是什么导致[* a]总体化?-Python 实用宝典

是什么导致[* a]总体化?

显然list(a)不是总归,[x for x in a]在某些时候[*a]总归,始终都是总归吗? 这是从0到12的大小n,以及三种方法的结果大小(以字节为单位): 0 56 56 56 1 64 88 88 2 72 88 96 3 80 88 104 4 88 88 112 5 96 120 120 6 104 120 128 7 112 120 136 8 120 120 152 9 128 184 184 10 136 184 192 …

问题:是什么导致[* a]总体化?

显然list(a)不是总归,[x for x in a]在某些时候[*a]总归,始终都是总归吗?

大小上限为n = 100

这是从0到12的大小n,以及三种方法的结果大小(以字节为单位):

0 56 56 56
1 64 88 88
2 72 88 96
3 80 88 104
4 88 88 112
5 96 120 120
6 104 120 128
7 112 120 136
8 120 120 152
9 128 184 184
10 136 184 192
11 144 184 200
12 152 184 208

这样计算,可以使用Python 3 在repl.it中重现。8

from sys import getsizeof

for n in range(13):
    a = [None] * n
    print(n, getsizeof(list(a)),
             getsizeof([x for x in a]),
             getsizeof([*a]))

因此:这是如何工作的?如何整体化[*a]?实际上,它使用什么机制从给定的输入创建结果列表?它是否使用了迭代器a并使用了类似的东西list.append?源代码在哪里?

产生图像的数据和代码协作。)

放大到较小的n:

大小上限为n = 40

缩小到较大的n:

大小最大为n = 1000

Apparently list(a) doesn't overallocate, [x for x in a] overallocates at some points, and [*a] overallocates all the time?

Sizes up to n=100

Here are sizes n from 0 to 12 and the resulting sizes in bytes for the three methods:

0 56 56 56
1 64 88 88
2 72 88 96
3 80 88 104
4 88 88 112
5 96 120 120
6 104 120 128
7 112 120 136
8 120 120 152
9 128 184 184
10 136 184 192
11 144 184 200
12 152 184 208

Computed like this, reproducable at repl.it, using Python 3.8:

from sys import getsizeof

for n in range(13):
    a = [None] * n
    print(n, getsizeof(list(a)),
             getsizeof([x for x in a]),
             getsizeof([*a]))

So: How does this work? How does [*a] overallocate? Actually, what mechanism does it use to create the result list from the given input? Does it use an iterator over a and use something like list.append? Where is the source code?

(Colab with data and code that produced the images.)

Zooming in to smaller n:

Sizes up to n=40

Zooming out to larger n:

Sizes up to n=1000


回答 0

[*a] 在内部执行C等效于

  1. 新建一个空的 list
  2. 呼叫 newlist.extend(a)
  3. 返回list

因此,如果您将测试扩展到:

from sys import getsizeof

for n in range(13):
    a = [None] * n
    l = []
    l.extend(a)
    print(n, getsizeof(list(a)),
             getsizeof([x for x in a]),
             getsizeof([*a]),
             getsizeof(l))

在线尝试!

你会看到的结果getsizeof([*a])l = []; l.extend(a); getsizeof(l)是相同的。

通常这是正确的做法;在extending时,您通常希望以后再添加,对于一般的解压缩,类似的情况是,假设要在一个接一个的基础上添加多个内容。[*a]这不是正常情况;Python假定list[*a, b, c, *d])中添加了多个项目或可迭代项,因此在常见情况下,过度分配可以节省工作。

相比之下,list由单个预先确定的可迭代大小(带有list())构成的结构在使用过程中可能不会增长或收缩,并且积算过早,除非另外证明。Python最近修复了一个错误,该错误使构造函数甚至可以对已知大小的输入进行整体分配

至于list理解,它们实际上等效于重复的appends,因此当您一次添加一个元素时,您会看到正常的过度分配增长模式的最终结果。

需要明确的是,这都不是语言保证。这就是CPython实现它的方式。Python语言规范通常与特定的增长模式无关list(除了保证从末期开始摊销O(1) appends和pops)。如评论中所述,具体实现在3.9中再次更改;尽管这不会影响[*a],但可能会影响其他情况,这些情况曾经是“构建tuple单个项目的临时内容,然后extend使用tuple”,现在变成的多个应用程序LIST_APPEND,当发生超额分配以及要计算的数字是多少时,该应用程序可能会发生变化。

[*a] is internally doing the C equivalent of:

  1. Make a new, empty list
  2. Call newlist.extend(a)
  3. Returns list.

So if you expand your test to:

from sys import getsizeof

for n in range(13):
    a = [None] * n
    l = []
    l.extend(a)
    print(n, getsizeof(list(a)),
             getsizeof([x for x in a]),
             getsizeof([*a]),
             getsizeof(l))

Try it online!

you'll see the results for getsizeof([*a]) and l = []; l.extend(a); getsizeof(l) are the same.

This is usually the right thing to do; when extending you're usually expecting to add more later, and similarly for generalized unpacking, it's assumed that multiple things will be added one after the other. [*a] is not the normal case; Python assumes there are multiple items or iterables being added to the list ([*a, b, c, *d]), so overallocation saves work in the common case.

By contrast, a list constructed from a single, presized iterable (with list()) may not grow or shrink during use, and overallocating is premature until proven otherwise; Python recently fixed a bug that made the constructor overallocate even for inputs with known size.

As for list comprehensions, they're effectively equivalent to repeated appends, so you're seeing the final result of the normal overallocation growth pattern when adding an element at a time.

To be clear, none of this is a language guarantee. It's just how CPython implements it. The Python language spec is generally unconcerned with specific growth patterns in list (aside from guaranteeing amortized O(1) appends and pops from the end). As noted in the comments, the specific implementation changes again in 3.9; while it won't affect [*a], it could affect other cases where what used to be "build a temporary tuple of individual items and then extend with the tuple" now becomes multiple applications of LIST_APPEND, which can change when the overallocation occurs and what numbers go into the calculation.


回答 1

的全貌是什么情况,建立在其他的答案和评论(尤其是ShadowRanger的答案,这也解释了为什么它这样做)。

拆卸显示已BUILD_LIST_UNPACK被使用:

>>> import dis
>>> dis.dis('[*a]')
  1           0 LOAD_NAME                0 (a)
              2 BUILD_LIST_UNPACK        1
              4 RETURN_VALUE

这是处理,它会构建一个空列表并将其扩展(带有a):

        case TARGET(BUILD_LIST_UNPACK): {
            ...
            PyObject *sum = PyList_New(0);
              ...
                none_val = _PyList_Extend((PyListObject *)sum, PEEK(i));

_PyList_Extend 用途 list_extend

_PyList_Extend(PyListObject *self, PyObject *iterable)
{
    return list_extend(self, iterable);
}

其中调用list_resize的总和为

list_extend(PyListObject *self, PyObject *iterable)
    ...
        n = PySequence_Fast_GET_SIZE(iterable);
        ...
        m = Py_SIZE(self);
        ...
        if (list_resize(self, m + n) < 0) {

overallocates如下:

list_resize(PyListObject *self, Py_ssize_t newsize)
{
  ...
    new_allocated = (size_t)newsize + (newsize >> 3) + (newsize < 9 ? 3 : 6);

让我们检查一下。用上面的公式计算预期的点数,并通过将预期的字节数乘以8(如我在此处使用64位Python)并添加一个空列表的字节数(即列表对象的固定开销)来计算预期的字节数:

from sys import getsizeof
for n in range(13):
    a = [None] * n
    expected_spots = n + (n >> 3) + (3 if n < 9 else 6)
    expected_bytesize = getsizeof([]) + expected_spots * 8
    real_bytesize = getsizeof([*a])
    print(n,
          expected_bytesize,
          real_bytesize,
          real_bytesize == expected_bytesize)

输出:

0 80 56 False
1 88 88 True
2 96 96 True
3 104 104 True
4 112 112 True
5 120 120 True
6 128 128 True
7 136 136 True
8 152 152 True
9 184 184 True
10 192 192 True
11 200 200 True
12 208 208 True

匹配,除了n = 0list_extend实际上是快捷方式,因此实际上也匹配:

        if (n == 0) {
            ...
            Py_RETURN_NONE;
        }
        ...
        if (list_resize(self, m + n) < 0) {

Full picture of what happens, building on the other answers and comments (especially ShadowRanger's answer, which also explains why it's done like that).

Disassembling shows that BUILD_LIST_UNPACK gets used:

>>> import dis
>>> dis.dis('[*a]')
  1           0 LOAD_NAME                0 (a)
              2 BUILD_LIST_UNPACK        1
              4 RETURN_VALUE

That's handled , which builds an empty list and extends it (with a):

        case TARGET(BUILD_LIST_UNPACK): {
            ...
            PyObject *sum = PyList_New(0);
              ...
                none_val = _PyList_Extend((PyListObject *)sum, PEEK(i));

_PyList_Extend uses list_extend:

_PyList_Extend(PyListObject *self, PyObject *iterable)
{
    return list_extend(self, iterable);
}

Which calls list_resize with the sum of the sizes:

list_extend(PyListObject *self, PyObject *iterable)
    ...
        n = PySequence_Fast_GET_SIZE(iterable);
        ...
        m = Py_SIZE(self);
        ...
        if (list_resize(self, m + n) < 0) {

And that overallocates as follows:

list_resize(PyListObject *self, Py_ssize_t newsize)
{
  ...
    new_allocated = (size_t)newsize + (newsize >> 3) + (newsize < 9 ? 3 : 6);

Let's check that. Compute the expected number of spots with the formula above, and compute the expected byte size by multiplying it with 8 (as I'm using 64-bit Python here) and adding an empty list's byte size (i.e., a list object's constant overhead):

from sys import getsizeof
for n in range(13):
    a = [None] * n
    expected_spots = n + (n >> 3) + (3 if n < 9 else 6)
    expected_bytesize = getsizeof([]) + expected_spots * 8
    real_bytesize = getsizeof([*a])
    print(n,
          expected_bytesize,
          real_bytesize,
          real_bytesize == expected_bytesize)

Output:

0 80 56 False
1 88 88 True
2 96 96 True
3 104 104 True
4 112 112 True
5 120 120 True
6 128 128 True
7 136 136 True
8 152 152 True
9 184 184 True
10 192 192 True
11 200 200 True
12 208 208 True

Matches except for n = 0, which list_extend actually shortcuts, so actually that matches, too:

        if (n == 0) {
            ...
            Py_RETURN_NONE;
        }
        ...
        if (list_resize(self, m + n) < 0) {

回答 2

这些将是CPython解释器的实现细节,因此在其他解释器之间可能不一致。

也就是说,您可以list(a)在这里看到理解和行为的位置:

https://github.com/python/cpython/blob/master/Objects/listobject.c#L36

专为理解:

 * The growth pattern is:  0, 4, 8, 16, 25, 35, 46, 58, 72, 88, ...
...

new_allocated = (size_t)newsize + (newsize >> 3) + (newsize < 9 ? 3 : 6);

在这些行的下面,有list_preallocate_exact调用时使用的行list(a)

These are going to be implementation details of the CPython interpreter, and so may not be consistent across other interpreters.

That said, you can see where the comprehension and list(a) behaviors come in here:

https://github.com/python/cpython/blob/master/Objects/listobject.c#L36

Specifically for the comprehension:

 * The growth pattern is:  0, 4, 8, 16, 25, 35, 46, 58, 72, 88, ...
...

new_allocated = (size_t)newsize + (newsize >> 3) + (newsize < 9 ? 3 : 6);

Just below those lines, there is list_preallocate_exact which is used when calling list(a).


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