为什么'x'中的'x'比'x'=='x'快？

Question

>>> timeit.timeit("'x' in ('x',)")
0.04869917374131205
>>> timeit.timeit("'x' == 'x'")
0.06144205736110564

也适用于具有多个元素的元组,两个版本似乎线性增长:

>>> timeit.timeit("'x' in ('x', 'y')")
0.04866674801541748
>>> timeit.timeit("'x' == 'x' or 'x' == 'y'")
0.06565782838087131
>>> timeit.timeit("'x' in ('y', 'x')")
0.08975995576448526
>>> timeit.timeit("'x' == 'y' or 'x' == 'y'")
0.12992391047427532

基于此,我认为我应该完全开始使用in到处而不是==!

Answer 1

正如我提到David Wolever所说的那样,除此之外还有更多的东西; 两种方法都派遣到is; 你可以通过这样做证明这一点

min(Timer("x == x", setup="x = 'a' * 1000000").repeat(10, 10000))
#>>> 0.00045456900261342525

min(Timer("x == y", setup="x = 'a' * 1000000; y = 'a' * 1000000").repeat(10, 10000))
#>>> 0.5256857610074803

第一个只能这么快,因为它按身份检查.

为了找出为什么人们需要比另一个更长的时间,让我们追溯执行.

它们都ceval.c从COMPARE_OP入门开始,因为这是涉及的字节码

TARGET(COMPARE_OP) {
    PyObject *right = POP();
    PyObject *left = TOP();
    PyObject *res = cmp_outcome(oparg, left, right);
    Py_DECREF(left);
    Py_DECREF(right);
    SET_TOP(res);
    if (res == NULL)
        goto error;
    PREDICT(POP_JUMP_IF_FALSE);
    PREDICT(POP_JUMP_IF_TRUE);
    DISPATCH();
}

这会弹出堆栈中的值(从技术上讲,它只会弹出一个)

PyObject *right = POP();
PyObject *left = TOP();

并运行比较:

PyObject *res = cmp_outcome(oparg, left, right);

cmp_outcome 这是:

static PyObject *
cmp_outcome(int op, PyObject *v, PyObject *w)
{
    int res = 0;
    switch (op) {
    case PyCmp_IS: ...
    case PyCmp_IS_NOT: ...
    case PyCmp_IN:
        res = PySequence_Contains(w, v);
        if (res < 0)
            return NULL;
        break;
    case PyCmp_NOT_IN: ...
    case PyCmp_EXC_MATCH: ...
    default:
        return PyObject_RichCompare(v, w, op);
    }
    v = res ? Py_True : Py_False;
    Py_INCREF(v);
    return v;
}

这是路径分裂的地方.该PyCmp_IN分支不

int
PySequence_Contains(PyObject *seq, PyObject *ob)
{
    Py_ssize_t result;
    PySequenceMethods *sqm = seq->ob_type->tp_as_sequence;
    if (sqm != NULL && sqm->sq_contains != NULL)
        return (*sqm->sq_contains)(seq, ob);
    result = _PySequence_IterSearch(seq, ob, PY_ITERSEARCH_CONTAINS);
    return Py_SAFE_DOWNCAST(result, Py_ssize_t, int);
}

请注意,元组定义为

static PySequenceMethods tuple_as_sequence = {
    ...
    (objobjproc)tuplecontains,                  /* sq_contains */
};

PyTypeObject PyTuple_Type = {
    ...
    &tuple_as_sequence,                         /* tp_as_sequence */
    ...
};

所以分支

if (sqm != NULL && sqm->sq_contains != NULL)

将被采取*sqm->sq_contains,这将是功能(objobjproc)tuplecontains,将采取.

这样做

static int
tuplecontains(PyTupleObject *a, PyObject *el)
{
    Py_ssize_t i;
    int cmp;

    for (i = 0, cmp = 0 ; cmp == 0 && i < Py_SIZE(a); ++i)
        cmp = PyObject_RichCompareBool(el, PyTuple_GET_ITEM(a, i),
                                           Py_EQ);
    return cmp;
}

......等等,是不是那个PyObject_RichCompareBool分支拿了什么？不,那是PyObject_RichCompare.

那条代码路径很短,所以它可能只是降低了这两条代码的速度.我们来比较吧.

int
PyObject_RichCompareBool(PyObject *v, PyObject *w, int op)
{
    PyObject *res;
    int ok;

    /* Quick result when objects are the same.
       Guarantees that identity implies equality. */
    if (v == w) {
        if (op == Py_EQ)
            return 1;
        else if (op == Py_NE)
            return 0;
    }

    ...
}

代码路径PyObject_RichCompareBool几乎立即终止.对于PyObject_RichCompare,确实如此

PyObject *
PyObject_RichCompare(PyObject *v, PyObject *w, int op)
{
    PyObject *res;

    assert(Py_LT <= op && op <= Py_GE);
    if (v == NULL || w == NULL) { ... }
    if (Py_EnterRecursiveCall(" in comparison"))
        return NULL;
    res = do_richcompare(v, w, op);
    Py_LeaveRecursiveCall();
    return res;
}

将Py_EnterRecursiveCall/ Py_LeaveRecursiveCall组合不采取在前面的路径,但这些都是比较快的宏将递增和递减一些全局后短路.

do_richcompare 作用:

static PyObject *
do_richcompare(PyObject *v, PyObject *w, int op)
{
    richcmpfunc f;
    PyObject *res;
    int checked_reverse_op = 0;

    if (v->ob_type != w->ob_type && ...) { ... }
    if ((f = v->ob_type->tp_richcompare) != NULL) {
        res = (*f)(v, w, op);
        if (res != Py_NotImplemented)
            return res;
        ...
    }
    ...
}

这会做一些快速检查来调用v->ob_type->tp_richcompare它

PyTypeObject PyUnicode_Type = {
    ...
    PyUnicode_RichCompare,      /* tp_richcompare */
    ...
};

哪个

PyObject *
PyUnicode_RichCompare(PyObject *left, PyObject *right, int op)
{
    int result;
    PyObject *v;

    if (!PyUnicode_Check(left) || !PyUnicode_Check(right))
        Py_RETURN_NOTIMPLEMENTED;

    if (PyUnicode_READY(left) == -1 ||
        PyUnicode_READY(right) == -1)
        return NULL;

    if (left == right) {
        switch (op) {
        case Py_EQ:
        case Py_LE:
        case Py_GE:
            /* a string is equal to itself */
            v = Py_True;
            break;
        case Py_NE:
        case Py_LT:
        case Py_GT:
            v = Py_False;
            break;
        default:
            ...
        }
    }
    else if (...) { ... }
    else { ...}
    Py_INCREF(v);
    return v;
}

也就是说,这个快捷方式left == right......但仅限于此之后

    if (!PyUnicode_Check(left) || !PyUnicode_Check(right))

    if (PyUnicode_READY(left) == -1 ||
        PyUnicode_READY(right) == -1)

总而言之,这些路径看起来像这样(手动递归内联,展开和修剪已知分支)

POP()                           # Stack stuff
TOP()                           #
                                #
case PyCmp_IN:                  # Dispatch on operation
                                #
sqm != NULL                     # Dispatch to builtin op
sqm->sq_contains != NULL        #
*sqm->sq_contains               #
                                #
cmp == 0                        # Do comparison in loop
i < Py_SIZE(a)                  #
v == w                          #
op == Py_EQ                     #
++i                             # 
cmp == 0                        #
                                #
res < 0                         # Convert to Python-space
res ? Py_True : Py_False        #
Py_INCREF(v)                    #
                                #
Py_DECREF(left)                 # Stack stuff
Py_DECREF(right)                #
SET_TOP(res)                    #
res == NULL                     #
DISPATCH()                      #

VS

POP()                           # Stack stuff
TOP()                           #
                                #
default:                        # Dispatch on operation
                                #
Py_LT <= op                     # Checking operation
op <= Py_GE                     #
v == NULL                       #
w == NULL                       #
Py_EnterRecursiveCall(...)      # Recursive check
                                #
v->ob_type != w->ob_type        # More operation checks
f = v->ob_type->tp_richcompare  # Dispatch to builtin op
f != NULL                       #
                                #
!PyUnicode_Check(left)          # ...More checks
!PyUnicode_Check(right))        #
PyUnicode_READY(left) == -1     #
PyUnicode_READY(right) == -1    #
left == right                   # Finally, doing comparison
case Py_EQ:                     # Immediately short circuit
Py_INCREF(v);                   #
                                #
res != Py_NotImplemented        #
                                #
Py_LeaveRecursiveCall()         # Recursive check
                                #
Py_DECREF(left)                 # Stack stuff
Py_DECREF(right)                #
SET_TOP(res)                    #
res == NULL                     #
DISPATCH()                      #

现在,PyUnicode_Check和PyUnicode_READY相当便宜,因为他们只检查了几个领域,但它应该是显而易见的是,上面一个是较小的代码路径,它具有较少的函数调用,只有一个开关语句是只是有点薄.

TL; DR:

两者都派遣到if (left_pointer == right_pointer); 不同之处在于他们为实现这一目标所做的工作量.in只是少做.

Answer 2

这里有三个因素共同产生了这种令人惊讶的行为.

第一:in运算符x is y在检查equality(x == y)之前采用快捷方式并检查identity ():

>>> n = float('nan')
>>> n in (n, )
True
>>> n == n
False
>>> n is n
True

第二:由于Python的字符串实习,两个"x"s "x" in ("x", )将是相同的:

>>> "x" is "x"
True

(大警告:这是实现特定的行为!is应该永远不会被用来比较字符串,因为它会给予有时令人惊讶的答案,例如"x" * 100 is "x" * 100 ==> False)

第三:在详细Veedrac的梦幻般的回答,tuple.__contains__(x in (y, )是大致相当于(y, ).__contains__(x))获取进行标识检查的速度比点str.__eq__(再次,x == y就是大致相当于x.__eq__(y))一样.

您可以看到这方面的证据,因为x in (y, )它明显慢于逻辑上等效的x == y:

In [18]: %timeit 'x' in ('x', )
10000000 loops, best of 3: 65.2 ns per loop

In [19]: %timeit 'x' == 'x'    
10000000 loops, best of 3: 68 ns per loop

In [20]: %timeit 'x' in ('y', ) 
10000000 loops, best of 3: 73.4 ns per loop

In [21]: %timeit 'x' == 'y'    
10000000 loops, best of 3: 56.2 ns per loop

这种x in (y, )情况比较慢,因为在is比较失败后,in操作员回退到正常的等式检查(即使用==),因此比较需要大约相同的时间==,因为创建元组的开销导致整个操作变慢,走其成员等

另请注意,a in (b, )在以下情况下更快a is b:

In [48]: a = 1             

In [49]: b = 2

In [50]: %timeit a is a or a == a
10000000 loops, best of 3: 95.1 ns per loop

In [51]: %timeit a in (a, )      
10000000 loops, best of 3: 140 ns per loop

In [52]: %timeit a is b or a == b
10000000 loops, best of 3: 177 ns per loop

In [53]: %timeit a in (b, )      
10000000 loops, best of 3: 169 ns per loop

(为什么a in (b, )比这更快a is b or a == b？我的猜测是虚拟机指令会更少 -  a in (b, )只有3条指令,其中a is b or a == b会有更多的VM指令)

Veedrac的回答 - /sf/answers/2022288691/ - 详细介绍了每一个期间发生的事情==,in并且非常值得一读.