Numpy中的数组搜索中np.where方法详细介绍

> aa = np.arange(10)
> np.where(aa,1,-1)
array([-1, 1, 1, 1, 1, 1, 1, 1, 1, 1]) # 0为False，所以第一个输出-1
> np.where(aa > 5,1,-1)
array([-1, -1, -1, -1, -1, -1, 1, 1, 1, 1])

> np.where([[True,False], [True,True]],  # 官网上的例子
  [[1,2], [3,4]],
       [[9,8], [7,6]])
array([[1, 8],
  [3, 4]])

> a = 10
> np.where([[a > 5,a < 5], [a == 10,a == 7]],
       [["chosen","not chosen"], ["chosen","not chosen"]],
       [["not chosen","chosen"], ["not chosen","chosen"]])

array([['chosen', 'chosen'],
    ['chosen', 'chosen']], dtype='<U10')

> a = np.array([2,4,6,8,10])
> np.where(a > 5)  # 返回索引
(array([2, 3, 4]),)  
> a[np.where(a > 5)]   # 等价于 a[a>5]
array([ 6, 8, 10])

> np.where([[0, 1], [1, 0]])
(array([0, 1]), array([1, 0]))

> a = np.arange(27).reshape(3,3,3)
> a
array([[[ 0, 1, 2],
    [ 3, 4, 5],
    [ 6, 7, 8]],

    [[ 9, 10, 11],
    [12, 13, 14],
    [15, 16, 17]],

    [[18, 19, 20],
    [21, 22, 23],
    [24, 25, 26]]])

> np.where(a > 5)
(array([0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2]),
 array([2, 2, 2, 0, 0, 0, 1, 1, 1, 2, 2, 2, 0, 0, 0, 1, 1, 1, 2, 2, 2]),
 array([0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2]))

# 符合条件的元素为
  [ 6, 7, 8]],

   [[ 9, 10, 11],
    [12, 13, 14],
    [15, 16, 17]],

   [[18, 19, 20],
    [21, 22, 23],
    [24, 25, 26]]]

import numpy as np
a = np.array([1, 2, 3, 4, 5])
b = np.where(a == 5)

print(b)

(array([4]),)

c = np.where(a%2 == 0)
print(c)

d = np.where(a%2 == 1)
print(d)

(array([1, 3]),)
(array([0, 2, 4]),)