numpy.array_equiv() in Python (original) (raw)

Last Updated : 29 Nov, 2018

numpy.array_equiv(arr1, arr2) : This logical function that checks if two arrays have the same elements and shape consistent.

Shape consistent means either they are having the same shape, or one input array can be broadcasted to create the same shape as the other one.

Parameters :

arr1 : [array_like]Input array, we need to test. arr2 : [array_like]Input array, we need to test.

Return :

True, if both arrays are equivalent; otherwise False

Code : Explaining Working

import numpy as np

arr1 = np.arange( 4 )

arr2 = [ 7 , 4 , 6 , 7 ]

print ( "arr1 : " , arr1)

print ( "arr2 : " , arr2)

print ( "\nResult : " , np.array_equiv(arr1, arr2))

arr1 = np.arange( 4 )

arr2 = np.arange( 4 )

print ( "\n\narr1 : " , arr1)

print ( "arr2 : " , arr2)

print ( "\nResult : " , np.array_equiv(arr1, arr2))

arr1 = np.arange( 4 )

arr2 = np.arange( 5 )

print ( "\n\narr1 : " , arr1)

print ( "arr2 : " , arr2)

print ( "\nResult : " , np.array_equiv(arr1, arr2))

a = np.array_equiv([ 1 , 2 ], [[ 1 , 2 , 1 , 2 ], [ 1 , 2 , 1 , 2 ]])

b = np.array_equiv([ 1 , 2 ], [[ 1 , 2 ], [ 1 , 2 ]])

print ( "\n\na : " , a)

print ( "\nb : " , b)

Output :

arr1 : [0 1 2 3] arr2 : [7, 4, 6, 7]

Result : False

arr1 : [0 1 2 3] arr2 : [0 1 2 3]

Result : True

arr1 : [0 1 2 3] arr2 : [0 1 2 3 4]

Result : False

a : False

b : True

References :
https://docs.scipy.org/doc/numpy-1.13.0/reference/generated/numpy.array_equiv.html#numpy.array_equiv
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