Python 中的 str 性能

在分析一段 Python 代码(从 python 2.6到 3.2)时，我发现 方法将一个对象(在我的例子中是一个整数)转换成一个字符串的数量级几乎比使用字符串格式要慢一个百分点。

这是基准

>>> from timeit import Timer
>>> Timer('str(100000)').timeit()
0.3145311339386332
>>> Timer('"%s"%100000').timeit()
0.03803517023435887

有人知道为什么会这样吗？ 我错过了什么吗？

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小开

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
小开
最佳答案

'%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.


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