5
\$\begingroup\$

I had a code challenge for a class I'm taking that built a NN algorithm. I got it to work but I used really basic methods for solving it. There are two 1D NP Arrays that have values 0-2 in them, both equal length. They represent two different trains and test data The output is a confusion matrix that shows which received the right predictions and which received the wrong (doesn't matter ;).

This code is correct - I just feel I took the lazy way out working with lists and then turning those lists into a ndarray. I would love to see if people have some tips on maybe utilizing Numpy for this? Anything Clever?

import numpy as np

x = [0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
     2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 2, 0, 0, 0, 0, 0, 1, 0]
y = [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
     2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

testy = np.array(x)
testy_fit = np.array(y)

row_no = [0,0,0]
row_dh = [0,0,0]
row_sl = [0,0,0]

# Code for the first row - NO
for i in range(len(testy)):
    if testy.item(i) == 0 and testy_fit.item(i) == 0:
        row_no[0] += 1
    elif testy.item(i) == 0 and testy_fit.item(i) == 1:
        row_no[1] += 1
    elif testy.item(i) == 0 and testy_fit.item(i) == 2:
        row_no[2] += 1

# Code for the second row - DH
for i in range(len(testy)):
    if testy.item(i) == 1 and testy_fit.item(i) == 0:
        row_dh[0] += 1
    elif testy.item(i) == 1 and testy_fit.item(i) == 1:
        row_dh[1] += 1
    elif testy.item(i) == 1 and testy_fit.item(i) == 2:
        row_dh[2] += 1

# Code for the third row - SL
for i in range(len(testy)):
    if testy.item(i) == 2 and testy_fit.item(i) == 0:
        row_sl[0] += 1
    elif testy.item(i) == 2 and testy_fit.item(i) == 1:
        row_sl[1] += 1
    elif testy.item(i) == 2 and testy_fit.item(i) == 2:
        row_sl[2] += 1

confusion = np.array([row_no,row_dh,row_sl])

print(confusion)

the result of the print is correct as follow:

[[16 10  0]
 [ 2 10  0]
 [ 2  0 22]]
\$\endgroup\$
1
  • 1
    \$\begingroup\$ Good thing this got an answer on SO before it was moved. Performance questions for numpy are routine on SO. \$\endgroup\$
    – hpaulj
    Commented May 6, 2019 at 0:15

2 Answers 2

5
\$\begingroup\$

This can be implemented concisely by using numpy.add.at:

In [2]: c = np.zeros((3, 3), dtype=int)                                         

In [3]: np.add.at(c, (x, y), 1)                                                 

In [4]: c                                                                       
Out[4]: 
array([[16, 10,  0],
       [ 2, 10,  0],
       [ 2,  0, 22]])
\$\endgroup\$
2
  • \$\begingroup\$ Oh my! I thought there would be something better but i didn't think 1 line of code! Wow. So glad I asked and thank you! \$\endgroup\$
    – broepke
    Commented May 6, 2019 at 2:04
  • 2
    \$\begingroup\$ Rule #1 of numpy is if you want to do something, check the docs first to check for a 1 line solution. \$\endgroup\$ Commented May 6, 2019 at 5:39
3
\$\begingroup\$

For now disregarding that there is a (way) better numpy solution to this, as explained in the answer by @WarrenWeckesser, here is a short code review of your actual code.

  • testy.item(i) is a very unusual way to say testy[i]. It is probably also slower as it involves an attribute lookup.
  • Don't repeat yourself. You test e.g. if testy.item(i) == 0 three times, each time with a different second condition. Just nest them in an if block:

    for i in range(len(testy)):
        if testy[i] == 0:
            if testy_fit[i] == 0:
                row_no[0] += 1
            elif testy_fit[i] == 1:
                row_no[1] += 1
            elif testy_fit[i] == 2:
                row_no[2] += 1
    
  • Loop like a native. Don't iterate over the indices of iterables, iterate over the iterable(s)! You can also use the fact that the value encodes the position you want to increment:

    for test, fit in zip(testy, testy_fit):
        if test == 0 and fit in {0, 1, 2}:
            row_no[fit] += 1
    
  • You can even use the fact that the first value encodes the list you want to use and iterate only once. Or even better, make it a list of lists right away:

    n = 3
    confusion_matrix = [[0] * n for _ in range(n)]
    for test, fit in zip(testy, testy_fit):
        confusion_matrix[test][fit] += 1
    
    print(np.array(confusion_matrix))
    
  • Don't put everything into the global space, to be run whenever you interact with the script at all. Put your code into functions, document them with a docstring, and execute them under a if __name__ == "__main__": guard, which allows you to import from this script from another script without your code running:

    def confusion_matrix(x, y):
        """Return the confusion matrix for two vectors `x` and `y`.
        x and y must only have values from 0 to n and 0 to m, respectively.
        """
        n, m = np.max(x) + 1, np.max(y) + 1
        matrix = [[0] * m for _ in range(n)]
        for a, b in zip(x, y):
            matrix[a][b] += 1
        return matrix
    
    if __name__ == "__main__":
        x = ...
        y = ...
        print(np.array(confusion_matrix(x, y)))
    

Once you have come this far, you can just swap the implementation of this function to the faster numpy one without changing anything (except that it then directly returns a numpy.array instead of a list of lists).

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.