efficiently checking that string consists of one character in Python

Question

This is by far the fastest, several times faster than even count(), just time it with that excellent mgilson's timing suite:

s == len(s) * s[0]

Here all the checking is done inside the Python C code which just:

• allocates len(s) characters;
• fills the space with the first character;
• compares two strings.

The longer the string is, the greater is time bonus. However, as mgilson writes, it creates a copy of the string, so if your string length is many millions of symbols, it may become a problem.

As we can see from timing results, generally the fastest ways to solve the task do not execute any Python code for each symbol. However, the set() solution also does all the job inside C code of the Python library, but it is still slow, probably because of operating string through Python object interface.

UPD: Concerning the empty string case. What to do with it strongly depends on the task. If the task is "check if all the symbols in a string are the same", s == len(s) * s[0] is a valid answer (no symbols mean an error, and exception is ok). If the task is "check if there is exactly one unique symbol", empty string should give us False, and the answer is s and s == len(s) * s[0], or bool(s) and s == len(s) * s[0] if you prefer receiving boolean values. Finally, if we understand the task as "check if there are no different symbols", the result for empty string is True, and the answer is not s or s == len(s) * s[0].

>>> s = 'AAAAAAAAAAAAAAAAAAA'
>>> s.count(s[0]) == len(s)
True

This doesn't short circuit. A version which does short-circuit would be:

>>> all(x == s[0] for x in s)
True

However, I have a feeling that due the the optimized C implementation, the non-short circuiting version will probably perform better on some strings (depending on size, etc)

---

Here's a simple timeit script to test some of the other options posted:

import timeit
import re

def test_regex(s,regex=re.compile(r'^(.)\1*$')):
    return bool(regex.match(s))

def test_all(s):
    return all(x == s[0] for x in s)

def test_count(s):
    return s.count(s[0]) == len(s)

def test_set(s):
    return len(set(s)) == 1

def test_replace(s):
    return not s.replace(s[0],'')

def test_translate(s):
    return not s.translate(None,s[0])

def test_strmul(s):
    return s == s[0]*len(s)

tests = ('test_all','test_count','test_set','test_replace','test_translate','test_strmul','test_regex')

print "WITH ALL EQUAL"
for test in tests:
    print test, timeit.timeit('%s(s)'%test,'from __main__ import %s; s="AAAAAAAAAAAAAAAAA"'%test)
    if globals()[test]("AAAAAAAAAAAAAAAAA") != True:
        print globals()[test]("AAAAAAAAAAAAAAAAA")
        raise AssertionError

print
print "WITH FIRST NON-EQUAL"
for test in tests:
    print test, timeit.timeit('%s(s)'%test,'from __main__ import %s; s="FAAAAAAAAAAAAAAAA"'%test)
    if globals()[test]("FAAAAAAAAAAAAAAAA") != False:
        print globals()[test]("FAAAAAAAAAAAAAAAA")
        raise AssertionError

On my machine (OS-X 10.5.8, core2duo, python2.7.3) with these contrived (short) strings, str.count smokes set and all, and beats str.replace by a little, but is edged out by str.translate and strmul is currently in the lead by a good margin:

WITH ALL EQUAL
test_all 5.83863711357
test_count 0.947771072388
test_set 2.01028490067
test_replace 1.24682998657
test_translate 0.941282987595
test_strmul 0.629556179047
test_regex 2.52913498878

WITH FIRST NON-EQUAL
test_all 2.41147494316
test_count 0.942595005035
test_set 2.00480484962
test_replace 0.960338115692
test_translate 0.924381017685
test_strmul 0.622269153595
test_regex 1.36632800102

The timings could be slightly (or even significantly?) different between different systems and with different strings, so that would be worth looking into with an actual string you're planning on passing.

Eventually, if you hit the best case for all enough, and your strings are long enough, you might want to consider that one. It's a better algorithm ... I would avoid the set solution though as I don't see any case where it could possibly beat out the count solution.

If memory could be an issue, you'll need to avoid str.translate, str.replace and strmul as those create a second string, but this isn't usually a concern these days.

You could convert to a set and check there is only one member:

len(set("AAAAAAAA"))

Try using the built-in function all:

all(c == 'A' for c in s)

Adding another solution to this problem

>>> not "AAAAAA".translate(None,"A")
True