Question

我需要根据元素出现在其他replacement数组

中的条件替换 Numpy 2D数组中的元素

例如：

>>> main = np.random.randint(5, size=(3, 4))
>>> main
array([[1, 2, 4, 2],
   [3, 2, 3, 2],
   [4, 4, 2, 3]])
>>> repl = [2,3]
>>> main[main in repl] = -1

我想将repl中的所有值更改为-1，因此我希望main为：

[[1, -1, 4, -1],
[-1, -1, -1, -1],
[4, 4, -1, -1]]

但是在尝试将ValueError置于替换条件中时会引发in

ValueError：具有多个元素的数组的真值是不明确的。使用a.any（）或a.all（）

Answer 1

我们可以使用np.in1d创建所有此类事件的拼合蒙版，并在展平输入中将其设置为-1，如此 -

main.ravel()[np.in1d(main, repl)] = -1

或者，我们可以使用np.putmask，从而避免np.ravel()避免显式展平，如此 -

np.putmask(main, np.in1d(main, repl), -1)

Answer 2

你可以制作一个布尔掩码并像这样使用它：

/* Selects the first <th> child of the second <tr>
   descendent of the <thead>, and also every first
   <td> child contained within the <tbody>: */
thead tr:nth-child(2) th:first-child,
tbody td:first-child {
  border-right-width: 0;
}

/* Selects the second <th> child of the second <tr>
   descendent of the <thead>, and also every second
   <td> child contained within the <tbody>: */
thead tr:nth-child(2) th:nth-child(2),
tbody td:nth-child(2) {
  border-left-width: 0;
}

/* Selects every <th> element that follows the second
   <th> child element of the second <tr> of the <thead>
   using the general sibling ('~') combinator; and also
   every <td> that appears after the second <td> of the
   <tbody>: */
thead tr:nth-child(2) th:nth-child(2) ~ th,
tbody td:nth-child(2) ~ td {
  border-left-width: 0;
}
/* Because we (presumably) want the final <th> and <td>
   of each <tr> to retain its right-border, we here select
   every <th> element which is not the last-child of its
   parent which follows the second-child of the parent, and
   similarly for the <td> elements: */
thead tr:nth-child(2) th:nth-child(2) ~ th:not(:last-child),
tbody td:nth-child(2) ~ td:not(:last-child) {
  border-right-width: 0;
}

Answer 3

不确定是否有任何本地numpy方法，但在旧式方式python中你可以这样做：

import numpy as np

main = np.random.randint(5, size=(3, 4))
repl = [2,3]
for k1, v1 in enumerate(main):
    for k2, v2 in enumerate(v1):
        if(v2 in repl):
            main[k1][k2] = -1
print(main)

根据另一个数组中的出现替换2D数组中的元素

3 个答案: