CODEWITHSUNDEEP
pythonarraysnumpyelementwise-operations
clstaudt
clstaudt

Reputation: 22438

Comparing two NumPy arrays for equality, element-wise

What is the simplest way to compare two NumPy arrays for equality (where equality is defined as: A = B iff for all indices i: A[i] == B[i])?

Simply using == gives me a boolean array:

 >>> numpy.array([1,1,1]) == numpy.array([1,1,1])

array([ True,  True,  True], dtype=bool)

Do I have to and the elements of this array to determine if the arrays are equal, or is there a simpler way to compare?

Upvotes: 455

Views: 777394

Answers (9)

Attila Korpos
Attila Korpos

Reputation: 1

A=np.array([1,2,3,4])
B=np.array([1,2,3,4])

sum(A!=B)==0

The idea is that the total number of unequal elements should be zero.

Upvotes: 0

rubengavidia0x
rubengavidia0x

Reputation: 590

Just for the sake of completeness. I will add the pandas approach for comparing two arrays:

import pandas as pd
import numpy as np
a = np.arange(0.0, 10.2, 0.12)
b = np.arange(0.0, 10.2, 0.12)
ap = pd.DataFrame(a)
bp = pd.DataFrame(b)

ap.equals(bp)
True

FYI: In case you are looking of How to compare Vectors, Arrays or Dataframes in R. You just you can use:

identical(iris1, iris2)
#[1] TRUE
all.equal(array1, array2)
#> [1] TRUE

Upvotes: 0

Sylvain
Sylvain

Reputation: 799

On top of the other answers, you can now use an assertion:

numpy.testing.assert_array_equal(x, y)

You also have similar function such as numpy.testing.assert_almost_equal()

https://numpy.org/doc/stable/reference/generated/numpy.testing.assert_array_equal.html

Upvotes: 2

keramat
keramat

Reputation: 4543

Now use np.array_equal. From documentation:

np.array_equal([1, 2], [1, 2])
True
np.array_equal(np.array([1, 2]), np.array([1, 2]))
True
np.array_equal([1, 2], [1, 2, 3])
False
np.array_equal([1, 2], [1, 4])
False

Upvotes: 5

Juh_
Juh_

Reputation: 15529

(A==B).all()

test if all values of array (A==B) are True.

Note: maybe you also want to test A and B shape, such as A.shape == B.shape

Special cases and alternatives (from dbaupp's answer and yoavram's comment)

It should be noted that:

  • this solution can have a strange behavior in a particular case: if either A or B is empty and the other one contains a single element, then it return True. For some reason, the comparison A==B returns an empty array, for which the all operator returns True.
  • Another risk is if A and B don't have the same shape and aren't broadcastable, then this approach will raise an error.

In conclusion, if you have a doubt about A and B shape or simply want to be safe: use one of the specialized functions:

np.array_equal(A,B)  # test if same shape, same elements values
np.array_equiv(A,B)  # test if broadcastable shape, same elements values
np.allclose(A,B,...) # test if same shape, elements have close enough values

Upvotes: 670

R Zhang
R Zhang

Reputation: 339

Usually two arrays will have some small numeric errors,

You can use numpy.allclose(A,B), instead of (A==B).all(). This returns a bool True/False

Upvotes: 10

user1767754
user1767754

Reputation: 25094

If you want to check if two arrays have the same shape AND elements you should use np.array_equal as it is the method recommended in the documentation.

Performance-wise don't expect that any equality check will beat another, as there is not much room to optimize comparing two elements. Just for the sake, i still did some tests. 

import numpy as np
import timeit

A = np.zeros((300, 300, 3))
B = np.zeros((300, 300, 3))
C = np.ones((300, 300, 3))

timeit.timeit(stmt='(A==B).all()', setup='from __main__ import A, B', number=10**5)
timeit.timeit(stmt='np.array_equal(A, B)', setup='from __main__ import A, B, np', number=10**5)
timeit.timeit(stmt='np.array_equiv(A, B)', setup='from __main__ import A, B, np', number=10**5)
> 51.5094
> 52.555
> 52.761

So pretty much equal, no need to talk about the speed.

The (A==B).all() behaves pretty much as the following code snippet:

x = [1,2,3]
y = [1,2,3]
print all([x[i]==y[i] for i in range(len(x))])
> True

Upvotes: 25

funk
funk

Reputation: 2287

Let's measure the performance by using the following piece of code.

import numpy as np
import time

exec_time0 = []
exec_time1 = []
exec_time2 = []

sizeOfArray = 5000
numOfIterations = 200

for i in xrange(numOfIterations):

    A = np.random.randint(0,255,(sizeOfArray,sizeOfArray))
    B = np.random.randint(0,255,(sizeOfArray,sizeOfArray))

    a = time.clock() 
    res = (A==B).all()
    b = time.clock()
    exec_time0.append( b - a )

    a = time.clock() 
    res = np.array_equal(A,B)
    b = time.clock()
    exec_time1.append( b - a )

    a = time.clock() 
    res = np.array_equiv(A,B)
    b = time.clock()
    exec_time2.append( b - a )

print 'Method: (A==B).all(),       ', np.mean(exec_time0)
print 'Method: np.array_equal(A,B),', np.mean(exec_time1)
print 'Method: np.array_equiv(A,B),', np.mean(exec_time2)

Output

Method: (A==B).all(),        0.03031857
Method: np.array_equal(A,B), 0.030025185
Method: np.array_equiv(A,B), 0.030141515

According to the results above, the numpy methods seem to be faster than the combination of the == operator and the all() method and by comparing the numpy methods the fastest one seems to be the numpy.array_equal method.

Upvotes: 18

huon
huon

Reputation: 102006

The (A==B).all() solution is very neat, but there are some built-in functions for this task. Namely array_equal, allclose and array_equiv.

(Although, some quick testing with timeit seems to indicate that the (A==B).all() method is the fastest, which is a little peculiar, given it has to allocate a whole new array.)

Upvotes: 138

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