divide cell array into date and time columns
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I have a cell array with 6 columns and 10000 rows. The first column contains date and time in the format eg. '31/12/2001 12:57:03'. I want to seperate the first column into two seperate columns, one for date and the other for time. Then I want to convert the date to dateserial number.
thanks
采纳的回答
Walter Roberson
2017-10-7
temp = regexp(YourCellArray(:,1), '\s+', 'split');
datecol = cellfun(@(C) C{1}, temp, 'uniform', 0);
timecol = cellfun(@(C) C{2}, temp, 'uniform', 0)
serialdates = datenum(datecol);
If you do not actually need the time column, then this can be abbreviated down to
serialdates = datenum( regexprep(YourCellArray(:,1), '\s.*', '') );
2 个评论
AA
2017-10-7
Walter Roberson
2017-10-8
YourCellArray{1,1} = '31/12/2001 12:57:03';
YourCellArray{2,1} = '30/12/2001 13:18:03';
then
temp = regexp(YourCellArray(:,1), '[\s:]+', 'split');
datecol = cellfun(@(C) C{1}, temp, 'uniform', 0);
hourcol = cellfun(@(C) C{2}, temp, 'uniform', 0);
mincol = cellfun(@(C) C{2}, temp, 'uniform', 0);
since_midnight = str2double(hourcol) * 60 + str2double(mincol);
... or,
temp = datevec(YourCellArray(:,1));
since_midnight = temp(:,4) * 60 + temp(:,5);
更多回答（1 个）
Peter Perkins
2017-10-13
Think about using datetime and duration instead of datenum:
>> dt = datetime({'30/12/2015 15:54:30';'30/12/2015 15:54:30';'30/12/2015 15:54:30'},'InputFormat','dd/MM/yyyy HH:mm:ss')
dt =
3×1 datetime array
30-Dec-2015 15:54:30
30-Dec-2015 15:54:30
30-Dec-2015 15:54:30
>> d = dateshift(dt,'start','day')
d =
3×1 datetime array
30-Dec-2015 00:00:00
30-Dec-2015 00:00:00
30-Dec-2015 00:00:00
>> t = timeofday(dt)
t =
3×1 duration array
15:54:30
15:54:30
15:54:30
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