count of months spanned

2 次查看（过去 30 天）
Leah
Leah 2013-8-15
I can't think of a way to do this without a loop. I have a bunch of start and end dates. Each row is a record and I want to be about to count how many months are in each date range. Here's some sample data startdates=[735406;735416;735404;735396;735363;735389]; enddates=[735433;735433;735433;735425;735416;735416];
So Jun 21-Jul 18 would be two months
>>countmonths =
2 1 2 2 3 2
My way is not very efficient
for i=1:length(startdates)
daterange=startdates(i):enddates(i);
dv=datevec(daterange);
countmonths(i)=size(unique(dv(:,[1 2]),'rows'),1);
end

请先登录，再回答此问题。

采纳的回答

Sven
Sven 2013-8-15
Hi Leah,
I think you can do this quite nicely as follows:
startvecs = datevec(startdates);
endvecs = datevec(enddates);
diffvecs = endvecs - startvecs;
countmonths = 12*diffvecs(:,1) + diffvecs(:,2)+1
Does that work for you? I'm not quite sure what you'd want to do between, say, Jan1 and Feb1, but I think if you take a look at diffvecs you'll understand what it contains and be able to make it work for your purposes.
  1 个评论
Leah
Leah 2013-8-19
Thanks, this does work for me. Jan1 and Feb1 would still count as two months, which is correct for my purposes. Thanks!

请先登录，再进行评论。

更多回答（1 个）

Jan
Jan 2013-8-15
编辑：Jan 2013-8-15
When you convert the serial date numbers to date vectors you get e.g.:
s = [2013 06 21 12 13 14; ...
2013 06 21 1 2 3]
e = [2013 07 18 12 13 14; ...
2014 06 21 1 2 3]
Now the distance in months is:
(e(:, 1) * 12 + e(:, 2)) - (s(:, 1) * 12 + s(:, 2)) + 1
Or more compact:
(e(:, 1:2) - s(:, 1:2)) * [12; 1] + 1
  1 个评论
Leah
Leah 2013-8-19
Thanks Jan! This was really similar to the first post they both work great

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Dates and Time 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by