Ignore case in Python strings

Asked 2008-09-15T12:57:10.677

What is the easiest way to compare strings in Python, ignoring case?

Of course one can do (str1.lower() <= str2.lower()), etc., but this created two additional temporary strings (with the obvious alloc/g-c overheads).

I guess I'm looking for an equivalent to C's stricmp().

[Some more context requested, so I'll demonstrate with a trivial example:]

Suppose you want to sort a looong list of strings. You simply do theList.sort(). This is O(n * log(n)) string comparisons and no memory management (since all strings and list elements are some sort of smart pointers). You are happy.

Now, you want to do the same, but ignore the case (let's simplify and say all strings are ascii, so locale issues can be ignored). You can do theList.sort(key=lambda s: s.lower()), but then you cause two new allocations per comparison, plus burden the garbage-collector with the duplicated (lowered) strings. Each such memory-management noise is orders-of-magnitude slower than simple string comparison.

Now, with an in-place stricmp()-like function, you do: theList.sort(cmp=stricmp) and it is as fast and as memory-friendly as theList.sort(). You are happy again.

The problem is any Python-based case-insensitive comparison involves implicit string duplications, so I was expecting to find a C-based comparisons (maybe in module string).

Could not find anything like that, hence the question here. (Hope this clarifies the question).

python string case-insensitive

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3 Answers

Are you using this compare in a very-frequently-executed path of a highly-performance-sensitive application? Alternatively, are you running this on strings which are megabytes in size? If not, then you shouldn't worry about the performance and just use the .lower() method.

The following code demonstrates that doing a case-insensitive compare by calling .lower() on two strings which are each almost a megabyte in size takes about 0.009 seconds on my 1.8GHz desktop computer:

from timeit import Timer

s1 = "1234567890" * 100000 + "a"
s2 = "1234567890" * 100000 + "B"

code = "s1.lower() < s2.lower()"
time = Timer(code, "from __main__ import s1, s2").timeit(1000)
print time / 1000   # 0.00920499992371 on my machine

If indeed this is an extremely significant, performance-critical section of code, then I recommend writing a function in C and calling it from your Python code, since that will allow you to do a truly efficient case-insensitive search. Details on writing C extension modules can be found here: https://docs.python.org/extending/extending.html

answered 2008-09-15T13:52:43.053

Your question implies that you don't need Unicode. Try the following code snippet; if it works for you, you're done:

Python 2.5.2 (r252:60911, Aug 22 2008, 02:34:17)
[GCC 4.3.1] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import locale
>>> locale.setlocale(locale.LC_COLLATE, "en_US")
'en_US'
>>> sorted("ABCabc", key=locale.strxfrm)
['a', 'A', 'b', 'B', 'c', 'C']
>>> sorted("ABCabc", cmp=locale.strcoll)
['a', 'A', 'b', 'B', 'c', 'C']

Clarification: in case it is not obvious at first sight, locale.strcoll seems to be the function you need, avoiding the str.lower or locale.strxfrm "duplicate" strings.

answered 2008-09-15T22:20:06.347

Just use the str().lower() method, unless high-performance is important - in which case write that sorting method as a C extension.

"How to write a Python Extension" seems like a decent intro..

More interestingly, This guide compares using the ctypes library vs writing an external C module (the ctype is quite-substantially slower than the C extension).

answered 2008-10-11T08:44:14.977

Ignore case in Python strings

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