Using RegEx (strrep) on CharArray with For Loop

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Hello. I have a series of dates that need to be reformatted as such:
2007-04-22 01:04:21.000
2007-05-10 01:08:48.000
2007-06-15 22:03:31.000
2007/04/22 01:04:21
2007/05/10 01:08:48
2007/06/15 22:03:31
I currently have this loop to try and replace each "-" with "/". (I also need to cut off the ".000", but I have not gotten to that yet.)
newTimes = get(c, 'START_STR');
for j = 1 : size(newTimes, 1);
fprintf('Looped'); %Just for testing
newsTimes(j) = strrep(newTimes(j), '-' ,'/')
However this does not work.
I am no longer getting an error that "Error using ==> strrep Input strings must have one row". Which I had before I put it in a loop. But my output is still wrong.
It does NOT make the swap of '-' for "/" and it also prints out the array for each time the loop runs, as opposed to printing one line for each run of the loop.
Any ideas? Ideally I would like to find a function similar to strrep that can be used in an array... otherwise a way to fix the loop would be great.
My Output looks Like this:
newstationtimes =
2007-04-22 01:04:21.000
2007-05-10 01:08:48.000
2007-06-15 22:03:31.000
... rest of list
Repeated like 40 times.
Cedric Wannaz
Cedric Wannaz on 29 Apr 2013
Edited: Cedric Wannaz on 29 Apr 2013
Well, I'd personally go for a reshape of the char. stream into a ?x25 array (or 23/24 if \r and or \n are not present), then a replacement like buffer(buffer=='-') = '/', and finally I'd kill columns 20 to 23.
Regexp are useless because of the rigid structure, but if the time format was more flexible or if there was text with variable length between the date and the time, one way to use pattern matching would be:
content = fileread('myFileWithTimeStamps.txt') ;
content_new = regexprep(content, '(.{4})-(..)-(.*?)\.000', '$1/$2/$3') ;

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Accepted Answer

Daniel Shub
Daniel Shub on 29 Apr 2013
What about
datestr(['2007-04-22 01:04:21.000';'2007-05-10 01:08:48.000';'2007-06-15 22:03:31.000'], 'yyyy/mm/dd HH:MM:SS')
I think some of the date functions are slow (if so Jan will probably chime in with the faster way) ...
  1 Comment
Brian Castro
Brian Castro on 29 Apr 2013
Thanks Strongbad!
errr... I mean Daniel.
That worked great. I don't anticipate it being too slow.

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More Answers (1)

Sean de Wolski
Sean de Wolski on 29 Apr 2013
Edited: Sean de Wolski on 29 Apr 2013
Two easy ways:
If the data is indeed all column-aligned, how about:
data(:,[5 8]) = '/'
data = data(:,1:end-4);
If the data is not perfectly column aligned:
  • Converting them to date numbers with datenum() and the format that you have?
  • Convert them back to a string using datestr() with your choice output format?
This surely won't be the fastest way.
  1 Comment
Brian Castro
Brian Castro on 29 Apr 2013
Thanks! Since Im using dates I went with datestr, but both of those should work

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