Index exceeds Matrix Dimension
이전 댓글 표시
I'm following a comparatively larger code and I'm stuck in this section.
T1=num2str(TTime);
if length(T1)==9
T1=['0',T1];
end
T11=datenum([str2double(T1(1,1:4)) str2double(T1(1,5:6)) str2double(T1(1,7:8)) str2double(T1(1,9:10)) str2double(T1(1,11:12)) str2double(T1(1,13:14))]);
Here, TTime is an array of numbers that specifies a date/time. (Ex: 20140705001529. i.e. year,month,date,hour,minutes,second) But I keep getting the error "Index exceeds matrix dimensions." Initially, str2double was replaced with str2num.
I know this snippet might not be enough to figure things out but any help from you guys is appreciated! If you have any suggestions, please do let me know.
Thanks in advance.
댓글 수: 6
Adam
2019년 12월 11일
Obvious question is are you sure T1 always has at least 14 elements?
Jake
2019년 12월 11일
"...are you sure T1 always has at least 14 elements?"
Judging by the code, sometimes it is expected to have 10 elements:
if length(T1)==9
T1=['0',T1];
end
"TTime is an array of numbers..."
Calling num2str on an array of numbers creates a character array whose size depends on the values in that numeric array, with multiple spaces separating the columns. It is not robust code to assume anything about its size.
Without an accurate description of TTime it is hard to advise what alternatives you could try. Best would be if you click the paperclip button to upload TTime in a .mat file.
Jake
2019년 12월 11일
Adam
2019년 12월 11일
You should be able to find out why you are getting the error trivially using the Pause on Errors option from the Run menu. It will stop the code at the line of the error. Then you can simply look at the relevant components in the workspace or on command line. How to fix it is another matter, but simply to know what the cause is should be very simple.
Jake
2019년 12월 12일
답변 (1개)
Simpler, more robust code:
>> N = 20140705001529; % Ugh... dates should not be stored like this!
>> S = sprintf('%d',N);
>> S(end+1:14) = '0'; % ensure 14 digits
>> T = datenum(S,'yyyymmddHHMMSS')
T =
7.357850107523148e+05
Checking:
>> datevec(T)
ans =
2014 7 5 0 15 29
Or even better using datetime:
>> T = datetime(S,'InputFormat','yyyyMMddHHmmss')
T =
05-Jul-2014 00:15:29
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