How to convert date variable with varying length?
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aaron Harvey
am 12 Feb. 2016
Kommentiert: aaron Harvey
am 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
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Fangjun Jiang
am 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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Jan
am 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;
Weitere Antworten (2)
Matthew Eicholtz
am 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 Kommentare
Jan
am 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
am 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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