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?

Answer 1

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]

Answer 2

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 '<string>' for i in '123')
# any(i in 'a' for i in '123') -> False
# any(i in 'b3' for i in '123') -> True

# And when looking for subsrings
any(i in '<string>' for i in ('11','22','33'))
# any(i in 'hello' for i in ('18','36','613')) -> False
# any(i in '613 mitzvahs' for i in ('18','36','613')) ->True

Answer 3

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!

Answer 4

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"

Answer 5

My simple, simple, simple approach! =D

Code

string_to_test = "The criminals stole $1,000,000 in jewels."
chars_to_check = ["$", ",", "0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
for char in chars_to_check:
    if char in string_to_test:
        print("Char \"" + char + "\" detected!")

Output

Char "$" detected!
Char "," detected!
Char "0" detected!
Char "1" detected!

Answer 6

Check if chars are in String:

parse_string = lambda chars, string: [char in string for char in chars]

example:

parse_string('$,x', 'The criminals stole $1,000,000 in ....') 

or

parse_string(['$', ',', 'x'], '..minals stole $1,000,000 i..')

output: [True, True, False]

Answer 7

Another approach, perhaps pythonic, is this:

aString = """The criminals stole $1,000,000 in jewels."""
#
if any(list(map(lambda char: char in aString, '0123456789,$')))
   print(True) # Do something.

Answer 8

Replace alphanumerics, spaces, hyphens and full-stops by a blank and then count the number of remaining characters:

import re
s = 'hello%world/'
specialCharacters = re.sub('[a-zA-Z0-9-\s\.()]',','',s)
print( "Special Characters:", specialCharacters )
print( "Length:", len( specialCharacters ) )

and the result is:

Special Characters: $/
Length: 2

and then you can just extend the regular expression if for instance you want to include quotes or questions marks as normal characters.

Answer 9

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")