在迭代中无需复制？ [英] Needless copying in iterations?
问题描述
大家好，
我正盯着今天看起来像这样的一段代码：
for stuff in stuff [x ：y]：
无论什么（东西）
并且想知道编译器是否真的从
制作了切片的副本代码似乎建议的东西，或者它是否找到了一些生成
迭代器的方法而无需复制（如果内容是内置的
序列类型）？或者更笨拙更有效率（在我的意见中，我的意见）：
$ x $ b for x in xrange（x，y）：
无论如何（东西[i]）
詹姆斯
Hello all,
I was staring at a segment of code that looked like this today:
for something in stuff[x:y]:
whatever(something)
and was wondering if the compiler really made a copy of the slice from
stuff as the code seems to suggest, or does it find some way to produce
an iterator without the need to make a copy (if stuff is a built-in
sequence type)? Or would it be more efficient to do the more clumsy (in
my opinion):
for i in xrange(x, y):
whatever(stuff[i])
James
推荐答案
9月15日，晚上11:58，James Stroud< jstr ... @ mbi.ucla.eduwrote：
On Sep 15, 11:58 pm, James Stroud <jstr...@mbi.ucla.eduwrote:
大家好，
我正盯着今天看起来像这样的一段代码：
用于东西的东西[x：y]：
无论什么（东西）
并且想知道编译器是否真的从
填充了片段的副本，因为代码似乎建议，或者它是否找到了生成<的方法br />
迭代器而不需要复制（如果东西是内置的
序列类型）？或者更笨拙更有效率（在我的意见中，我的意见）：
$ x $ b for x in xrange（x，y）：
无论什么（东西[i]）
詹姆斯
Hello all,
I was staring at a segment of code that looked like this today:
for something in stuff[x:y]:
whatever(something)
and was wondering if the compiler really made a copy of the slice from
stuff as the code seems to suggest, or does it find some way to produce
an iterator without the need to make a copy (if stuff is a built-in
sequence type)? Or would it be more efficient to do the more clumsy (in
my opinion):
for i in xrange(x, y):
whatever(stuff[i])
James
itertools.islice做你想要的事情
导入itertools
用于itertools.islice中的东西（stuff，x，y）：
无论什么（东西）
itertools.islice does what you want
import itertools
for something in itertools.islice(stuff, x, y):
whatever(something)
buffi写道：
buffi wrote:
9月15日晚上11:58，James Stroud< jstr .. 。@ mbi.ucla.eduwrote：
On Sep 15, 11:58 pm, James Stroud <jstr...@mbi.ucla.eduwrote:
>大家好，
我正盯着今天看起来像这样的一段代码：
东西[x：y]：
什么（东西）
并且想知道编译器是否真的从<代码似乎建议的东西，或者它是否找到了一些方法来生成一个迭代器而不需要复制（如果内容是内置的序列） ce型）？或者更笨拙（在我的意见中）更有效率：
我在xrange（x，y）：
无论什么（东西[i]）
James
>Hello all,
I was staring at a segment of code that looked like this today:
for something in stuff[x:y]:
whatever(something)
and was wondering if the compiler really made a copy of the slice from
stuff as the code seems to suggest, or does it find some way to produce
an iterator without the need to make a copy (if stuff is a built-in
sequence type)? Or would it be more efficient to do the more clumsy (in
my opinion):
for i in xrange(x, y):
whatever(stuff[i])
James
itertools.islice做你想做的事
import itertools
for itertools.islice（stuff，x，y）：
无论什么（东西）
itertools.islice does what you want
import itertools
for something in itertools.islice(stuff, x, y):
whatever(something)
谢谢buffi！
所以我猜解释器后者没有优化？
James
Thanks buffi!
So I guess the interpreter does no optimization in the latter?
James
9月16日上午12:20，James Stroud< jstr ... @ mbi.ucla.eduwrote：
On Sep 16, 12:20 am, James Stroud <jstr...@mbi.ucla.eduwrote:
buffi写道：
buffi wrote:
9月15日晚上11:58，James Stroud< jstr ... @ mbi.ucla.eduwrote：
On Sep 15, 11:58 pm, James Stroud <jstr...@mbi.ucla.eduwrote:
大家好，
Hello all,
我正盯着看上去的一段代码今天这个：
I was staring at a segment of code that looked like this today:
for stuff [x：y]：
无论什么（某事）
for something in stuff[x:y]:
whatever(something)
并且想知道编译器是否真的制作了副本代码似乎暗示了来自
的切片，或者它是否找到某种方式来生成一个迭代器而不需要复制（如果东西是一个副本）内置
序列类型）？或者更笨拙更有效率（我的意见是：
）：
and was wondering if the compiler really made a copy of the slice from
stuff as the code seems to suggest, or does it find some way to produce
an iterator without the need to make a copy (if stuff is a built-in
sequence type)? Or would it be more efficient to do the more clumsy (in
my opinion):
for x in xrange（x，y）：
what（stuff [i]）
for i in xrange(x, y):
whatever(stuff[i])
James
James
itertools.islice做你想要的
itertools.islice does what you want
import itertools
for itertools.islice（stuff，x，y）：
（无论如何）某事）
import itertools
for something in itertools.islice(stuff, x, y):
whatever(something)
谢谢buffi！
所以我想解释器在后者中没有优化？
詹姆斯
Thanks buffi!
So I guess the interpreter does no optimization in the latter?
James
不，据我所知，当你这样做时，它会从切片中创建一个新列表
喜欢
的东西[x：y]
- Bj？rnKempén
No, as far as I know it makes a new list out of the slice when you do
it like
for something in stuff[x:y]
- Bj?rn Kempén
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