how to vectorize the following code?

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UCL student
UCL student 2014-10-23
评论： Star Strider 2014-10-23
Could anyone explain the steps?
DelayStep = 60;%in seconds
MaxStep = 86400;
for j=dStart:dEnd
currDay = datevec(j);
for k=0:DelayStep:MaxStep
hms = [0 0 0 fix(mod(k, [0, 3600, 60]) ./ [3600, 60, 1])];
UTCTimetoCalculate = datestr(hms+currDay);
end
end
  2 个评论
Patrik Ek
Patrik Ek 2014-10-23
Well, what have you tried? How about vectorizing k for the innermost loop by the way?
UCL student
UCL student 2014-10-23
I tried this in the inner section, but it doesn't seem to work:
k=0:DelayStep:MaxStep
hms = [0 0 0 fix(mod(k(:), [0, 3600, 60]) ./ [3600, 60, 1])];
UTCTimetoCalculate = datestr(hms+currDay)

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采纳的回答

Star Strider
Star Strider 2014-10-23
You can replace the inner ‘k’ loop with:
currDay = datevec(j);
hms = bsxfun(@plus, currDay, zeros(1440,6));
hms(:,5) = 1:1440;
TCTimetoCalculate = datestr(datenum(hms));
since you are actually incrementing every minute (with 1440 minutes/day)|.
  2 个评论
UCL student
UCL student 2014-10-23
It looks good, thanks. But how to vectorize the outer part?
Star Strider
Star Strider 2014-10-23
This will work, assuming I understand what you want to do. You can vectorise both loops in four statements:
dStart = datenum(now);
dEnd = datenum(now+7); % Spans One Week
Minutes = fix(dEnd-dStart+1)*1440;
hms = bsxfun(@plus, datevec(dStart), zeros(Minutes,6));
hms(:,5) = 1:Minutes;
TCTimetoCalculate = datestr(datenum(hms));
I created ‘dStart’ and ‘dEnd’ to test it, and it works assuming they are both date numbers. If they aren’t convert them to date numbers to use my code.
To be sure it is producing the results you want, this line lets you look at the first five and last five entries of ‘TCTimetoCalculate’:
TCTimetoCalculate([1:5 end-5:end],:)

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