If Regexp matches return 1 otherwise 0 syntax

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)

[Azzi Abdelmalek]

What do you mean?

Answer 1

[Walter Roberson]

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.

[Michael]

Doesn't seem to work with other boolean operators, try:

 regexp('aa00aa00','[0-9]+') && 1

Because two returns are given, it is unhappy, thus the size function, which seems awkward.

[Walter Roberson]

~isempty(regexp('aa00aa00', '\d+')) && 1

or

any(regexp('aa00aa00', '\d+')) && 1

Note: all() instead of any() will not work.

[Michael]

Thanks

Answer 2

[gwoo]

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'));

[James Van Zandt]

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

Answer 3

[Sean de Wolski]

doc strfind %?

Answer 4

[Dani A]

a more modern solution:

% for reproducibility
rng(0);
% 10 strings
strSize = [10 1];
% 10 characters in each string
numChars = 10;
% only lowercase Latin characters
unicodeBnds = [97 122];

%
myInput = arrayfun(@(a) string(char(randi(unicodeBnds,[1 numChars]))),ones(strSize))

myInput = 10×1 string array
    "vxdxqchoyz"
    "ezymudkxuy"
    "rawyrttkre"
    "sahbcvsiya"
    "ljtuemlqst"
    "hrredmyipf"
    "tgnsxyoddg"
    "vgvgyjfgqm"
    "jvpoxhttjo"
    "bbnuydomai"

myPattern = "[abc]";

contains(myInput,regexpPattern(myPattern))

ans = 10×1 logical array
   1
   0
   1
   1
   0
   0
   0
   0
   0
   1

[gwoo]

this is nice but doesn't work on a cell array unless you use:

cellfun(@(x) contains(x, regexpPattern(myPattern)), myInput)

Unrecognized function or variable 'myInput'.

[Walter Roberson]

@gwoo

The original solution (size(regexp(myInput,myPattern),1)>0) was also not generalizable to cell arrays myInput so I do not think this was a consideration for the original poster.

[Dani A]

@gwoo if `myInput` is a properly formatted cell string, then my solution should work. I suspect that you had a cell array of 1-by-1 strings, rather than a cell array of character vectors.

% for reproducibility
rng(0);
% 10 strings
cellSize = [10 1];
% 10 characters in each string
numChars = 10;
% only lowercase Latin characters
unicodeBnds = [97 122];
%
myInput = cellfun(@(c) char(randi(unicodeBnds,[1 numChars])),repmat({1},cellSize),'UniformOutput',false)

myInput = 10×1 cell array
    {'vxdxqchoyz'}
    {'ezymudkxuy'}
    {'rawyrttkre'}
    {'sahbcvsiya'}
    {'ljtuemlqst'}
    {'hrredmyipf'}
    {'tgnsxyoddg'}
    {'vgvgyjfgqm'}
    {'jvpoxhttjo'}
    {'bbnuydomai'}

myPattern = '[abc]';
contains(myInput,regexpPattern(myPattern))

ans = 10×1 logical array
   1
   0
   1
   1
   0
   0
   0
   0
   0
   1

[gwoo]

@Walter Roberson yes that's a good point.

@Dani A wasn't even thinking of that. thank you!

Answer 5

[Azzi Abdelmalek]

Maybe you want to use isequal or strcmp