If string "x" contains any letter or number, print that string. How to do that using regular expressions? The code below is wrong
if re.search('^[A-Z]?[a-z]?[0-9]?', i):
print i
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5 Answers
re — Regular expression operations
This question is actually rather tricky. Unfortunately \w includes _ and [a-z] solutions assume a 26-letter alphabet. With the below solution please read the pydoc where it talks about LOCALE and UNICODE.
"[^_\\W]"
Note that since you are only testing for existence, no quantifiers need to be used -- and in fact, using quantifiers that may match 0 times will returns false positives.
4 Comments
Andre Miller
To get rid of the double \\, you can use a raw string: r'[^_\W]'
hasen
It just occured to me that
[^_\\] looks like a pirate
Bart Kiers
That would also match string consisting solely of punctuation marks.
Matthieu M.
How ? \W matches punctuation marks, so its negation will not.
You want
if re.search('[A-Za-z0-9]+', i):
print i
2 Comments
Paul McMillan
Removed the +. It's habit since an expression like that invariably becomes more complex in my own code, and I usually do want the + if there's anything else to it.
I suggest that you check out RegexBuddy. It can explain regexes well.
Comments
[A-Z]?[a-z]?[0-9]? matches an optional upper case letter, followed by an optional lower case letter, followed by an optional digit. So, it also matches an empty string. What you're looking for is this: [a-zA-Z0-9] which will match a single digit, lower- or upper case letter.
And if you need to check for letter (and digits) outside of the ascii range, use this if your regex flavour supports it: [\p{L}\p{N}]. Where \p{L} matches any letter and \p{N} any number.
Comments
don't need regex.
>>> a="abc123"
>>> if True in map(str.isdigit,list(a)):
... print a
...
abc123
>>> if True in map(str.isalpha,list(a)):
... print a
...
abc123
>>> a="##@%$#%#^!"
>>> if True in map(str.isdigit,list(a)):
... print a
...
>>> if True in map(str.isalpha,list(a)):
... print a
...
