Join the list on the pipe character |, which represents different options in regex.

string_lst = ['fun', 'dum', 'sun', 'gum']
x="I love to have fun."

print re.findall(r"(?=("+'|'.join(string_lst)+r"))", x)

Output: ['fun']

You cannot use match as it will match from start. Using search you will get only the first match. So use findall instead.

Also use lookahead if you have overlapping matches not starting at the same point.

The Machine Learning Lab at the University of Trieste, Italy, wrote a web app to solve this exact problem, available at: Automatic Generation of Regular Expressions from Examples based on "genetic programming" and published a paper about it. Based on the fact that it had to come from the Machine Learning Lab of a university in Italy, and was worth publishing a paper about, it is probably a pretty hard problem to solve. Doesn't sound like the type of question for Code Golf SE, but I'm new here, so I wouldn't be the expert on that.

Assuming you're using POSIX regcomp/regexec, each call to regexec will only find the first match in the string. To find the next, use the end position of the first match (the 0th entry of the regmatch_t array filled by regexec) as an offset to apply to the string before searching it again. Repeat until you have no more matches. You can write a function to do this if you want.

I'm a little unsure of what your actual question is. To match "foo" or "bar", you'd simply want "foo|bar" for your pattern. If you want to do this to a list of strings, you'd likely want to check each string individually—you could join the strings first and check that, but I'm not sure this would be of much use. If you want to get the exact text that matched your pattern, you should surround the pattern in parentheses—such as "([fg]oo|[bt]ar)", which would match "foo", "goo", "bar", or "tar"—then use the Groups property of the Match object to retrieve these captures, so you can determine exactly which word matched. Groups[1] is the first captured value (that is, the value in the first set of parentheses in your pattern). Groups[0] is the entire match. You can also name your captures—"(?<word>[fg]oo|[bt]ar)"—and refer to them by name—Groups["word"]. I would recommend reading through the documentation on regular expression language elements.

As for sanitizing the input, there is no input that will "break" the regex. It might prevent a match, but that's really kinda what regexes are all about, isn't it?

You probably want to use a regular expression for this if your patterns are going to be complex....

you could either use a proper regular expression as your filter (e.g for your specific example it would be new Regex(@"^.*_Test\.txt$") or you could apply a conversion algorithm.

Either way you could then just use linq to apply the regex.

for example

var myRegex=new Regex(@"^.*_Test\.txt$");
List<string> resultList=files.Where(myRegex.IsMatch).ToList();

Some people may think the above answer is incorrect, but you can use a method group instead of a lambda. If you wish the full lamda you would use:

var myRegex=new Regex(@"^.*_Test\.txt$");
List<string> resultList=files.Where(f => myRegex.IsMatch(f)).ToList();

or non Linq

List<string> resultList=files.FindAll(delegate(string s) { return myRegex.IsMatch(s);});

if you were converting the filter a simple conversion would be

 var myFilter="*_Test.txt";
 var myRegex=new Regex("^" + myFilter.Replace("*",".*") +"$");

You could then also have filters like "*Test*.txt" with this method.

However, if you went down this conversion route you would need to make sure you escaped out all the special regular expression chars e.g. "." becomes @".", "(" becomes @"(" etc.......

Edit -- The example replace is TOO simple because it doesn't convert the . so it would find "fish_Textxtxt" so escape atleast the .

so

string myFilter="*_Test.txt";
foreach(char x in @"\+?|{[()^$.#") {
  myFilter = myFilter.Replace(x.ToString(),@"\"+x.ToString());
}
Regex myRegex=new Regex(string.Format("^{0}$",myFilter.Replace("*",".*")));

Since your capture groups define explicitly one character on either side of the common word, it's looking for space word space and then when it doesn't find another space, it fails.

In this case, since you don't want to match all the characters word boundary's would catch (period, apostrophe, etc.) you need to use a bit of trickery with lookaheads, lookbehinds, and non-capture groups. Try this:

(?:^|(?<= ))(one|common|word|or|another)(?:(?= )|$)

http://regex101.com/r/cM9hD8

Word boundaries are still simpler to implement, so for reference sake, you could also do this (though it would include ', ., etc.).

\b(one|common|word|or|another)\b

Regular expressions work on strings, not on a "string list" and not multiple string lists. Wherever you need to process more than one string, you will typically need some environmental code to do the processing. For your example, this code has to apply the regex to every element of the first list, then collect the results and use this results to process the second list.

Said that, the usual approach to apply a regexp to a list of strings is to concatenate them by a separator character like "newline". To concatenate two lists and distinguish them, you would need at least a special "magic" character or word for separating the first list from the second, which is not part of the list. Using such magic can cause some maintenance headaches if you are not very careful, nevertheless by combining this with backreferences, this can be used to solve your problem.

For example, numbered backreferences like \1 to \9 refer to other capturing groups found before. Lets assume you used "###" as a separator for the two lists, a regexp along the lines of

  ^([A-Z])\W*([0-9]+)$.*###.*^(\1\W*\+\+\W*\2)$
    ^         ^          ^     ^           ^
    |         |          |     |           |
   first    second       |    backrefs to first
   group    group        |    or second group
                       lists
                       separator

might be a first approximation for what you are looking for (beware of bugs, I did not test it). Put this into a global regexp search, then it should produce all pairs of matches which fit to your constraints.

As a final remark: the resulting code may be very compact, nevertheless harder to maintain (and probably slower) than a more explicit solution where you process the two lists individually.

There are quite a lot of regular expression packages, but yours seems to match the one in POSIX: regcomp() etc.

The two structures it defines in <regex.h> are:

• regex_t containing at least size_t re_nsub, the number of parenthesized subexpressions.

• regmatch_t containing at least regoff_t rm_so, the byte offset from start of string to start of substring, and regoff_t rm_eo, the byte offset from start of string of the first character after the end of substring.

Note that 'offsets' are not pointers but indexes into the character array.

The execution function is:

• int regexec(const regex_t *restrict preg, const char *restrict string, size_t nmatch, regmatch_t pmatch[restrict], int eflags);

Your printing code should be:

for (int i = 0; i <= r.re_nsub; i++)
{
    int start = m[i].rm_so;
    int finish = m[i].rm_eo;
//  strcpy(matches[ind], ("%.*s\n", (finish - start), p + start));  // Based on question
    sprintf(matches[ind], "%.*s\n", (finish - start), p + start);   // More plausible code
    printf("Storing:  %.*s\n", (finish - start), matches[ind]);     // Print once
    ind++;
    printf("%.*s\n", (finish - start), p + start);                  // Why print twice?
}

Note that the code should be upgraded to ensure that the string copy (via sprintf()) does not overflow the target string — maybe by using snprintf() instead of sprintf(). It is also a good idea to mark the start and end of a string in the printing. For example:

    printf("<<%.*s>>\n", (finish - start), p + start);

This makes it a whole heap easier to see spaces etc.

[In future, please attempt to provide an MCVE (Minimal, Complete, Verifiable Example) or SSCCE (Short, Self-Contained, Correct Example) so that people can help more easily.]

This is an SSCCE that I created, probably in response to another SO question in 2010. It is one of a number of programs I keep that I call 'vignettes'; little programs that show the essence of some feature (such as POSIX regexes, in this case). I find them useful as memory joggers.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <errno.h>
#include <regex.h>

#define tofind    "^DAEMONS=\\(([^)]*)\\)[ \t]*$"

int main(int argc, char **argv)
{
    FILE *fp;
    char line[1024];
    int retval = 0;
    regex_t re;
    regmatch_t rm[2];
    //this file has this line "DAEMONS=(sysklogd network sshd !netfs !crond)"
    const char *filename = "/etc/rc.conf";

    if (argc > 1)
        filename = argv[1];

    if (regcomp(&re, tofind, REG_EXTENDED) != 0)
    {
        fprintf(stderr, "Failed to compile regex '%s'\n", tofind);
        return EXIT_FAILURE;
    }
    printf("Regex: %s\n", tofind);
    printf("Number of captured expressions: %zu\n", re.re_nsub);

    fp = fopen(filename, "r");
    if (fp == 0)
    {
        fprintf(stderr, "Failed to open file %s (%d: %s)\n", filename, errno, strerror(errno));
        return EXIT_FAILURE;
    }

    while ((fgets(line, 1024, fp)) != NULL)
    {
        line[strcspn(line, "\n")] = '\0';
        if ((retval = regexec(&re, line, 2, rm, 0)) == 0)
        {
            printf("<<%s>>\n", line);
            // Complete match
            printf("Line: <<%.*s>>\n", (int)(rm[0].rm_eo - rm[0].rm_so), line + rm[0].rm_so);
            // Match captured in (...) - the \( and \) match literal parenthesis
            printf("Text: <<%.*s>>\n", (int)(rm[1].rm_eo - rm[1].rm_so), line + rm[1].rm_so);
            char *src = line + rm[1].rm_so;
            char *end = line + rm[1].rm_eo;
            while (src < end)
            {
                size_t len = strcspn(src, " ");
                if (src + len > end)
                    len = end - src;
                printf("Name: <<%.*s>>\n", (int)len, src);
                src += len;
                src += strspn(src, " ");
            }
        }
    } 
    return EXIT_SUCCESS;
}

This was designed to find a particular line starting DAEMONS= in a file /etc/rc.conf (but you can specify an alternative file name on the command line). You can adapt it to your purposes easily enough.

I assume your regex_match is some combination of regcomp and regexec. To enable grouping, you need to call regcomp with the REG_EXTENDED flag, but without the REG_NOSUB flag (in the third argument).

regex_t compiled;
regcomp(&compiled, "(match1)|(match2)|(match3)", REG_EXTENDED);

Then allocate space for the groups. The number of groups is stored in compiled.re_nsub. Pass this number to regexec:

size_t ngroups = compiled.re_nsub + 1;
regmatch_t *groups = malloc(ngroups * sizeof(regmatch_t));
regexec(&compiled, str, ngroups, groups, 0);

Now, the first invalid group is the one with a -1 value in both its rm_so and rm_eo fields:

size_t nmatched;
for (nmatched = 0; nmatched < ngroups; nmatched++)
    if (groups[nmatched].rm_so == (size_t)(-1))
        break;

nmatched is the number of parenthesized subexpressions (groups) matched. Add your own error checking.

Regular expressions actually aren't part of ANSI C. It sounds like you might be talking about the POSIX regular expression library, which comes with most (all?) *nixes. Here's an example of using POSIX regexes in C (based on this):

#include <regex.h>        
regex_t regex;
int reti;
char msgbuf[100];

/* Compile regular expression */
reti = regcomp(&regex, "^a[[:alnum:]]", 0);
if (reti) {
    fprintf(stderr, "Could not compile regex\n");
    exit(1);
}

/* Execute regular expression */
reti = regexec(&regex, "abc", 0, NULL, 0);
if (!reti) {
    puts("Match");
}
else if (reti == REG_NOMATCH) {
    puts("No match");
}
else {
    regerror(reti, &regex, msgbuf, sizeof(msgbuf));
    fprintf(stderr, "Regex match failed: %s\n", msgbuf);
    exit(1);
}

/* Free memory allocated to the pattern buffer by regcomp() */
regfree(&regex);

Alternatively, you may want to check out PCRE, a library for Perl-compatible regular expressions in C. The Perl syntax is pretty much that same syntax used in Java, Python, and a number of other languages. The POSIX syntax is the syntax used by grep, sed, vi, etc.