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numpy.array_equiv() in Python

Last Updated : 29 Nov, 2018
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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 Python 
# Python program explaining
# array_equiv() function
import numpy as np

# input
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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