For loop or Array?

2015 7 29
2 个回答
更新时间：2015 7 31
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0 个投票

Have a matrix [0 0 0 0 0 0; 0 0 0 1 1 1; 0 0 0 0 1 1; 0 0 0 0 1 1; 0 0 0 0 0 1; 0 0 0 1 1 1]. Want Matlab, in any row it encounters 1, to replace the first three zeros before the 1 with 1 so that I will have something like this [0 0 0 0 0 0; 1 1 1 1 1 1; 0 1 1 1 1 1; 0 1 1 1 1 1; 0 0 1 1 1 1; 1 1 1 1 1 1]. How do I go about this? I have a very big matrix I am dealing with.

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Williams Ackaah
Williams Ackaah 2015-7-29
Dear Andrei Bobrov, thanks for your brilliance. Have learnt some new functions. However, with the data I am working with, replacing the first three zeros can be at different sections of the matrix. See the example below A = [0 0 0 0 1 1 0 0 0 0 0 1 1; 0 0 0 1 1 1 0 0 0 0 1 1 1; 0 0 0 1 1 1 0 0 0 0 1 1 1; 0 0 0 0 0 1 0 0 0 0 0 0 1] and this is what I expect to generate Ans = [0 1 1 1 1 1 0 0 1 1 1 1 1; 1 1 1 1 1 1 0 1 1 1 1 1 1; 1 1 1 1 1 1 0 1 1 1 1 1 1; 0 0 1 1 1 1 0 0 0 1 1 1 1]. Thanks
Andrei Bobrov
Andrei Bobrov 2015-7-29
编辑：Andrei Bobrov 2015-7-29
see part of my answer after 'add'
Williams Ackaah
Williams Ackaah 2015-7-31
编辑：Williams Ackaah 2015-7-31
Dear Stephen Cobeldick, Thank you for your answer. As I mentioned, my matrix is very big (24 by 1440) and the position(s) of where the ones are first encountered is not fixed (may be different for each row). Is a for loop the way to go?
Stephen23
Stephen23 2015-7-31
Using loops seems reasonable to me.
Your requirements exclude the use of circshift and cumsum in a trivial combination (e.g. Andrei Bobrov's initial answer). While there might be more compact solutions, using loops is likely the least obfuscated and yet also reasonably fast.

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回答（2 个）

Andrei Bobrov
Andrei Bobrov 2015-7-29
编辑：Andrei Bobrov 2015-7-29

1 个投票

out = cumsum(circshift(A,[0,3]),2)>0;
add
m = 3
b = [zeros(size(A,1),1),diff(A,[],2)]==1;
out = cumsum(b(:,[m+1:end,1:m])-b,2)+A;
Stephen23
Stephen23 2015-7-29
编辑：Stephen23 2015-7-29

0 个投票

>> A = [0 0 0 0 1 1 0 0 0 0 0 1 1; 0 0 0 1 1 1 0 0 0 0 1 1 1; 0 0 0 1 1 1 0 0 0 0 1 1 1; 0 0 0 0 0 1 0 0 0 0 0 0 1]
A =
0 0 0 0 1 1 0 0 0 0 0 1 1
0 0 0 1 1 1 0 0 0 0 1 1 1
0 0 0 1 1 1 0 0 0 0 1 1 1
0 0 0 0 0 1 0 0 0 0 0 0 1
>> B = A;
>> for k = 1:3, B(:,4-k:end-k) = B(:,4-k:end-k) | A(:,4:end); end
>> B
B =
0 1 1 1 1 1 0 0 1 1 1 1 1
1 1 1 1 1 1 0 1 1 1 1 1 1
1 1 1 1 1 1 0 1 1 1 1 1 1
0 0 1 1 1 1 0 0 0 1 1 1 1

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