int .__ mul__,执行速度比operator.mul慢2倍

Question

如果你看下面的时间:

C:\Users\Henry>python -m timeit -s "mul = int.__mul__" "reduce(mul,range(10000))"
1000 loops, best of 3: 908 usec per loop

C:\Users\Henry>python -m timeit -s "from operator import mul" "reduce(mul,range(10000))"
1000 loops, best of 3: 410 usec per loop

两者之间的执行速度存在显着差异

reduce(int.__mul__,range(10000))并且reduce(mul,range(10000))后者更快.

使用该dis模块查看发生的情况:

使用int.__mul__方法:

C:\Users\Henry>python
Python 2.7.4 (default, Apr  6 2013, 19:55:15) [MSC v.1500 64 bit (AMD64)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> mul = int.__mul__
>>> def test():
...     mul(1,2)
...
>>> import dis
>>> dis.dis(test)
  2           0 LOAD_GLOBAL              0 (mul)
              3 LOAD_CONST               1 (1)
              6 LOAD_CONST               2 (2)
              9 CALL_FUNCTION            2
             12 POP_TOP
             13 LOAD_CONST               0 (None)
             16 RETURN_VALUE
>>>

和操作员mul方法

C:\Users\Henry>python
Python 2.7.4 (default, Apr  6 2013, 19:55:15) [MSC v.1500 64 bit (AMD64)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> from operator import mul
>>> def test():
...     mul(1,2)
...
>>> import dis
>>> dis.dis(test)
  2           0 LOAD_GLOBAL              0 (mul)
              3 LOAD_CONST               1 (1)
              6 LOAD_CONST               2 (2)
              9 CALL_FUNCTION            2
             12 POP_TOP
             13 LOAD_CONST               0 (None)
             16 RETURN_VALUE
>>>

它们看起来是一样的,为什么执行速度有差异？我指的是Python的CPython实现

---

在python3上也是如此:

$ python3 -m timeit -s 'mul=int.__mul__;from functools import reduce' 'reduce(mul, range(10000))'
1000 loops, best of 3: 1.18 msec per loop
$ python3 -m timeit -s 'from operator import mul;from functools import reduce' 'reduce(mul, range(10000))'
1000 loops, best of 3: 643 usec per loop
$ python3 -m timeit -s 'mul=lambda x,y:x*y;from functools import reduce' 'reduce(mul, range(10000))'
1000 loops, best of 3: 1.26 msec per loop

Answer 1

int.__mul__是一个插槽包装器,即PyWrapperDescrObject,同时operator.mul是一个buit -in函数.我认为相反的执行速度是由这种差异引起的.

>>> int.__mul__
<slot wrapper '__mul__' of 'int' objects>
>>> operator.mul
<built-in function mul>

当我们调用PyWrapperDescrObject时,wrapperdescr_call会调用它.

static PyObject *
wrapperdescr_call(PyWrapperDescrObject *descr, PyObject *args, PyObject *kwds)
{
    Py_ssize_t argc;
    PyObject *self, *func, *result;

    /* Make sure that the first argument is acceptable as 'self' */
    assert(PyTuple_Check(args));
    argc = PyTuple_GET_SIZE(args);
    if (argc d_type->tp_name);
        return NULL;
    }
    self = PyTuple_GET_ITEM(args, 0);
    if (!_PyObject_RealIsSubclass((PyObject *)Py_TYPE(self),
                                  (PyObject *)(descr->d_type))) {
        PyErr_Format(PyExc_TypeError,
                     "descriptor '%.200s' "
                     "requires a '%.100s' object "
                     "but received a '%.100s'",
                     descr_name((PyDescrObject *)descr),
                     descr->d_type->tp_name,
                     self->ob_type->tp_name);
        return NULL;
    }

    func = PyWrapper_New((PyObject *)descr, self);
    if (func == NULL)
        return NULL;
    args = PyTuple_GetSlice(args, 1, argc);
    if (args == NULL) {
        Py_DECREF(func);
        return NULL;
    }
    result = PyEval_CallObjectWithKeywords(func, args, kwds);
    Py_DECREF(args);
    Py_DECREF(func);
    return result;
}

让我们看看我们发现了什么!

func = PyWrapper_New((PyObject *)descr, self);

已经构造了一个新的PyWrapper对象.它会显着降低执行速度.有时,创建新对象比运行简单函数需要更多时间.
因此,比它int.__mul__慢的并不奇怪operator.mul.