在numpy数组中获得第k个维度的第i个切片 [英] get the i-th slice of the k-th dimension in a numpy array

问题描述

我有一个n维的numpy数组，我想获得第k个维的第i个切片.一定有比...更好的东西

I have an n-dimensional numpy array, and I'd like to get the i-th slice of the k-th dimension. There must be something better than

# ... 
elif k == 5:
    b = a[:, :, :, :, :, i, ...]
# ...

推荐答案

b = a[(slice(None),) * k + (i,)]

手动构建索引元组.

Python语言参考中所述，表格

a[:, :, :, :, :, i]

转换为

a[(slice(None), slice(None), slice(None), slice(None), slice(None), i)]

我们可以直接构建该元组而不使用切片符号来达到相同的效果. (有一点需要注意的是，构建元组直接生成a[(i,)]而不是k=0的a[i]，但是NumPy对于标量i的处理相同.)

We can achieve the same effect by building that tuple directly instead of using slicing notation. (There's the minor caveat that building the tuple directly produces a[(i,)] instead of a[i] for k=0, but NumPy handles these the same for scalar i.)

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