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numpy.array_equiv () in Python

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Consistent form means that either they have the same form, or one input array can be passed to create the same form as another.

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 Operation

# 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 ( "Result:" , np.array_equiv (arr1, arr2))

 

arr1 = np.arange ( 4 )

arr2 = np.arange ( 4 )

print ( "arr1:" , arr1)

print ( " arr2: " , arr2)

 

print ( "Result:" , np.array_equiv (arr1, arr2))

 

arr1 = np.arange ( 4 )

arr2 = np.arange ( 5 )

print ( "arr1:" , arr1)

print ( "arr2:" , arr2)

< code class = "undefined spaces">  

print ( " Result: " , 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 ( "a:" , a)

print ( "b:" , 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: Tru e arr1: [0 1 2 3] arr2: [0 1 2 3 4] Result: False a: False b: True 

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