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Replacing each array element with a series of values

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Varun Nair
Varun Nair 2017-4-25
评论： Varun Nair 2017-4-29
Hi, I have a a 1 x N matrix of numbers. For example, a matrix T:
T = [1 4 7]
I would like to convert T to the new matrix as follows:
T_new = [1 2 3 4 5 6 7 8 9]
Basically, each element of T is replaced by an arithmetic series of 3 values with that element as the starting element and common difference = 1.
I tried
T_new = [T:T+3];
But it only updates the first element (I get T_new = [1 2 3 4]). Is there any quick way to do this without using a for loop ?
Thanks.

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采纳的回答

Stephen23
Stephen23 2017-4-25
>> T = [1,4,7];
>> cell2mat(arrayfun(@(n)n+(0:2),T,'uni',0))
ans =
1 2 3 4 5 6 7 8 9

更多回答（2 个）

Andrei Bobrov
Andrei Bobrov 2017-4-25
编辑：Andrei Bobrov 2017-4-25
T_new = reshape(T(:)' + (0:2)',1,[])
  3 个评论
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Andrei Bobrov
Andrei Bobrov 2017-4-26
You are using an older version of MATLAB.
Variant for you:
T_new = reshape(bsxfun(@plus,T(:)',(0:2)'),1,[])

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dbmn
dbmn 2017-4-25
One possible solution would be:
0. having
T = [1 4 7];
1. create a temporary matrix T_temp that looks like (this could be done in a loop very easily)
T_new = [1 2 3;
4 5 6;
7 8 9 ];
2. then reshape that matrix using
T_new=reshape(T_tmp', 1, numel(T_tmp));
1 2 3 4 5 6 7 8 9
  1 个评论
Varun Nair
Varun Nair 2017-4-25
编辑：Varun Nair 2017-4-25
Thanks, but, as I mentioned in the question, I was looking for a way to do it without using a for loop (i.e. without having to loop through all elements).

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