How to convert date variable with varying length?

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Hi there, I have a date variable with thousands of dates in the format mmddyy i would like to convert it to a yyyymmdd format. A problem i am having is whenever the month is bellow 10 the 0 is missing e.g. rather than reading 012988 it reads 12988. Thank-you
  1 Comment
Fangjun Jiang
Fangjun Jiang on 12 Feb 2016
where is your source data come from, a text file, or already stored in a matrix, in what format?

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Accepted Answer

Jan
Jan on 13 Feb 2016
I'd avoud the time consuming indirection over CHAR strings, but convert the doubles directly:
d = [120316, 12988];
year = rem(d, 100);
day = rem(floor(d / 100), 100);
month = rem(floor(d / 10000), 100);
% Decide how to convert the 2 digits year to a 4 digits year:
year = 1900 + year + 100 * (year < 1960);
result = year * 10000 + month * 100 + day;

More Answers (2)

Matthew Eicholtz
Matthew Eicholtz on 12 Feb 2016
If your data is stored in a cell array, such as
d = {'21216','112515','91101','122515','70487'}; %random dates in mmddyy format (with leading '0' missing)
then add the leading '0' to months Jan-Sep,
d = cellfun(@(x) sprintf('%06s',x),d,'uni',0)
and use the built-in date functions to convert to the desired format,
newdates = num2cell(datestr(datenum(d,'mmddyy'),'yyyymmdd'),2);
  2 Comments
Jan
Jan on 13 Feb 2016
@Aaron: You see that it would have been useful to post some example data in the question to clarify the type.

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Azzi Abdelmalek
Azzi Abdelmalek on 12 Feb 2016
You have to indicate the format of your data. Suppose your data are like below:
d=[12988 112589 52214]
e=arrayfun(@(x) num2str(x),d,'un',0)
f=cellfun(@(x) [num2str(zeros(1,6-numel(x))) x],e,'un',0)
out=cellfun(@(x) datestr(datenum(x,'mmddyy'),'yyyymmdd'),f,'un',0)

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