Comparing two numpy arrays to each other

Question

I have two equally sized numpy arrays (they happen to be 48x365) where every element is either -1, 0, or 1. I want to compare the two and see how many times they are both the same and how many times they are different while discounting all the times where at least one of the arrays has a zero as no data. For instance:

for x in range(48):
    for y in range(365):
        if array1[x][y] != 0:
            if array2[x][y] != 0:
                if array1[x][y] == array2[x][y]:
                    score = score + 1
                else:
                    score = score - 1
return score

This takes a very long time. I was thinking to take advantage of the fact that multiplying the elements together and summing all the answers may give the same outcome, and I'm looking for a special numpy function to help with that. I'm not really sure what unusual numpy function are out there.

Paul · Accepted Answer · 2011-07-14 18:29:35Z

12

Simpy do not iterate. Iterating over a numpy array defeats the purpose of using the tool.

ans = np.logical_and(
    np.logical_and(array1 != 0, array2 != 0),
    array1 == array2 )

should give the correct solution.

answered Jul 14, 2011 at 18:29

Paul

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2 Comments

Double AA Over a year ago

Good idea! But this gives me a boolean array. I still need to sum up all the True's to get a score. Is there a numpy-thonic way to do that?

ahelm Over a year ago

you can also use np.sum(array1[ans]) or np.sum(array2[ans]) if you want sum by itself. everytime you have a false as an entry it will not take the value into account.

prideout · Accepted Answer · 2012-10-21 20:43:45Z

7

For me the easiest way is to do this :

A = numpy.array()
B = numpy.array()

T = A - B
max = numpy.max(numpy.abs(T))

epsilon = 1e-6
if max > epsilon:
    raise Exception("Not matching arrays")

It allow to know quickly if arrays are the same and allow to compare float values !!

edited Oct 21, 2012 at 20:43

prideout

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answered Jul 14, 2011 at 22:32

ykatchou

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1 Comment

petr Over a year ago

A bit more general solution than the OP asked for but very useful indeed!

eat · Accepted Answer · 2011-07-14 19:44:36Z

1

Simple calculations along the following lines, will help you to select the most suitable way to handle your case:

In []: A, B= randint(-1, 2, size= (48, 365)), randint(-1, 2, size= (48, 365))
In []: ignore= (0== A)| (0== B)
In []: valid= ~ignore

In []: (A[valid]== B[valid]).sum()
Out[]: 3841
In []: (A[valid]!= B[valid]).sum()
Out[]: 3849
In []: ignore.sum()
Out[]: 9830

Ensuring that the calculations are valid:

In []: 3841+ 3849+ 9830== 48* 365
Out[]: True

Therefore your score (with these random values) would be:

In []: a, b= A[valid], B[valid]
In []: score= (a== b).sum()- (a!= b).sum()
In []: score
Out[]: -8

edited Jul 14, 2011 at 19:44

answered Jul 14, 2011 at 19:02

eat

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Comments

wollez · Accepted Answer · 2013-09-11 12:58:49Z

0

import numpy as np

A = np.array()
B = np.array()
...
Z = np.array()

to_test = np.array([A, B, .., Z])

# compare linewise if all lines are equal 
np.all(map(lambda x: np.all(x==to_test[0,:]), to_test[1:,:]))

answered Sep 11, 2013 at 12:58

wollez

1

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