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Indexing portions of a vector with array

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Mohammed Kagalwala
Mohammed Kagalwala am 10 Dez. 2019
Kommentiert: Jeremy am 10 Dez. 2019
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.
  4 Kommentare
Jeremy
Jeremy am 10 Dez. 2019
What exactly is the problem? The code you provided appears to do what you are asking for
Mohammed Kagalwala
Mohammed Kagalwala am 10 Dez. 2019
Bearbeitet: Mohammed Kagalwala am 10 Dez. 2019
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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Antworten (1)

Jeremy
Jeremy am 10 Dez. 2019
Bearbeitet: Jeremy am 10 Dez. 2019
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 Kommentare
Mohammed Kagalwala
Mohammed Kagalwala am 10 Dez. 2019
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 am 10 Dez. 2019
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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