How to check a string for specific characters

Question

How can I check if a string has several specific characters in it using Python 2   For example  given the following string      The criminals stole  1 000 000 in jewels    How do I detect if it has dollar signs        commas        and numbers

User · Answer

Quick comparison of timings in response to the post by Abbafei:

import timeit

def func1():
    phrase = 'Lucky Dog'
    return any(i in 'LD' for i in phrase)

def func2():
    phrase = 'Lucky Dog'
    if ('L' in phrase) or ('D' in phrase):
        return True
    else:
        return False

if __name__ == '__main__': 
    func1_time = timeit.timeit(func1, number=100000)
    func2_time = timeit.timeit(func2, number=100000)
    print('Func1 Time: {0}\nFunc2 Time: {1}'.format(func1_time, func2_time))

Output:

Func1 Time: 0.0737484362111
Func2 Time: 0.0125144964371

So the code is more compact with any, but faster with the conditional.

EDIT : TL;DR -- For long strings, if-then is still much faster than any!

I decided to compare the timing for a long random string based on some of the valid points raised in the comments:

# Tested in Python 2.7.14

import timeit
from string import ascii_letters
from random import choice

def create_random_string(length=1000):
    random_list = [choice(ascii_letters) for x in range(length)]
    return ''.join(random_list)

def function_using_any(phrase):
    return any(i in 'LD' for i in phrase)

def function_using_if_then(phrase):
    if ('L' in phrase) or ('D' in phrase):
        return True
    else:
        return False

if __name__ == '__main__':
    random_string = create_random_string(length=2000)
    func1_time = timeit.timeit(stmt="function_using_any(random_string)",
                               setup="from __main__ import function_using_any, random_string",
                               number=200000)
    func2_time = timeit.timeit(stmt="function_using_if_then(random_string)",
                               setup="from __main__ import function_using_if_then, random_string",
                               number=200000)
    print('Time for function using any: {0}\nTime for function using if-then: {1}'.format(func1_time, func2_time))

Output:

Time for function using any: 0.1342546
Time for function using if-then: 0.0201827

If-then is almost an order of magnitude faster than any!

User · Answer

This will test if strings are made up of some combination or digits  the dollar sign  and a commas  Is that what you re looking for    import re  s1    Testing string  s2    1234 12345    regex   re compile   0-9         if   regex match s1        print  s1 matched  else     print  s1 didn t match   if   regex match s2        print  s2 matched  else     print  s2 didn t match

User · Answer

Assuming your string is s       in s          found     not in s      not found    original answer given  but less Pythonic than the above    s find       -1   not found s find       -1   found   And so on for other characters       or  pattern   re compile r  d      if pattern findall s       print  Found   else     print  Not found         or  chars   set  0123456789     if any  c in chars  for c in s       print  Found   else      print  Not Found      Edit  added the     in s answers

User · Answer

My simple  simple  simple approach   D Code string to test    quot The criminals stole  1 000 000 in jewels  quot  chars to check     quot   quot    quot   quot    quot 0 quot    quot 1 quot    quot 2 quot    quot 3 quot    quot 4 quot    quot 5 quot    quot 6 quot    quot 7 quot    quot 8 quot    quot 9 quot   for char in chars to check      if char in string to test          print  quot Char   quot  quot    char    quot   quot  detected  quot    Output Char  quot   quot  detected  Char  quot   quot  detected  Char  quot 0 quot  detected  Char  quot 1 quot  detected   Thanks

User · Answer

s input  Enter any character       if s isalnum          print  Alpha Numeric Character         if s isalpha              print  Alphabet character             if s islower                print  Lower case alphabet character             else              print  Upper case alphabet character         else          print  it is a digit      elif s isspace           print  It is space character        else       print  Non Space Special Character

User · Answer

user Jochen Ritzel said this in a comment to an answer to this question from user dappawit  It should work     1  in var  and   2  in var  and   3  in var         1    2   etc  should be replaced with the characters you are looking for   See this page in the Python 2 7 documentation for some information on strings  including about using the in operator for substring tests   Update  This does the same job as my above suggestion with less repetition     When looking for single characters  this checks for any of the characters         since strings are collections of characters any i in   lt string gt   for i in  123     any i in  a  for i in  123   - gt  False   any i in  b3  for i in  123   - gt  True    And when looking for subsrings any i in   lt string gt   for i in   11   22   33      any i in  hello  for i in   18   36   613    - gt  False   any i in  613 mitzvahs  for i in   18   36   613    - gt True

[python] How to check a string for specific characters?

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