numpy.array_equiv () in Python

Michael Zippo 18.07.2021

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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numpy.array_equiv () in Python

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