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Regular expression to search for Gadaffi
I'm trying to search for the word Gadaffi. What's the best regular expression to search for this?
15 Answers
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Can I squash commits in Mercurial?
I have a pair of commits that should really be just one. If I was using git, I would use:
8 Answers
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Python - Check If Word Is In A String
... answered Mar 16 '11 at 1:13
fabrizioMfabrizioM
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Are there pronounceable names for common Haskell operators? [closed]
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Here is how I pronounce them:
>>= bind
>> then
*> then
-> to a -> b: a to b
<- bind (as it desugars to >>=)
<$> (f)map
<$ map-replace by 0 <$ f: "f map-replace by ...
How to run a single test with Mocha?
...a to test my JavaScript stuff. My test file contains 5 tests. Is that possible to run a specific test (or set of tests) rather than all the tests in the file?
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Convert the values in a column into row names in an existing data frame
...he values in a column of an existing data frame into row names. Is is possible to do this without exporting the data frame and then reimporting it with a row.names = call?
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How can I strip all punctuation from a string in JavaScript using regex?
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If you want to remove specific punctuation from a string, it will probably be best to explicitly remove exactly what you want like
replace(/[.,\/#!$%\^&\*;:{}=\-_`~()]/g,"")
Doing the above still doesn't return the string as you have specified it. If you want to remove any extra spaces ...
How do I find the duplicates in a list and create another list with them?
..., 2, 5]
Note that Counter is not particularly efficient (timings) and probably overkill here. set will perform better. This code computes a list of unique elements in the source order:
seen = set()
uniq = []
for x in a:
if x not in seen:
uniq.append(x)
seen.add(x)
or, more c...
Representing and solving a maze given an image
What is the best way to represent and solve a maze given an image?
10 Answers
10
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How to replace all dots in a string using JavaScript
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You need to escape the . because it has the meaning of "an arbitrary character" in a regular expression.
mystring = mystring.replace(/\./g,' ')
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