将Python生成器解压缩为参数-内存有效吗? [英] Unpacking a Python generator into arguments - memory efficient?

问题描述

说我有一个sets的生成器:

def f(n) :
  for i in xrange(n) :
    yield set(xrange(i) )

>>> for s in f(5) :
      print s

set([])
set([0])
set([0, 1])
set([0, 1, 2])
set([0, 1, 2, 3])

现在我要union它们.我可以创建一组临时列表，然后将该列表解压缩为union的参数:

Now I want to union them. I can create a temporary list of sets, and unpack that list into arguments for union:

>>> set.union( * list( f(5) ) )
set([0, 1, 2, 3])

我还可以将生成器本身分配给union:

I can also give the generator itself to union:

>>> set.union( * f(5) )
set([0, 1, 2, 3])

第二种方法是否像第一种方法一样创建完整的临时列表?哪种方法对内存有效?

Does the second approach create the full temporary list like the first one? Which approach is memory efficient?

推荐答案

在将生成器用作参数时，Python首先扩展生成器 first .这两个选项中，在调用发生之前，生成器生成的所有值都将被加载到内存中.

Python expands a generator first when applying it as arguments; all values produced by the generator are loaded into memory before the call takes place, in both options.

您可以改用 reduce()函数调用:

You could use a reduce() function call instead:

from functools import reduce  # Python 3 forward compatibility

reduce(set.union, f(5))

这一步一步地遍历了f(5)产生的值，而无需先建立它们的序列.

This iterates over the values produced by f(5) one by one without building up a sequence of them first.

演示:

>>> def f(n):
...     for i in xrange(n):
...         yield set(xrange(i))
... 
>>> reduce(set.union, f(5))
set([0, 1, 2, 3])

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