问题描述
我发现我的一个非常类似的问题，但不完全一样。
这其中：here
然而，在ntimes的情况下，数组的大小相匹配的尺寸元组点的数量。
在我来说，我有一个4维数组和二维元组，就像这样：
I found a very similar question to mine, but not exactly the same.This one: hereHowever in ntimes's case the size of the array matches the number of the dimensions the tuple is point at.In my case I have a 4-dimensional array and a 2-dimensional tuple, just like this:
from numpy.random import rand
big_array=rand(3,3,4,5)
tup=(2,2)
我要使用的元组作为索引到第一两维，并手动索引的最后两个。是这样的：
I want to use the tuple as an index to the first two dimensions, and manually index the last two. Something like:
big_array[tup,3,2]
不过，我获得与指数= 2的第一个维度的重复，沿着第四维（因为它在技术上并没有被索引）。这是因为这个索引是除preting双索引的第一尺寸，而不是一个值对每个维，
However, I obtain a repetition of the first dimension with index=2, along the fourth dimension( since it technically hasn't been indexed). That is because this indexing is interpreting a double indexing to the first dimension instead of one value for each dimension,
eg.
| dim 0:(index 2 AND index 2) , dim 1:(index 3), dim 2:(index 2), dim 3:(no index)|
instead of
|dim 0(index 2), dim 1(index 2), dim 2:(index 3), dim 3:(index 2)|.
我怎样才能解开''这个元组呢？有任何想法吗？
谢谢！
How can I 'unpack' this tuple then? Any ideas?thanks!
推荐答案
您还可以通过在你的第一个元组单独获得感兴趣的片，然后再建立索引seprately：
You can also pass in your first tuple alone to get the slice of interest, then index it seprately:
from numpy.random import rand
big_array=rand(3,3,4,5)
chosen_slice = (2,2)
>>> big_array[ chosen_slice ]
array([[ 0.96281602, 0.38296561, 0.59362615, 0.74032818, 0.88169483],
[ 0.54893771, 0.33640089, 0.53352849, 0.75534718, 0.38815883],
[ 0.85247424, 0.9441886 , 0.74682007, 0.87371017, 0.68644639],
[ 0.52858188, 0.74717948, 0.76120181, 0.08314177, 0.99557654]])
>>> chosen_part = (1,1)
>>> big_array[ chosen_slice ][ chosen_part ]
0.33640088565877657
这可能是一些用户稍微更具可读性，但除此之外，我对mgilson的解决方案倾斜。
That may be slightly more readable for some users, but otherwise I'd lean towards mgilson's solution.
这篇关于元组多维数组的索引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！