enumerate()函数是否提前计算元素？

Question

为了支持对集合的索引,Python包含enumerate()函数.它提供了收集索引.

for index, item in enumerate(list):
    # do domething
    print index

在我的情况下,我有一个巨大的列表,并想知道使用enumerate()手动创建索引是否更快？例如

index = 0
for item in list:
    # do something
    print index
    index = index + 1

Answer 1

该enumerate功能是内置的; 它不会先验地计算元素.以下是C代码实现:

static PyObject *
enum_next(enumobject *en)
{
    PyObject *next_index;
    PyObject *next_item;
    PyObject *result = en->en_result;
    PyObject *it = en->en_sit;

    next_item = (*it->ob_type->tp_iternext)(it);
    if (next_item == NULL)
        return NULL;

    next_index = PyInt_FromLong(en->en_index);
    if (next_index == NULL) {
        Py_DECREF(next_item);
        return NULL;
    }
    en->en_index++; 

    if (result->ob_refcnt == 1) {
        Py_INCREF(result);
        Py_DECREF(PyTuple_GET_ITEM(result, 0));
        Py_DECREF(PyTuple_GET_ITEM(result, 1));
    } else {
        result = PyTuple_New(2);
        if (result == NULL) {
            Py_DECREF(next_index);
            Py_DECREF(next_item);
            return NULL;
        }
    }
    PyTuple_SET_ITEM(result, 0, next_index);
    PyTuple_SET_ITEM(result, 1, next_item);
    return result;
}

因此,该函数动态产生一个next en整数.