How to convert date variable with varying length?

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aaron Harvey 2016-2-12

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评论： aaron Harvey 2016-2-13

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 2016-2-12

where is your source data come from, a text file, or already stored in a matrix, in what format?

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Answer 1

Jan 2016-2-13

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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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aaron Harvey 2016-2-13

thank-you that worked a treat

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Answer 2

Matthew Eicholtz 2016-2-12

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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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aaron Harvey 2016-2-13

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Thank-you for the reply, I have tried your code and a few slight adjustments as well but with no success.

HOT.date = cellfun(@(x) sprintf('%06s',x),HOT.date,'uni',0);
HOT.date = datestr(datenum(num2str(HOT.date(i)),'mmddyy'),'yyyymmdd');

Where HOT.date is my variable. I think its not working as my dates are in a double not a cell array, i would preferably like to keep them as a double as well as all the corresponding variables are, but i don't have enough MATLAB know-how to know how.

Jan 2016-2-13

@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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Answer 3

Azzi Abdelmalek 2016-2-12

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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)