9
\$\begingroup\$

Background

I have a string in Python which I want to convert to an integer. Normally, I would just use int:

>>> int("123")
123

Unfortunately, this method is not very robust, as it only accepts strings that match -?[0-9]+ (after removing any leading or trailing whitespace). For example, it can't handle input with a decimal point:

>>> int("123.45")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: invalid literal for int() with base 10: '123.45'

And it certainly can't handle this:

>>> int("123abc?!")

On the other hand, exactly this behavior can be had without any fuss in Perl, PHP, and even the humble QBasic:

INT(VAL("123abc"))   ' 123

Question

Here's my shortest effort at this "generalized int" in Python. It's 50 bytes, assuming that the original string is in s and the result should end up in i:

n="";i=0
for c in s:
 n+=c
 try:i=int(n)
 except:0

Fairly straightforward, but the try/except bit is ugly and long. Is there any way to shorten it?

Details

Answers need to do all of the following:

  • Start with a string in s; end with its integer value in i.
  • The integer is the first run of digits in the string. Everything after that is ignored, including other digits if they come after non-digits.
  • Leading zeros in the input are valid.
  • Any string that does not start with a valid integer has a value of 0.

The following features are preferred, though not required:

  • A single - sign immediately before the digits makes the integer negative.
  • Ignores whitespace before and after the number.
  • Works equally well in Python 2 or 3.

(Note: my code above meets all of these criteria.)

Test cases

"0123"   -> 123
"123abc" -> 123
"123.45" -> 123
"abc123" -> 0
"-123"   -> -123 (or 0 if negatives not handled)
"-1-2"   -> -1 (or 0 if negatives not handled)
"--1"    -> 0
""       -> 0
\$\endgroup\$
5
  • \$\begingroup\$ Somewhat related: codegolf.stackexchange.com/questions/28783/… (but there it was explicitly stated that input would be properly-formed integers). \$\endgroup\$
    – DLosc
    May 27, 2015 at 2:57
  • 1
    \$\begingroup\$ What should "12abc3" give? \$\endgroup\$
    – orlp
    May 27, 2015 at 6:10
  • \$\begingroup\$ @orlp 12--it's analogous to the "123.45" case. \$\endgroup\$
    – DLosc
    May 27, 2015 at 6:38
  • \$\begingroup\$ (lambda(x)(or(parse-integer x :junk-allowed t)0)) (Common Lisp, 49 bytes) -- Only posted as a comment since it is built-in. \$\endgroup\$
    – coredump
    May 27, 2015 at 11:28
  • 1
    \$\begingroup\$ @coredump :junk-allowed--ha, that's great! I would have made this a general golf challenge, were it not for the fact that the answer in many languages is trivial. But thanks for the Lisp. :^) \$\endgroup\$
    – DLosc
    May 27, 2015 at 14:24

2 Answers 2

6
\$\begingroup\$

Python 2, 47, 46

It's not as short as using regex, but I thought it was entertainingly obscure.

i=int(('0%sx'%s)[:~len(s.lstrip(str(1<<68)))])

-1 due to KSab – str with some large integer works better than the repr operator since it does not put an L on the end.

\$\endgroup\$
3
  • 2
    \$\begingroup\$ you can shave off a byte by using str(1<<68) inside the lstrip \$\endgroup\$
    – KSab
    May 27, 2015 at 8:28
  • \$\begingroup\$ Wow. Entertainingly obscure is right! (This only handles nonnegative numbers, correct?) \$\endgroup\$
    – DLosc
    May 27, 2015 at 14:41
  • \$\begingroup\$ Another bonus of @KSab's suggestion is Python 3 compatibility. \$\endgroup\$
    – DLosc
    May 27, 2015 at 14:42
4
\$\begingroup\$

40 bytes

import re;i=int("0"+re.split("\D",s)[0])

and you can do negatives for 8 characters more:

import re;i=int((re.findall("^-?\d+",s)+[0])[0])
\$\endgroup\$
2
  • \$\begingroup\$ @DLosc Ah you're right, didn't test the second one well enough apparently. The 'aha' moment was when I realized some python regex functions return strings not MatchObjects \$\endgroup\$
    – KSab
    May 27, 2015 at 7:00
  • 1
    \$\begingroup\$ import re;i=int((re.findall("^-?\d+",s)+[0])[0]) works, for 48 bytes. \$\endgroup\$
    – DLosc
    May 27, 2015 at 14:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.