用列表的值替换numpy索引数组的值 [英] Replace values of a numpy index array with values of a list

问题描述

假设您有一个numpy数组和一个列表：

Suppose you have a numpy array and a list:

>>> a = np.array([1,2,2,1]).reshape(2,2)
>>> a
array([[1, 2],
       [2, 1]])
>>> b = [0, 10]

我想替换数组中的值，以便1取而代之的是0和2乘10。

I'd like to replace values in an array, so that 1 is replaced by 0, and 2 by 10.

我在这里发现了类似的问题 - http://mail.python.org/pipermail//tutor/2011-September/085392.html

I found a similar problem here - http://mail.python.org/pipermail//tutor/2011-September/085392.html

但是使用这个解决方案：

But using this solution:

for x in np.nditer(a):
    if x==1:
        x[...]=x=0
    elif x==2:
        x[...]=x=10

给我一​​个错误：

ValueError: assignment destination is read-only

我想这是因为我无法真正写入numpy数组。

I guess that's because I can't really write into a numpy array.

PS numpy数组的实际大小为514乘504，列表为8.

P.S. The actual size of the numpy array is 514 by 504 and of the list is 8.

推荐答案

而不是一个一个地替换值一，可以像这样重新映射整个数组：

Instead of replacing the values one by one, it is possible to remap the entire array like this:

import numpy as np
a = np.array([1,2,2,1]).reshape(2,2)
# palette must be given in sorted order
palette = [1, 2]
# key gives the new values you wish palette to be mapped to.
key = np.array([0, 10])
index = np.digitize(a.ravel(), palette, right=True)
print(key[index].reshape(a.shape))

收益率

[[ 0 10]
 [10  0]]

---

上述想法归功于@JoshAdel 。它明显快于我原来的答案：

---

Credit for the above idea goes to @JoshAdel. It is significantly faster than my original answer:

import numpy as np
import random
palette = np.arange(8)
key = palette**2
a = np.array([random.choice(palette) for i in range(514*504)]).reshape(514,504)

def using_unique():
    palette, index = np.unique(a, return_inverse=True)
    return key[index].reshape(a.shape)

def using_digitize():
    index = np.digitize(a.ravel(), palette, right=True)
    return key[index].reshape(a.shape)

if __name__ == '__main__':
    assert np.allclose(using_unique(), using_digitize())

我用这种方式对这两个版本进行基准测试：

I benchmarked the two versions this way:

In [107]: %timeit using_unique()
10 loops, best of 3: 35.6 ms per loop
In [112]: %timeit using_digitize()
100 loops, best of 3: 5.14 ms per loop

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