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
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Fangjun Jiang
2016년 2월 12일
where is your source data come from, a text file, or already stored in a matrix, in what format?
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Jan
2016년 2월 13일
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;
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Matthew Eicholtz
2016년 2월 12일
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);
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Jan
2016년 2월 13일
@Aaron: You see that it would have been useful to post some example data in the question to clarify the type.
Azzi Abdelmalek
2016년 2월 12일
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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