based on the ratio of the area of a square with sides 2 units long (\(A_s = 2 \times 2 = 4\)) to a circle with radius of 1 unit (\(A_c = \pi r^2 = \pi\)), being \((A_s / A_c) = 4/\pi\), is:
# import numpy for random number generation
import numpy as np
# create random number generator
rstate = np.random.default_rng()
# set the number of samples to draw for estimation
nsamples = 10000
# draw nsamples samples in x and y uniformly from the square between -1 and +1
samples = rstate.uniform(-1, 1, (nsamples, 2))
# get "magnitude" of each point sqrt(x^2 + y^2)
radius = np.sqrt(samples[:, 0] ** 2 + samples[:, 1] ** 2)
# radius = np.linalg.norm(samples, axis=1) # another option
# work out how many samples are within the unit circle (i.e., radius < 1)
numincirc = np.sum(radius < 1)
# get estimate of pi
estpi = 4 * (numincirc / nsamples)
print(estpi)
Exercise 28¶
Question
Write a function that implements the bisection method to find the root of a continuous function in a provided interval. Test your function by solving the equation \(\exp(-x)(x^2+5x+2) + 1 = 0\) between -1 and 0.
Solution
A potential solution is:
#!/usr/bin/env python3
import math
import textwrap as tw
def test_function(x):
"""Function whose roots we seek."""
return math.e**(-x) * (x*(x + 5) + 2.0) + 1.0
def bisection(fun, xrange, toll = 1e-12, niter = 1000):
"""
A function that implements the bisection method to find the root of a
function in a given interval.
Parameters:
-----------
fun: callable
the function to be solved
xrange: list
list containing the interval where the 0 of the function lies
toll: float
tolerance between two iterations, defaults to 1e-12
niter: integer
maximum number of iterations, defaults to 1000
Returns:
--------
c: float
root of the given equation with toll precision
fc: float
value of the given function at c (approximately 0.0)
n: integer
number of iterations done to reach the given solution
"""
a = xrange[0]
fa = fun(a)
b = xrange[1]
fb = fun(b)
if fa * fb >= 0.0:
raise ValueError("The provided function does not contain zeros in the "
"given interval")
eps = 10000.0
n = 0
c_old = a
fc = fa
while eps > toll and n <= niter and fc != 0.0:
c = (a + b) / 2
fc = fun(c)
if fa * fc < 0.0:
b = c
fb = fc
elif fb * fc < 0.0:
a = c
fa = fc
else:
raise ValueError("Oh oh, you missed the zero")
eps = abs(c - c_old)
c_old = c
n += 1
if n == niter:
print(f"niter, n = {n}")
raise ValueError("Maximum number of iterations reached")
return c, fc, n
xrange = [-1.0, 0.0]
sol, fsol, niter = bisection(test_function, xrange)
print(tw.fill(f"The root of your function is approximately x = {sol}, where "
f"the function value is {fsol}. This solution has been reached "
f"after {niter} bisection iterations."))
Advanced exercises¶
These exercises are very much just for fun if you fancy something a bit more challenging!
Exercise 29¶
Question
Write a class that implements a noughts-and-crosses game.
Solution
A potential solution is:
#!/usr/bin/env python3
"""
A class to create a tic-tac-toe game. The class can either be imported and run
in a Python terminal, or the script can be run from the command line.
"""
import sys
import numpy as np
class TicTacToe:
"""
A tic-tac-toe game. After creating the object, the game can be played
by calling it, e.g.
>>> game = TicTacToe()
>>> game()
"""
def __init__(self):
player1 = input("Input the name of the first player: ")
player2 = input("Input the name of the second player: ")
self.players = [player1, player2]
try:
self.gridsize = int(input("Set the grid size (integer): "))
except ValueError:
raise ValueError("Grid size must be an integer")
# initialise grid as empty single space strings
self.grid = np.full((self.gridsize, self.gridsize), " ")
self.linecount = 3 # number of lines to rewind by
def __call__(self):
# start game
self.drawgrid()
self.currentplayer = False # boolean to flip between players
while not self.checkstate():
# get player one's turn
self.getturn()
# check if game has been completed
if self.checkstate():
break
# get player two's turn
self.getturn()
self.showwinner()
play = __call__
def drawgrid(self):
"""
Draw the current state of the grid.
"""
# rewind to overwrite previous grid (see, e.g.,
# https://stackoverflow.com/a/59147732/1862861)
for _ in range(self.linecount):
sys.stdout.write("\x1b[1A\x1b[2K")
print()
for i, row in enumerate(self.grid):
rowstr = "|".join([f" {x} " for x in row])
print(rowstr)
if i < self.gridsize - 1:
print("-" * (self.gridsize * 3 + (self.gridsize - 1)))
print()
self.linecount = 2 * self.gridsize + 1
def getturn(self):
"""
Ask player for input coordinates and re-draw grid.
"""
counters = ["✕", "○"]
playeridx = int(self.currentplayer)
# make sure coordinates are valid
while 1:
coords = input(
f"{self.players[playeridx]}: Input the grid coordinates 'x y'"
f" (between 1 and {self.gridsize}) for your move: "
)
self.linecount += 1
x, y = [int(coord.strip()) for coord in coords.split()]
# check grid coordinates are valid
if not (1 <= x <= self.gridsize) or not (1 <= y <= self.gridsize):
print("x and y coordinates must be between 1 and "
f"{self.gridsize}")
self.linecount += 1
continue
# check grid cooridate has not already been used
if self.grid[self.gridsize - y][x - 1] != " ":
print("That grid point has already be used. Try again.")
self.linecount += 1
continue
break
# fill in grid
self.grid[self.gridsize - y][x - 1] = counters[playeridx]
# draw the grid
self.drawgrid()
# flip player
self.currentplayer = not self.currentplayer
def checkstate(self):
"""
Check whether anyone has won.
"""
# check if any rows are completed
complete = False
for row in (
[np.diag(self.grid), np.diag(np.fliplr(self.grid))]
+ list(self.grid)
+ list(np.transpose(self.grid))
):
# ignore "empty" rows
if not np.all(row == " "):
# check if all values in row are the same
if np.all(row == row[0]):
complete = True
break
return complete
def showwinner(self):
"""
Show the winner.
"""
playeridx = int(not self.currentplayer)
print(f"The winner is {self.players[playeridx]}!")
# run the game if calling the code directly
if __name__ == "__main__":
game = TicTacToe()
game()