Calculate duration from cell with date hours with Matlab 2013a

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Christine Abou Nasr
评论： Christine Abou Nasr 2017-4-25
I have a cell array of more than 16700 rows and 3 column as in the example below:
'01/01/2014' '12:00:00' 'AM'
'01/01/2014' '06:00:00' 'AM'
'01/01/2014' '10:00:00' 'AM'
'01/01/2014' '12:00:00' 'PM'
'09/04/2014' '11:00:00' 'PM'
I would like to calculate the duration of two chosen dates and time (for example the last row-first raw=number of days, hours and seconds). My big problem is that I have a R2013a version of Matlab. So I can't use DateTime, I tried this:
newA = cellfun(@(s) strsplit(s, ' '), DateTime, 'UniformOutput', false);
DateTimeSep = vertcat(newA{:});
%%Calculs de durée et récupération de données
date=datenum(DateTimeSep(:,1),'mm/dd/yyyy');
d=datestr(date(1256,1)-date(1,1), 'dd')
Time=DateTimeSep(10,2)-DateTimeSep(1,2)% I didn't try to consider the AM PM problem yet....
I get d=04 (which is false) and for time I get: "Undefined function 'minus' for input arguments of type 'cell'"

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Jan
Jan 2017-4-3
编辑：Jan 2017-4-8
% [Don't use this - see [EDITED 2] for a nicer solution!]
D = {'01/01/2014' '12:00:00' 'AM'; ...
'01/01/2014' '06:00:00' 'AM'; ...
'01/01/2014' '10:00:00' 'AM'; ...
'01/01/2014' '12:00:00' 'PM'; ...
'09/04/2014' '11:00:00' 'PM'};
DateV = datevec(strcat(D(:, 1), '#', D(:, 2)), 'mm/dd/yyyy#HH:MM:SS');
IsPM = strcmpi(D(:, 3), 'PM');
pstPM = (DateV(:, 4) < 12 & IsPM);
DateV(pstPM, 4) = DateV(pstPM, 4) + 12; % 12:01PM ==> 12:01, 1:01PM ==> 13:01
midAM = (DateV(:, 4) == 12 & ~IsPM);
DateV(midAM, 4) = 0; % 12:01AM ==> 00:01
DateN = datenum(DateV);
Now the time difference can be calculated by a simple subtraction.
datevec(DateN(end) - DateN(1)) % [EDITED, DateTimeN -> DateN]
[EDITED 2] No, the conversion of the time difference to months and years is not meaningful: It is not clear if the "month" has 28, 29, 30 or 31 days, if it concerns the difference between dates. Better:
Duration = DateN(end) - DateN(1);
DurationDays = fix(Duration);
Tmp = rem(Duration, 1); % Time
DurationHour = floor(Tmp * 24);
DurationMin = floor(rem(Tmp * 1440, 60));
DurationSec = round(rem(Tmp * 86400, 60));
Sorry, this is too ugly! Nicer:
D = {'02/14/2014', '10:02:04', 'AM'; ...
'02/26/2014', '01:00:00', 'PM'};
DateN = datenum(strcat(D(:, 1), '#', D(:, 2), '#', D(:, 3)), ...
'mm/dd/yyyy#HH:MM:SS#AM');
Duration = DateN(end) - DateN(1);
DDays = fix(Duration);
[~,~,~, Dhour, Dmin, Dsec] = datevec(rem(Duration, 1));
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dpb
dpb 2017-4-25
Well, that way splits on a a blank instead of parsing the date/time fields. Apparently there's one (or more) record that aren't all the same length. After the split, look at
l=cellfun(@length, newA, 'UniformOutput', false);
and use min() max() or unique() to see who's the odd man out.
That will tell you what is the problem in concatenating. But, the real answer is to use the correct format string and convert directly.

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dpb
dpb 2017-4-3
编辑：dpb 2017-4-3
If using version prior to introduction of datetime, just use date numbers directly...
>> dn=datenum([char(A{:,1}) char(A{:,2}) char(A{:,3})],'mm/dd/yyyyHH:MM:SSAM'); % to datenumbers
>> dt=dn(end)-dn(1); % a difference first-last
>> [y,m,d,h,m,s]=datevec(dt) % numeric difference in y,m,d,...
y =
0
m =
9
d =
2
h =
23
m =
0
s =
0
>> datestr(dt,'dd HH:MM:SS') % as string in days HH:MM:SS
ans =
02 23:00:00
>>
ADDENDUM
Hmmm...I thought datestr and datevec were just a little smarter than are--I was thinking they would take the value 246 days and reflect that in the datestr conversion if used a field of days only. But, they wrap internally to months so if want days and hours of elapsed time use
days=fix(dt);
then can convert from
fracdays=dt=days;
with datevec datestr to get the time fields in whichever form desired.

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