Question

我有一个这样的数组：

[[0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0]
 [0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0]
 [0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
 [0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
 [0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
 [0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
 [0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0]
 [0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0]
 [0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0]
 [0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0]
 [0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0]
 [0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0]
 [1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0]
 [1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0]
 [0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0]
 [0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0]
 [0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0]]

我喜欢遍历数组。在数组值为1的任何地方，我都希望得到数组的索引并执行操作。

大概是这样的：

for value in array:
   if value ==1:
     print arrayIndexX, arrayIndexY

Answer 1

您可以在此numpy.where使用numpy.column_stack。例如：

>>> import numpy as np
>>> a = np.array([[1, 0, 0], [0, 1, 0], [1, 1, 1]])
>>> np.column_stack(np.where(a==1))
array([[0, 0],
       [1, 1],
       [2, 0],
       [2, 1],
       [2, 2]])

Answer 2

使用enumerate，以及通过行和列的嵌套循环：

for y, row in enumerate(array):
    for x, val in enumerate(row):
        if val == 1:
            print x, y

Answer 3

for r, row in enumerate(array):
    for c, val in enumerate(row):
        if val == 1:
            print r,c

或者，您可以构建包含所需坐标值的列表：

[(r,c) for r,row in enumerate(array) for c,val in enumerate(row) if val==1]

Answer 4

在这种情况下，您还可以使用np.nonzero( YourArray )功能，这将为您提供您想要的功能。

Answer 5

也许这可能有所帮助，其中“a”是你的阵列：

for i in range(len(a)):
    for j in range(len(a[i])):
        if(a[i][j]==1):
            print i,j

Answer 6

可能numpy.nonzero是最简单的......

b = np.array(
[[0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0],
 [0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1],
 [0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1],
 [0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1],
 [0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1],
 [0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0],
 [0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0],
 [0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0],
 [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0],
 [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0],
 [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0],
 [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0],
 [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0],
 [0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0]])

ind = np.nonzero(b)
print(ind)
(array([ 0,  0,  1,  1,  1,  1,  1,  2,  2,  2,  2,  2,  2,  2,  2,  2,  3,
    3,  3,  3,  3,  3,  3,  3,  3,  3,  3,  3,  3,  4,  4,  4,  4,  4,
    4,  4,  4,  4,  4,  4,  4,  4,  4,  4,  4,  5,  5,  5,  5,  5,  5,
    5,  5,  5,  5,  5,  5,  5,  5,  5,  5,  5,  5,  5,  6,  6,  6,  6,
    6,  6,  6,  6,  6,  6,  6,  6,  6,  6,  6,  6,  6,  6,  6,  6,  6,
    6,  6,  6,  7,  7,  7,  7,  7,  7,  7,  7,  7,  7,  7,  7,  7,  7,
    7,  7,  7,  7,  7,  7,  7,  7,  7,  7,  8,  8,  8,  8,  8,  8,  8,
    8,  8,  8,  8,  8,  8,  8,  8,  8,  8,  8,  8,  8,  8,  8,  8,  8,
    9,  9,  9,  9,  9,  9,  9,  9,  9,  9,  9,  9,  9,  9,  9,  9,  9,
    9,  9,  9,  9,  9,  9,  9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10,
   10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11,
   11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11,
   11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12,
   12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13,
   13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13,
   13, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14,
   14, 14, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 15,
   15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 16, 16,
   16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16,
   16, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17,
   17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 18, 18, 18, 18,
   18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18,
   18, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19,
   19, 19, 19, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20,
   21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 22, 22, 22, 22, 22, 22,
   22, 22, 23, 23, 23, 23]), array([ 5,  6,  5,  6,  7,  8,  9,  5,  6,  7,  8,  9, 10, 11, 12, 13,  4,
    5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16,  4,  5,  6,  7,  8,
    9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19,  4,  5,  6,  7,  8,  9,
   10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22,  3,  4,  5,  6,
    7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23,
   24, 25, 27,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16,
   17, 18, 19, 20, 21, 22, 23, 24, 25, 27,  3,  4,  5,  6,  7,  8,  9,
   10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 27,
    3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19,
   20, 21, 22, 23, 24, 25, 27,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11,
   12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25,  2,  3,  4,
    5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21,
   22, 23, 24, 25,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14,
   15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25,  1,  2,  3,  4,  5,  6,
    7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23,
   24,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16,
   17, 18, 19, 20, 21, 22, 23, 24,  1,  2,  3,  4,  5,  6,  7,  8,  9,
   10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24,  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,  0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10,
   11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23,  2,  3,  4,  5,
    6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22,
   23,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20,
   21, 22, 23,  9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22,
   12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 15, 16, 17, 18, 19, 20,
   21, 22, 18, 19, 20, 21]))

Answer 7

但是，我认为使用蒙面数组可能要好得多......

marray = np.ma.array(b, mask=(b == 0))
print(marray)
[[-- -- -- -- -- 1 1 -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --]
 [-- -- -- -- -- 1 1 1 1 1 -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --]
 [-- -- -- -- -- 1 1 1 1 1 1 1 1 1 -- -- -- -- -- -- -- -- -- -- -- -- -- --]
 [-- -- -- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 -- -- -- -- -- -- -- -- -- -- --]
 [-- -- -- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- -- -- -- -- -- -- --]
 [-- -- -- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- -- -- -- --]
 [-- -- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- 1]
 [-- -- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- 1]
 [-- -- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- 1]
 [-- -- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- 1]
 [-- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- --]
 [-- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- --]
 [-- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- --]
 [-- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- -- --]
 [-- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- -- --]
 [-- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- -- --]
 [1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- -- --]
 [1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- -- -- --]
 [-- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- -- -- --]
 [-- -- -- -- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- -- -- --]
 [-- -- -- -- -- -- -- -- -- 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -- -- -- -- --]
 [-- -- -- -- -- -- -- -- -- -- -- -- 1 1 1 1 1 1 1 1 1 1 1 -- -- -- -- --]
 [-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- 1 1 1 1 1 1 1 1 -- -- -- -- --]
 [-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- 1 1 1 1 -- -- -- -- -- --]]

使用蒙面数组，您可以只进行所需的操作，并且使用的唯一元素是未屏蔽的元素。

Answer 8

答案取决于您要执行的操作。对于索引和值的标准枚举，有一个内置的 numpy 迭代器，即 numpy.ndenumerate，它与维度无关。

import numpy as np

rng = np.random.default_rng()
a = rng.integers(2, size=(5,5))
print(a)

for ind, val in np.ndenumerate(a):
    if val == 1:
        print(ind)

给予

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

请注意，掩码数组通常是对特定条目集进行操作的更好方法。

获取特定值python的数组索引

8 个答案: