如何有条件地合并两个相同形状的Numpy数组 [英] How to conditionally combine two numpy arrays of the same shape

问题描述

这听起来很简单，我想我的想法太复杂了.

This sounds simple, and I think I'm overcomplicating this in my mind.

我想制作一个数组，其元素由两个形状相同的源数组生成，具体取决于源数组中哪个元素更大.

I want to make an array whose elements are generated from two source arrays of the same shape, depending on which element in the source arrays is greater.

说明:

import numpy as np
array1 = np.array((2,3,0))
array2 = np.array((1,5,0))

array3 = (insert magic)
>> array([2, 5, 0))

我不知道如何产生一个将array1和array2的元素组合在一起的array3，以产生一个只使用两个array1/array2元素值中的较大者的数组.

I can't work out how to produce an array3 that combines the elements of array1 and array2 to produce an array where only the greater of the two array1/array2 element values is taken.

任何帮助将不胜感激.谢谢.

Any help would be much appreciated. Thanks.

推荐答案

我们可以使用内置的NumPy

We could use NumPy built-in np.maximum, made exactly for that purpose -

np.maximum(array1, array2)

另一种方法是使用NumPy ufunc np.max 在2D堆叠数组上，并且max-reduce沿第一个轴(axis=0)-

Another way would be to use the NumPy ufunc np.max on a 2D stacked array and max-reduce along the first axis (axis=0) -

np.max([array1,array2],axis=0)

一百万个数据集的时间-

Timings on 1 million datasets -

In [271]: array1 = np.random.randint(0,9,(1000000))

In [272]: array2 = np.random.randint(0,9,(1000000))

In [274]: %timeit np.maximum(array1, array2)
1000 loops, best of 3: 1.25 ms per loop

In [275]: %timeit np.max([array1, array2],axis=0)
100 loops, best of 3: 3.31 ms per loop

# @Eric Duminil's soln1
In [276]: %timeit np.where( array1 > array2, array1, array2)
100 loops, best of 3: 5.15 ms per loop

# @Eric Duminil's soln2
In [277]: magic = lambda x,y : np.where(x > y , x, y)

In [278]: %timeit magic(array1, array2)
100 loops, best of 3: 5.13 ms per loop

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扩展到其他支持的功能

类似地，还有np.minimum，用于在两个相同或可广播形状的数组之间查找元素方式的最小值.因此，要找到array1和array2之间的逐元素最小值，我们将有:

Similarly, there's np.minimum for finding element-wise minimum values between two arrays of same or broadcastable shapes. So, to find element-wise minimum between array1 and array2, we would have :

np.minimum(array1, array2)

有关支持此功能的ufuncs的完整列表，请参考 docs 并查找关键字:element-wise. Grep-为此，我得到了以下ufuncs:

For a complete list of ufuncs that support this feature, please refer to the docs and look for the keyword : element-wise. Grep-ing for those, I got the following ufuncs :

加，减，乘，除，logaddexp，logaddexp2，true_divide， floor_divide，功率，余数，mod，fmod，divmod，heaviside，gcd， lcm，arctan2，hypo，bitwise_and，bitwise_or，bitwise_xor，left_shift， right_shift，更大，更大等于，更少，更少等于，不等于， 等于，逻辑与，逻辑或，逻辑异或，最大值，最小值，fmax， fmin，copysign，nextafter，ldexp，fmod

add, subtract, multiply, divide, logaddexp, logaddexp2, true_divide, floor_divide, power, remainder, mod, fmod, divmod, heaviside, gcd, lcm, arctan2, hypot, bitwise_and, bitwise_or, bitwise_xor, left_shift, right_shift, greater, greater_equal, less, less_equal, not_equal, equal, logical_and, logical_or, logical_xor, maximum, minimum, fmax, fmin, copysign, nextafter, ldexp, fmod

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