How do I remove leading characters from a string?

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pruth
pruth 2019 年 9 月 18 日
コメント済み: Walter Roberson 2019 年 9 月 19 日
I read one ascii file. and created this dummy mat file (attached ).
To make simple matrix out of it i need to split the string
i used this code (below) but it coudnt split the first string which is a date and time with some leading number since there is no space in between. There is this number before the date and time, i want remove those numbers so can use datenum on that string !
for i = 1:len_data
segments = regexp(dummy{1,1}{i,1},' ', 'split');
end

採用された回答

Walter Roberson
Walter Roberson 2019 年 9 月 18 日
segments = regexp(dummy{1,1}{i,1}, '\t', 'split');
Perhaps you could have used readtable() on the ascii file: it would likely have detected the tab delimiter.
  3 件のコメント
Walter Roberson
Walter Roberson 2019 年 9 月 18 日
strjoin(segments(2:3), ' ')
Note that we no longer recommend using datenum: we recommend using datetime()
pruth
pruth 2019 年 9 月 18 日
that works !

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その他の回答 (1 件)

Ankit
Ankit 2019 年 9 月 18 日
Hello Pruth,
One possibility is to use extractBetween command.
a = extractBetween(dummy{1, 1}, 10,20)
where 10 is the start position and 20 is the end position.
Based on your requirement you can also try extractAfter, extractBefore
Cheers
Ankit
  3 件のコメント
Ankit
Ankit 2019 年 9 月 18 日
hello pruth,
cell data can be converted to double using str2double. But you can't use it to convert date, it will return a NaN value.
thanks
Ankit
Walter Roberson
Walter Roberson 2019 年 9 月 19 日
However, datetime() is happy to process cell array of character vectors ;-)
Also, in newer releases, duration() is happy to process cell array of character vectors. This was not possible in older releases.

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