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?

8 Answers

(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 

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.)

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 

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.

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

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 

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()

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

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