- How to match, but not capture, part of a regex? - Stack Overflow
The key observation here is that when you have either "apple" or "banana", you must also have the trailing hyphen, but you don't want to match it And when you're matching the blank string, you must not have the trailing hyphen A regex that encapsulates this assertion will be the right one, I think
- C# Regex Validation Rule using Regex. Match() - Stack Overflow
aaaa999999 matches aaaa9999999 matches aaaa99999999 doesn't match aaa999999 doesn't match Try it as
- If two cells match, return value from third - Stack Overflow
=INDEX(B:B,MATCH(C2,A:A,0)) I should mention that MATCH checks the position at which the value can be found within A:A (given the 0, or FALSE, parameter, it looks only for an exact match and given its nature, only the first instance found) then INDEX returns the value at that position within B:B
- mongodb - $match in $lookup result - Stack Overflow
Below answer is for mongoDB 3 6 or later Given that: You have a collection users with a field CompanyID and a collection of companies with a field CompanyID
- regex - Python extract pattern matches - Stack Overflow
import re s = #that big string # the parenthesis create a group with what was matched # and '\w' matches only alphanumeric charactes p = re compile("name +(\w+) +is valid", re flags) # use search(), so the match doesn't have to happen # at the beginning of "big string" m = p search(s) # search() returns a Match object with information about
- Regex: ignore case sensitivity - Stack Overflow
G[a-b] * i string match("G[a-b] *", "i") Check the documentation for your language platform tool to find how the matching modes are specified If you want only part of the regex to be case insensitive (as my original answer presumed), then you have two options:
- OR condition in Regex - Stack Overflow
For example, ab|de would match either side of the expression However, for something like your case you might want to use the ? quantifier, which will match the previous expression exactly 0 or 1 times (1 times preferred; i e it's a "greedy" match) Another (probably more relyable) alternative would be using a custom character group:
- How can I compare two lists in python and return matches
A quick performance test showing Lutz's solution is the best: import time def speed_test(func): def wrapper(*args, **kwargs): t1 = time time() for x in xrange(5000): results = func(*args, **kwargs) t2 = time time() print '%s took %0 3f ms' % (func func_name, (t2-t1)*1000 0) return results return wrapper @speed_test def compare_bitwise(x, y): set_x = frozenset(x) set_y = frozenset(y) return set
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