numpy.ma.choose()函数Python

numpy.ma.choose()函数使用索引数组从一组选择中构造一个新数组。给定一个整数数组和一组 n 个选择数组，此方法将创建一个新数组，该数组合并每个选择数组。如果 arr 中的 arr 值为 i，则新数组将具有choices[i] 在同一位置包含的值。

Syntax : numpy.ma.choose(arr, choices, out = None, mode = ‘raise’)

Parameters :
arr : [ndarray of ints] This array must contain integers in [0, n-1], where n is the number of choices.
choices : [sequence of arrays] Choice arrays. The index array and all of the choices should be broadcastable to the same shape.
out : [array, optional] If provided, the result will be inserted into this array. It should be of the appropriate shape and dtype.
mode : [{‘raise’, ‘wrap’, ‘clip’}, optional] Specifies how out-of-bounds indices will behave. ‘raise’: raise an error. ‘wrap’: wrap around. ‘clip’: clip to the range.

Return : A new array that merges each of the choice arrays.

代码#1：

# Python program explaining
# numpy.ma.choose() function
    
# importing numpy as geek   
# and numpy.ma module as ma  
import numpy as geek  
import numpy.ma as ma
   
choice = geek.array([[1, 1, 1], [2, 2, 2], [3, 3, 3]])
arr = geek.array([2, 1, 0])
  
gfg = geek.ma.choose(arr, choice)
  
print (gfg)

输出 ：

[3 2 1]

代码#2：

# Python program explaining
# numpy.ma.choose() function
    
# importing numpy as geek   
# and numpy.ma module as ma  
import numpy as geek  
import numpy.ma as ma
   
choice = geek.array([[1, 1, 1], [2, 2, 2], [3, 3, 3]])
arr = geek.array([0, 1, 2])
  
gfg = geek.ma.choose(arr, choice)
  
print (gfg)

输出 ：

[1 2 3]