Python：当更改相同时，是否有更快的方法来修改相同形状的多个数组？

vnjpjtjt 于 2022-12-17 发布在 Python

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我有多个numpy数组，它们都是相同的形状，我有两个面具也是相同的形状。
现在我一次改变一个数组，有没有更快的方法？
这是我正在做的一个小规模的例子。

start = time.time()

mask1 = np.array([[1,0,1],[0,1,1],[1,0,1]])
mask2 = np.array([[0,0,0],[1,0,1],[0,0,1]])

arr1 = np.array([[20,10,51],[21,1,2],[25,23,38]])
arr2 = np.array([[99,1,6],[66,54,11],[22,21,1]])
arr3 = np.array([[23,2,3],[55,2,16],[90,37,1]])
arr4 = np.array([[81,25,22],[1,63,24],[47,58,1]])

arr1[(mask1 == 1) & (mask2 == 0)] = 9999
arr2[(mask1 == 1) & (mask2 == 0)] = 9999
arr3[(mask1 == 1) & (mask2 == 0)] = 9999
arr4[(mask1 == 1) & (mask2 == 0)] = 9999

print(time.time() - start)

python

来源：https://stackoverflow.com/questions/74819916/python-numpy-is-there-a-faster-way-to-modify-multiple-arrays-of-the-same-shape

2条答案

按热度按时间

t1qtbnec1#

在@hpaulj的建议下，我做了以下几点，看到了一些改进！

import timeit

def func1():
    mask1 = np.array([[1,0,1],[0,1,1],[1,0,1]])
    mask2 = np.array([[0,0,0],[1,0,1],[0,0,1]])
    arr1 = np.array([[20,10,51],[21,1,2],[25,23,38]])
    arr2 = np.array([[99,1,6],[66,54,11],[22,21,1]])
    arr3 = np.array([[23,2,3],[55,2,16],[90,37,1]])
    arr4 = np.array([[81,25,22],[1,63,24],[47,58,1]])
    arr1[(mask1 == 1) & (mask2 == 0)] = 9999
    arr2[(mask1 == 1) & (mask2 == 0)] = 9999
    arr3[(mask1 == 1) & (mask2 == 0)] = 9999
    arr4[(mask1 == 1) & (mask2 == 0)] = 9999

def func2():
    mask1 = np.array([[1,0,1],[0,1,1],[1,0,1]])
    mask2 = np.array([[0,0,0],[1,0,1],[0,0,1]])
    mask1[(mask1==1) & (mask2 != 0)] = 0
    arr1 = np.array([[20,10,51],[21,1,2],[25,23,38]])
    arr2 = np.array([[99,1,6],[66,54,11],[22,21,1]])
    arr3 = np.array([[23,2,3],[55,2,16],[90,37,1]])
    arr4 = np.array([[81,25,22],[1,63,24],[47,58,1]])
    arr1[(mask1 == 1)] = 9999
    arr2[(mask1 == 1)] = 9999
    arr3[(mask1 == 1)] = 9999
    arr4[(mask1 == 1)] = 9999

print(timeit.timeit(func1, number=1000000))
#17.574998669995693

print(timeit.timeit(func2, number=1000000))
#15.318040108977584

赞(0）回复(0）举报 2022-12-17

2uluyalo2#

我想说的是：

mask3 = (mask1 == 1) & (mask2 == 0)    
arr1[mask3] = 9999
arr2[mask3] = 9999
arr3[mask3] = 9999
arr4[mask3] = 9999

赞(0）回复(0）举报 2022-12-17

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Python：当更改相同时，是否有更快的方法来修改相同形状的多个数组？

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