why does this not work?
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>> x='0:07.50 2:03.91 3:57.36 5:09.44 6:32.90 7:43.03 9:43.45';
>> datetime(x,'InputFormat','m:ss.SS')
Error using datetime (line 257)
Unable to convert '0:07.50 2:03.91 3:57.36 5:09.44 6:32.90 7:43.03 9:43.45' to
datetime using the format 'm:ss.SS'.
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추가 답변 (3개)
It failed because datetime did not recognize that string as a list of distinct times. Just because you know what you want code to do, does not mean it will read your mind. Code is pretty dumb in general.
x='0:07.50 2:03.91 3:57.36 5:09.44 6:32.90 7:43.03 9:43.45';
xspl = strsplit(x,' ')
datetime(xspl,'InputFormat','m:ss.SS')
need to separate the elements of the array (that I changed to a cell array here).
Then, it works —
x = {'0:07.50' '2:03.91' '3:57.36' '5:09.44' '6:32.90' '7:43.03' '9:43.45'};
datetime(x,'InputFormat','m:ss.SS')
.
EDIT — (27 Mar 2025 at 21:13)
Perhaps this —
x = {'0:07.50' '2:03.91' '3:57.36' '5:09.44' '6:32.90' '7:43.03' '9:43.45'};
dt = datetime(x,'InputFormat','m:ss.SS')
dtm = seconds(timeofday(dt))
This works, and there may also be other ways of doing that conversion.
.
댓글 수: 3
S = [ "0:07.50" "2:03.91" "3:57.36"];
DT = datetime(S, "InputFormat", "m:ss.SS")
D = seconds(DT - dateshift(DT, 'start', 'day'))
Or you can create a duration instead of a datetime and avoid having to do the dateshift.
S = [ "0:07.50" "2:03.91" "3:57.36"];
DT = duration(S, "InputFormat", "mm:ss.SS")
D = seconds(DT)
Walter Roberson
2025년 3월 27일
Ah, when I tested with duration, I missed that you need to use mm instead of m
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