添加不同形状的numpy数组 [英] adding numpy arrays of differing shapes

问题描述

我想添加两个不同形状的numpy数组，但不进行广播，而是将缺失"值视为零.像这样的示例可能最简单

I'd like to add two numpy arrays of different shapes, but without broadcasting, rather the "missing" values are treated as zeros. Probably easiest with an example like

[1, 2, 3] + [2] -> [3, 2, 3]

或

[1, 2, 3] + [[2], [1]] -> [[3, 2, 3], [1, 0, 0]]

我事先不知道形状.

我正在弄乱每个np.shape的输出，试图找到容纳两个形状的最小形状，将每个形状嵌入到该形状的零位数组中，然后添加它们.但这似乎需要大量工作，是否有更简单的方法?

I'm messing around with the output of np.shape for each, trying to find the smallest shape which holds both of them, embedding each in a zero-ed array of that shape and then adding them. But it seems rather a lot of work, is there an easier way?

提前谢谢！

很多工作"我的意思是很多工作对我来说"而不是对机器而言，我追求的是优雅而不是效率:我努力使两者都保持最小的形状是

edit: by "a lot of work" I meant "a lot of work for me" rather than for the machine, I seek elegance rather than efficiency: my effort getting the smallest shape holding them both is

def pad(a, b) :
    sa, sb = map(np.shape, [a, b])
    N = np.max([len(sa),len(sb)])
    sap, sbp = map(lambda x : x + (1,)*(N-len(x)), [sa, sb])
    sp = np.amax( np.array([ tuple(sap), tuple(sbp) ]), 1)

不漂亮:-/

推荐答案

这是我能想到的最好的方法:

This is the best I could come up with:

import numpy as np

def magic_add(*args):
    n = max(a.ndim for a in args)
    args = [a.reshape((n - a.ndim)*(1,) + a.shape) for a in args]
    shape = np.max([a.shape for a in args], 0)
    result = np.zeros(shape)

    for a in args:
        idx = tuple(slice(i) for i in a.shape)
        result[idx] += a
    return result

如果您知道期望结果有多少个维度，可以稍微清理一下for循环，例如:

You can clean up the for loop a little if you know how many dimensions you expect on result, something like:

for a in args:
    i, j = a.shape
    result[:i, :j] += a

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