What is best pure Python implementation to check if a string contains ANY letters from the alphabet?
string_1 = "(555).555-5555"
string_2 = "(555) 555 - 5555 ext. 5555
Where string_1 would return False for having no letters of the alphabet in it and string_2 would return True for having letter.
Regex should be a fast approach:
re.search('[a-zA-Z]', the_string)
any(c.isalpha() for c in string_1) is deliciously Pythonic.
isalpha even means? This will have totally different behaviors when comparing Python 2 with Python 3. Is Chinese part of the alphabet? If not, you are blindly matching it with your generator on Python 3 (Or Python 2 for unicode strings!). If you want Pythonic, here it is: Simple is better than complex.. And check OP's comment above: He wants only the roman alphabet to be matched.
Match object if there is a match, or None if there isn't. So this is compatible with a if re.search(... pattern.
re module (Python 2.7 to 3.9.5).How about:
>>> string_1 = "(555).555-5555"
>>> string_2 = "(555) 555 - 5555 ext. 5555"
>>> any(c.isalpha() for c in string_1)
False
>>> any(c.isalpha() for c in string_2)
True
set(string_1) be more efficent?
set may or may not reduce function calls, but adds some overhead.
You can use islower() on your string to see if it contains some lowercase letters (amongst other characters). or it with isupper() to also check if contains some uppercase letters:
below: letters in the string: test yields true
>>> z = "(555) 555 - 5555 ext. 5555"
>>> z.isupper() or z.islower()
True
below: no letters in the string: test yields false.
>>> z= "(555).555-5555"
>>> z.isupper() or z.islower()
False
>>>
Not to be mixed up with isalpha() which returns True only if all characters are letters, which isn't what you want.
Note that Barm's answer completes mine nicely, since mine doesn't handle the mixed case well.
I liked the answer provided by @jean-françois-fabre, but it is incomplete.
His approach will work, but only if the text contains purely lower- or uppercase letters:
>>> text = "(555).555-5555 extA. 5555"
>>> text.islower()
False
>>> text.isupper()
False
The better approach is to first upper- or lowercase your string and then check.
>>> string1 = "(555).555-5555 extA. 5555"
>>> string2 = '555 (234) - 123.32 21'
>>> string1.upper().isupper()
True
>>> string2.upper().isupper()
False
I tested each of the above methods for finding if any alphabets are contained in a given string and found out average processing time per string on a standard computer.
~250 ns for
import re
~3 µs for
re.search('[a-zA-Z]', string)
~6 µs for
any(c.isalpha() for c in string)
~850 ns for
string.upper().isupper()
Opposite to as alleged, importing re takes negligible time, and searching with re takes just about half time as compared to iterating isalpha() even for a relatively small string.
Hence for larger strings and greater counts, re would be significantly more efficient.
But converting string to a case and checking case (i.e. any of upper().isupper() or lower().islower() ) wins here. In every loop it is significantly faster than re.search() and it doesn't even require any additional imports.
You can use regular expression like this:
import re
print re.search('[a-zA-Z]+',string)
You can also do this in addition
import re
string='24234ww'
val = re.search('[a-zA-Z]+',string)
val[0].isalpha() # returns True if the variable is an alphabet
print(val[0]) # this will print the first instance of the matching value
Also note that if variable val returns None. That means the search did not find a match
