Indexing portions of a vector with array

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Mohammed Kagalwala
Mohammed Kagalwala 2019-12-10
评论： Jeremy 2019-12-10
Hi,
I currently have a vector a that is 12 x1. I want to index it such that I get the following vector q = [a(4:7),a(4:7)]. I can create an array [4 4] and [7 7], when I try to index a([4 4], [7 7]). I only get q = a(4:7). Can someone help me with the correct syntax here? Is this even possible with this sort of indexing ?
Sample input:
a = [1 2 3 4 5 6 7 8 9 10 11 12];
Sample output:
q = [a(4:7),a(4:7)] = [4 5 6 7 4 5 6 7]
The size of a can vary and the index 4:7 can vary
Thank you.
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Jeremy
Jeremy 2019-12-10
What exactly is the problem? The code you provided appears to do what you are asking for
Mohammed Kagalwala
Mohammed Kagalwala 2019-12-10
编辑：Mohammed Kagalwala 2019-12-10
I want to perform this action in general where all I'm given is an array of start indecies and an array of stop indecies. In the example, provided above this would [4 4], [7 7]. This could be of size N. I want to create q using these two arrays, WITHOUT using for loops.

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回答（1 个）

Jeremy
Jeremy 2019-12-10
编辑：Jeremy 2019-12-10
I had to think about this for a while, but I think that this accomplishes the task you are looking for
a = 1:12;
id_start = [4 4]';
id_stop = [7 7]';
c = arrayfun(@colon,id_start,id_stop,'UniformOutput',false);
d = cell2mat(c);
p = reshape(d',[1 numel(d)]);
q = a(p);
  2 个评论
Mohammed Kagalwala
Mohammed Kagalwala 2019-12-10
Thanks ! Works just as I described. Though I did a timing test against a for-loop which showed the for-loop was the faster method. I'm curious if there's a way to beat a for-loops speed.
Jeremy
Jeremy 2019-12-10
I think arrayfun is, in general, not great if you are looking to optimize efficiency. But you asked for no loops and I wanted to solve the puzzle - I wasn't able to get the colon operator to work on the array of start and stop indices without it. Maybe there is another way...

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