当x（x）时，x为xrange（x）中的x [英] [ x for x in xrange(10) when p(x) ]

问题描述

我想知道是否有这样的事情，因为python正在远离

FP函数，如dropwhile / takewhile等。

解决方案

bo **** @ gmail。 com < bo **** @ gmail.com>写道：

我想知道是否有这样的事情，因为python正在远离


这通过写''如果'而不是

''''来成为有效的列表理解。

FP函数如dropwhile / takewhile等。

没办法 - itertools模块是并且仍然是一个珍贵的资源。如果

你想要一个迭代器而不是一个列表，那么itertools.ifilter在这里是相当的

。

Alex

" Alex Martelli"写道：

bon ... @ gmail.com< bon ... @ gmail.com>写道：

FP函数如dropwhile / takewhile等。

没办法 - itertools模块是并且仍然是一个珍贵的资源。如果你想要一个迭代器而不是一个列表，那么itertools.ifilter非常适合这里。

python 3K中itertools的未来怎么样？ IIRC，目前返回列表的几个

函数和方法将返回

迭代器。这可能意味着itertools。（imap / ifilter / izip）将取代地图/过滤器/拉链作为内置物吗？


George

Alex Martelli写道：

通过编写if来改为有效的列表理解
''when''。

有效，是的。效率很高，我不确定。


[x代表x x（10000000）如果p（x）]


意味着我需要即使p = lambda x：x

Hi,

I am wondering if there is such a thing, as python is moving away from
FP functions like dropwhile/takewhile etc.

解决方案

bo****@gmail.com <bo****@gmail.com> wrote:

Hi,

I am wondering if there is such a thing, as python is moving away from
This becomes a valid list comprehension by writing ''if'' instead of
''when''.
FP functions like dropwhile/takewhile etc.

No way -- the itertools module is and remains a PRECIOUS resource. If
you want an iterator rather than a list, itertools.ifilter is quite
appropriate here.
Alex

"Alex Martelli" wrote:

[e-mail] <[e-mail]> wrote:

FP functions like dropwhile/takewhile etc.

No way -- the itertools module is and remains a PRECIOUS resource.
If you want an iterator rather than a list, itertools.ifilter is quite
appropriate here.

What about the future of itertools in python 3K ? IIRC, several
functions and methods that currently return lists are going to return
iterators. Could this imply that itertools.(imap/ifilter/izip) will
take the place of map/filter/zip as builtins ?

George

Alex Martelli wrote:

This becomes a valid list comprehension by writing ''if'' instead of
''when''.

valid, yes. efficient, I am not sure.

[ x for x in xrange(10000000) if p(x) ]

means I need to go through the whole range even if p = lambda x: x < 2.

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