numpy.array_equiv() in Python
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)
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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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