# How to convert date variable with varying length?

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aaron Harvey on 12 Feb 2016
Commented: aaron Harvey on 13 Feb 2016
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
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?

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;
aaron Harvey on 13 Feb 2016
thank-you that worked a treat

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)
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 CommentsShowHide 1 older comment
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.

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)