安裝中文字典英文字典辭典工具!
安裝中文字典英文字典辭典工具!
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- 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
- 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
- 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:
- C# Regex Validation Rule using Regex. Match() - Stack Overflow
aaaa999999 matches aaaa9999999 matches aaaa99999999 doesn't match aaa999999 doesn't match Try it as
- How do if statements differ from match case statments in Python?
This question asks for a switch case or match case equivalent in Python It seems since Python 3 10 we can now use match case statement I cannot see and understand the difference between match case and an if, elif statement other than the syntactical differences!
- 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:
- 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
- How does the regular expression (aa)+\1 match aaaaaa?
Again (aa)+ can match the remaining string So it matches the next aa Remaining string - "" Remember, the quantifiers by default act greedy They will match as much as they can Now, (aa)+ can't match any further Next in the pattern is \1 But there is nothing left to match Backtrack the last pattern matched by (aa)+ Remaining string - "aa"
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