str performance in python

'%s' % 100000 is evaluated by the compiler and is equivalent to a constant at run-time.

>>> import dis>>> dis.dis(lambda: str(100000))  8           0 LOAD_GLOBAL              0 (str)              3 LOAD_CONST               1 (100000)              6 CALL_FUNCTION            1              9 RETURN_VALUE        >>> dis.dis(lambda: '%s' % 100000)  9           0 LOAD_CONST               3 ('100000')              3 RETURN_VALUE        

% with a run-time expression is not (significantly) faster than str:

>>> Timer('str(x)', 'x=100').timeit()0.25641703605651855>>> Timer('"%s" % x', 'x=100').timeit()0.2169809341430664

Do note that str is still slightly slower, as @DietrichEpp said, this is because str involves lookup and function call operations, while % compiles to a single immediate bytecode:

>>> dis.dis(lambda x: str(x))  9           0 LOAD_GLOBAL              0 (str)              3 LOAD_FAST                0 (x)              6 CALL_FUNCTION            1              9 RETURN_VALUE        >>> dis.dis(lambda x: '%s' % x) 10           0 LOAD_CONST               1 ('%s')              3 LOAD_FAST                0 (x)              6 BINARY_MODULO                     7 RETURN_VALUE        

Of course the above is true for the system I tested on (CPython 2.7); other implementations may differ.

One reason that comes to mind is the fact that str(100000) involves a global lookup, but "%s"%100000 does not. The str global has to be looked up in the global scope. This does not account for the entire difference:

>>> Timer('str(100000)').timeit()0.2941889762878418>>> Timer('x(100000)', 'x=str').timeit()0.24904918670654297

As noted by thg435,

>>> Timer('"%s"%100000',).timeit()0.034214019775390625>>> Timer('"%s"%x','x=100000').timeit()0.2940788269042969