Overcoming 'index exceeds matrix dimensions' without changing methodology of code

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Cary
Cary 2015-8-19
回答： Star Strider 2015-8-19
This matrix below is a 25x7 matrix. Basically what I'm doing is taking a start date and an end date, and adding 1 to the start date, and subtracting 1 from the end date. The problem is when I get to the last (25th) iteration, where my index exceeds the matrix dimensions. Here the start date is 20081210 and I need to get 20081211. How can I do so without changing the methodology of my code? Thank you.
for i = 1:length(matrix)
plus1=matrix(i+1,1);
minus1=matrix(i,2)-1;
[~,startIdx]=ismember(plus1,date); % index days in between entry date and exit date
[~,cutoffIdx]=ismember(minus1,date); % index days in between entry date and exit date
j=date(startIdx:cutoffIdx);
end
  1 个评论
dpb
dpb 2015-8-19
The loop would run w/o error if you simply used length(matrix)-1 as the upper limit.
But, it doesn't look like this makes any sense to be using a loop here as you are unless this is a shortened example and there's something else not shown as all the values of j other than the last are going to waste.
What's the real objective?; in most instances like this with Matlab you can likely do away with the loop entirely--again unless there's much else that isn't shown in play.

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Star Strider
Star Strider 2015-8-19
I don’t entirely undrestand what you’re doing, or what you’re adding or subtracting 1 from. Here are two ways of dealing with dates, the first using datenum and the second creating date vectors:
datevcts = [2008*ones(25,1), 12*ones(25,1), 11*ones(25,1), [[1:25]' zeros(25,2)]];
matrix_dn = datenum(datevcts);
matrix_dv = datevec(matrix_dn);
That may give us a place to begin sorting this.

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