在Python中创建此numpy数组 [英] Creating this numpy array in Python

问题描述

我有以下numpy数组

I have the following numpy array

import numpy as np
a = np.array([1,2,6,8])

我想从a创建另一个numpy数组，以使其包含a的两个元素的所有不同可能和.然后很容易表明存在int(a.size*(a.size-1)/2)个不同的和，它们由以下组成:

I want to create another numpy array from a such that it contains all the different possible sums of TWO elements of a. It's easy to show then that there are int(a.size*(a.size-1)/2) different possible sums, composed from:

a[0] + a[1]
a[0] + a[2]
a[0] + a[3]
a[1] + a[2]
a[1] + a[3]
a[2] + a[3]

如何在不使用double for循环的情况下构造具有上述总和的numpy数组(我可以想到的唯一方法).对于上面的示例，输出应为[3,7,9,8,10,14]

How can I construct a numpy array with the above sums as elements without using a double for loop (the only way I can think of it). For the above example, the output should be [3,7,9,8,10,14]

MWE

eff = int(a.size*(a.size-1)/2)
c = np.empty((0, eff))

推荐答案

您可以使用triu_indices:

i0,i1 = np.triu_indices(4,1)
a[i0]
# array([1, 1, 1, 2, 2, 6])
a[i1]
# array([2, 6, 8, 6, 8, 8])
a[i0]+a[i1]
# array([ 3,  7,  9,  8, 10, 14])

有关更多术语，我们需要构建自己的"nd_triu_idx".这是在5个列表中的3个词中执行此操作的方法:

For more terms we need to build our own "nd_triu_idx". Here is how to do it for 3 terms out of a list of 5:

n = 5
full = np.mgrid[:n,:n,:n]
nd_triu_idx = full[:,(np.diff(full,axis=0)>0).all(axis=0)]
nd_triu_idx
# array([[0, 0, 0, 0, 0, 0, 1, 1, 1, 2],
#        [1, 1, 1, 2, 2, 3, 2, 2, 3, 3],
#        [2, 3, 4, 3, 4, 4, 3, 4, 4, 4]])

要完全概括术语的数量，请使用类似的

To fully generalize the number of terms use something like

k = 4
full = np.mgrid[k*(slice(n),)]

等

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