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Replacing numerics in text using regular expressions.

Asked by D. Plotnick on 28 Sep 2017
Latest activity Commented on by Cedric Wannaz
on 28 Sep 2017
Hello, I am trying to figure out whether it is possible to dynamically replace numeric values in a long text block using regular expressions. Here is an example from a made-up xml file.
str = '<document><placemark><when>5</when><lat>41</lat></placemark><placemark><when>11</when></placemark></document>';
Now, I want to perform some numerical function on all of the outputs of <when>, lets say subtract 3 so that the string will read
'<document><placemark><when>2</when><lat>41</lat></placemark><placemark><when>8</when></placemark></document>';
I can already find the locations using
exp='<when>(\d+)</when>'
but I don't know how to
  1. extract the actual numerical value at that location,
  2. perform some arbitrary function on that value (subtraction, addition, division, anything)
  3. write that new value back into the string so it reads <when>newValue</when>
If I was certain that the number of characters would stay the same, I could do a for-loop with some pretty gross indexing. However, as in the above example the length of the charstring representing the numeric value might change as a result of my function (11 became 8).
I suspect there is either a really elegant regexp solution, or it is not possible at all. Hoping for the former. Cheers, Dan

  3 Comments

Apologies on the inadvertent "broken links". The angle brackets have a different meaning here than in my example. A side question, is there a way to suppress that behavior?
Sometimes for presentation purposes here you need to change < to &lt; which shows up like <
Please see my last edit with a more "classical" approach.

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2 Answers

Answer by Cedric Wannaz
on 28 Sep 2017
Edited by Cedric Wannaz
on 28 Sep 2017
 Accepted Answer

Here is a small trick assuming that no character in your string interferes with formatSpec stuff:
pattern = '(?<=<when>)\d+' ;
values = str2double( regexp( str, pattern, 'match' )) ;
values = values - 3 ; % Some operation.
fSpec = regexprep( str, pattern, '%d' ) ; % Make input str a formatSpec ;)
newStr = sprintf( fSpec, values ) ;
Got to run, I will answer more tonight if you don't get a better answer.
Ok, got 5 more minutes, otherwise a good alternative was mentioned by Walter and is based on the fact that you can run MATLAB code within the replacement pattern:
repFun = @(s) sprintf( '%d', sscanf( s,'%d' ) - 3 ) ; % Update function.
newStr = regexprep( str, '(?<=<when>)\d+', '${repFun($0)}' ) ;
Finally a more "classical" approach, that matches and splits the input string, replaces the matches and rebuilds the output.
[numbers, parts] = regexp( str, '(?<=<when>)\d+', 'match', 'split' ) ;
numbers = arrayfun( @(x) sprintf( '%d', x ), str2double( numbers ) - 3, ...
'UniformOutput', false ) ;
buffer = [parts; [numbers, {''}]] ;
newStr = sprintf( '%s', buffer{:} ) ;

  2 Comments

I updated the answer after you accepted it (@ 20:58 UTC), adding a more classical approach.
Thanks! Both to you and Walter. This worked very well, and I was hoping that I would be able to run MATLAB functions in this context, so that opens up a whole bunch of additional ways I can use this method.
Thanks again, Dan

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Answer by Walter Roberson
on 28 Sep 2017

Yes. If you can devise a regexp pattern to isolate the number, then you can use regexprep with the ${cmd} replacement. Arguments to the commands will be passed as strings. Values can be returned as strings or as integers that will be converted to strings.
For example,
str = '<document><placemark><when>5</when><lat>41</lat></placemark><placemark><when>11</when></placemark></document>';
regexprep(str, '\d+', '${$0 - 2}')
I did not test this code (my system is busy at the moment)

  2 Comments

This unfortunately had some odd behavior (Evaluation of '$0 - 2' did not produce a char vector or scalar string.) and changing this to
regexprep(str, '\d+', '${num2str($0 - 2)}')
did in fact return a modified string, but nothing like what I expected. Cedric seemed inspired by your idea however, and his solution worked quite well.
Danke, Dan
You could have done it by coding the conversion to double as well. $0 refers to the match, which is a string. It must be converted to double before you can do math. Instead of loading the replacement string with commands, I created a function repFun that we call, and this function does the double conversion string-num-string.

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