Checking string for conditions

Question

Let's say I have a string, and I want to check a lot of conditions on this string. For example:

1. It is of size x
2. Has no white spaces
3. The last letter is a numeric character

All of the conditions from 1-3 It can be done in a regular standard way in a function (if this and that etc')

But... How would I do that in one line in a good python style way?

• mgilson: There are certainly other ways to do it, but I don't think anything will be as clear as the "if this and that etc."
• NullUserException: Just use a regex, \S{x-1}\d$, (replacing x-1 with the actual number)
• mayhewr: Now you have 2 problems.
• NullUserException: @mayhewr Do enlighten us. Is there a justification to this vitriol against regexes besides "I don't understand them"?
• mayhewr: To me, regex, while totally awesome, powerful, and admittedly appropriate in OP's situation, is not the "pythonic" way of solving this problem, which is what OP asked for. The "pythonic" way is what @mgilson said IMO. Also, I think it's funny.

Answer 1

You can use a simple if condition with any:

s='fdsfgsgsfds9'
if len(s)==7 and not any(c.isspace() for c in s) and s[-1].isdigit():
 pass

• hexparrot: +1 for the non-regex solution that is far more readable for all users, especially those not regex-inclined.
• DSM: I would write this as if len(s) == 7 and not any(c.isspace() for c in s) and s[-1].isdigit(): somewhat easier to read, I think.
• Ashwini Chaudhary: @DSM thanks for teaching me about any(),solution edited. :)

Answer 2

It may be more complicated than it's worth, but you can check all of those conditions with a regular expression.

For example, if the size you wanted was 8 characters, you could use the following to check all three conditions:

if re.match(r'\S{7}\d$', text):
 print 'all conditions match'

Answer 3

One way would be to use a regular expression (assuming an x of 10):

if re.match(r"\S{10}(?<=\d)$", mystring):
 # Success!

• Tim Pietzcker: @PaulBeckingham: No, it isn't. You do know about lookaround assertions?
• Paul Beckingham: 10 + the digit, so your example is for 11, not 10.
• NullUserException: @PaulBeckingham That's a zero-width assertion.
• Jeff Tratner: Why not change this to use a format string and use any number. This would be valid, right? "\\S{%d}(?<=\\d)$" % x (not sure if you can do this with raw string)
• Tim Pietzcker: @Jeff: Good idea. And yes, it also works with a raw string: re.match(r"\S{%d}(?<=\d)$" % x, mystring)

Answer 4

Try something like

import re
def test(s):
 return len(s)>=x and re.match("^\S*\d$", s)

This will test whether the string has length of at least x and that it is a sequence of non-space-characters followed by a digit character at the end.