Comparing strings containing numbers in Python

Question

I have a list of strings and I would like to verify some conditions on the strings. For example:

String_1: 'The price is 15 euros'. 
String_2: 'The price is 14 euros'. 
Condition: The price is > 14 --> OK 

How can I verify it? I'm actually doing like this:

if ('price is 13' in string): 
    print('ok')

and I'm writing all the valid cases. I would like to have just one condition.

Answer 1

You can just compare strings if they differ only by number and numbers have the same digits count. I.e.:

String_1 = 'The price is 15 euros'
String_2 = 'The price is 14 euros'
String_3 = 'The price is 37 EUR'

The will be naturally sorted as String_3 > String_1 > String_2

But will NOT work for:

String_4 = 'The price is 114 euros'

it has 3 digits instead of 2 and it will be String_4 < String_3 thus

So, the better, if you can extract number from the string, like following:

import re
def get_price(s):
    m = re.match("The price is ([0-9]*)", s)
    if m:
        return = int(m.group(1))
    return 0

Now you can compare prices as integer:

price = get_price(String_1)
if price > 14:
    print ("Okay!")

. . .

if get_price(String_1) > 14:
    print ("Okay!")

([0-9]*) - is the capturing group of the regular expression, all defined in the round parenthesis will be returned in group(1) method of the Python match object. You can extend this simple regular expression [0-9]* further for your needs.

If you have list of strings:

string_list = [String_1, String_2, String_3, String_4]

for s in string_list:
    if get_price(s) > 14:
        print ("'{}' is okay!".format(s))