If Regexp matches return 1 otherwise 0 syntax
43 ビュー (過去 30 日間)
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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)
採用された回答
Walter Roberson
2013 年 3 月 25 日
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 件のコメント
Walter Roberson
2013 年 3 月 25 日
編集済み: Walter Roberson
2013 年 3 月 25 日
~isempty(regexp('aa00aa00', '\d+')) && 1
or
any(regexp('aa00aa00', '\d+')) && 1
Note: all() instead of any() will not work.
その他の回答 (3 件)
gwoo
2021 年 8 月 23 日
編集済み: gwoo
2023 年 2 月 8 日
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'));
1 件のコメント
James Van Zandt
2022 年 5 月 12 日
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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