Question

我只是想弄清楚这些结果背后的逻辑：

>>>nan = float('nan')
>>>nan == nan
False 
# I understand that this is because the __eq__ method is defined this way
>>>nan in [nan]
True 
# This is because the __contains__ method for list is defined to compare the identity first then the content?

但在这两种情况下，我认为在幕后的函数PyObject_RichCompareBool被称为正确？为什么会有区别？他们不应该有相同的行为吗？

Answer 1

但在这两种情况下我都认为幕后功能 PyObject_RichCompareBool被称为对吗？为什么会有区别？他们不应该有相同的行为吗？

==从不直接在float对象上调用PyObject_RichCompareBool，浮点数有自己的rich_compare方法（调用__eq__），可能会调用PyObject_RichCompareBool也可能不调用/* Comparison is pretty much a nightmare. When comparing float to float, * we do it as straightforwardly (and long-windedly) as conceivable, so * that, e.g., Python x == y delivers the same result as the platform * C x == y when x and/or y is a NaN. * When mixing float with an integer type, there's no good *uniform* approach. * Converting the double to an integer obviously doesn't work, since we * may lose info from fractional bits. Converting the integer to a double * also has two failure modes: (1) a long int may trigger overflow (too * large to fit in the dynamic range of a C double); (2) even a C long may have * more bits than fit in a C double (e.g., on a a 64-bit box long may have * 63 bits of precision, but a C double probably has only 53), and then * we can falsely claim equality when low-order integer bits are lost by * coercion to double. So this part is painful too. */ static PyObject* float_richcompare(PyObject *v, PyObject *w, int op) { double i, j; int r = 0; assert(PyFloat_Check(v)); i = PyFloat_AS_DOUBLE(v); /* Switch on the type of w. Set i and j to doubles to be compared, * and op to the richcomp to use. */ if (PyFloat_Check(w)) j = PyFloat_AS_DOUBLE(w); else if (!Py_IS_FINITE(i)) { if (PyInt_Check(w) || PyLong_Check(w)) /* If i is an infinity, its magnitude exceeds any * finite integer, so it doesn't matter which int we * compare i with. If i is a NaN, similarly. */ j = 0.0; else goto Unimplemented; } ...取决于传递给它的参数。

PyObject_RichCompareBool

另一方面，list_contains直接调用项目__eq__，因此在第二种情况下你会得到True。

请注意，这仅适用于CPython，PyPy的list.__contains__方法似乎只是通过调用$~/pypy-2.4.0-linux64/bin# ./pypy Python 2.7.8 (f5dcc2477b97, Sep 18 2014, 11:33:30) [PyPy 2.4.0 with GCC 4.6.3] on linux2 Type "help", "copyright", "credits" or "license" for more information. >>>> nan = float('nan') >>>> nan == nan False >>>> nan is nan True >>>> nan in [nan] False方法来比较项目：

{{1}}

Answer 2

您说PyObject_RichCompareBool被调用是对的，请参阅list_contains中的listobject.c函数。

文档说：

这相当于Python表达式o1 op o2，其中op是对应于opid的运算符。

然而，这似乎并不完全正确。

在cpython源代码中我们有这一部分：

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;
    }

在这种情况下，由于对象是相同的，我们有平等。

Answer 3

在数学上，将无穷大与无穷大相比较并不会产生sense。这就是为什么没有为nan定义平等的原因。

对于nan in [nan]的情况，引用了不可变变量。但要小心::

>>> nan is nan
True

>>> float('nan') is float('nan')
False

在第一种情况下，引用了不可变变量。在第二个中，创建并比较了两个不同的浮点数。

Python：列表中的Nan是否相等？

3 个答案: