If Regexp matches return 1 otherwise 0 syntax
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Michael
am 25 Mär. 2013
Bearbeitet: gwoo
am 8 Feb. 2023
Hi,
Right now I'm using the following to get a boolean result from a regexp. It just doesn't feel right - is there a better way to do this?
(size(regexp(myInput,myPattern),1)>0)
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Walter Roberson
am 25 Mär. 2013
if regexp(myInput,myPattern)
regexp() by default returns a list of indices upon a match, and [] if there are no matches. The list of indices will all be non-zero numbers, and "if" applied to an array of non-zero numbers is considered to be true, just as if all() had been applied to the list. "if" applied to the empty matrix is false. So, you do not need to do any conversion: you can just test regexp() result directly.
3 Kommentare
Walter Roberson
am 25 Mär. 2013
Bearbeitet: Walter Roberson
am 25 Mär. 2013
~isempty(regexp('aa00aa00', '\d+')) && 1
or
any(regexp('aa00aa00', '\d+')) && 1
Note: all() instead of any() will not work.
Weitere Antworten (3)
gwoo
am 23 Aug. 2021
Bearbeitet: gwoo
am 8 Feb. 2023
If your input to regex is a cell array (say from collecting from a struct or something), then your output will be a cell array which is not immediately able to be used for logical indexing. You need to convert it from a cell array to a logical array. But many times you'll get empty cells so you can't just use cell2mat because that will implicitly ditch the empty cells and only leave you with the non-empty which doesn't help for indexing. Therefore, I use this following approach to go from an input of a cell array to an output of a logical array.
This is how I get a logical array out of regex:
logicalMatches = ~cellfun('isempty', regexpi({filesInDir.name}, stringToBeFound, 'once'));
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James Van Zandt
am 12 Mai 2022
I have a cell array of strings to test, so I used this method to collect the matches.
K>> ca={'able','baker','charlie','delta','echo','fox','golf','hotel'}
ca =
1×8 cell array
{'able'} {'baker'} {'charlie'} {'delta'} {'echo'} {'fox'} {'golf'} {'hotel'}
K>> regexp(ca,'a')
ans =
1×8 cell array
{[1]} {[2]} {[3]} {[5]} {0×0 double} {0×0 double} {0×0 double} {0×0 double}
K>> ~cellfun('isempty',regexp(ca,'a'))
ans =
1×8 logical array
1 1 1 1 0 0 0 0
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