how to vectorize the following code?

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UCL student
UCL student am 23 Okt. 2014
Kommentiert: Star Strider am 23 Okt. 2014
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 Kommentare
Patrik Ek
Patrik Ek am 23 Okt. 2014
Well, what have you tried? How about vectorizing k for the innermost loop by the way?
UCL student
UCL student am 23 Okt. 2014
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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Star Strider
Star Strider am 23 Okt. 2014
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 Kommentare
UCL student
UCL student am 23 Okt. 2014
It looks good, thanks. But how to vectorize the outer part?
Star Strider
Star Strider am 23 Okt. 2014
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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