根据 2D numpy 数组过滤 3D numpy 数组 [英] filtering a 3D numpy array according to 2D numpy array

问题描述

我有一个形状为 (3024, 4032) 的二维 numpy 数组.

I have a 2D numpy array with the shape (3024, 4032).

我有一个形状为 (3024, 4032, 3) 的 3D numpy 数组.

I have a 3D numpy array with the shape (3024, 4032, 3).

2D numpy 数组由 0 和 1 填充.

2D numpy array is filled with 0s and 1s.

3D numpy 数组填充了 0 到 255 之间的值.

3D numpy array is filled with values between 0 and 255.

通过查看 2D 数组值，我想更改 3D 数组中的值.如果 2D 数组中的值为 0，我会将 3D 数组中的所有 3 个像素值沿最后一个轴更改为 0.如果二维数组中的值是 1，我不会改变它.

By looking at the 2D array values, I want to change the values in 3D array. If a value in 2D array is 0, I will change the all 3 pixel values in 3D array into 0 along the last axes. If a value in 2D array is 1, I won't change it.

我查了这个问题，如何过滤具有另一个数组值的 numpy 数组，但它适用于具有相同维度的 2 个数组.就我而言，尺寸不同.

I have checked this question, How to filter a numpy array with another array's values, but it applies for 2 arrays which have same dimensions. In my case, dimensions are different.

如何在两个数组中应用过滤，在两个维度上具有相同的大小，但在最后一个维度上没有大小?

How the filtering is applied in two arrays, with same size on 2 dimensions, but not size on the last dimension?

推荐答案

好的，我会回答这个问题以突出一个关于缺失"维度的特殊性.让我们假设 a.shape==(5,4,3) 和 b.shape==(5,4)

Ok, I'll answer this to highlight one pecularity regarding "missing" dimensions. Lets' assume a.shape==(5,4,3) and b.shape==(5,4)

当索引时，现有维度左对齐，这就是@Divakar的解决方案a[b == 0] = 0有效的原因.

When indexing, existing dimensions are left aligned which is why @Divakar's solution a[b == 0] = 0 works.

当广播时，现有尺寸右对齐，这就是@InvaderZim的a*b不起作用的原因.您需要做的是 a*b[..., None] 在右侧插入一个可广播维度

When broadcasting, existing dimensions are right aligned which is why @InvaderZim's a*b does not work. What you need to do is a*b[..., None] which inserts a broadcastable dimension at the right

这篇关于根据 2D numpy 数组过滤 3D numpy 数组的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！