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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






# 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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