Python 3过滤器 - 错误或功能？ [英] Python 3 filter - Bug or Feature?

问题描述

好吧，我是一个Python的完全新手 - 和stackoverflow。我来自ksh和perl的背景。

以下是与Python 2.7的交互式会话：

<$ p

$ b Python 2.7.3（默认，2013年1月2日，16:53:07）
[gcc 4.7.2] on linux2
键入help，版权，信用或许可证以获取更多信息。
>>> import re
>>> KEY =REC_PAPER
>>> VALIDVALUES = filter（lambda x：re.search（r'^'+ KEY +'\ = '，x），[
...REC_METAL = | YES | NO |，
...REC_PAPER = | YES | NO |，
...REC_GLASS = | YES | NO |，
...REC_PLAST = | YES | NO |，
...DEBUG_FLAG = | 0 | 1 |
...]）＃结束一般清单。
>>> print（VALIDVALUES）
['REC_PAPER = | YES | NO |']
>>>

这是我期望VALIDVALUES返回。然而，Python 3.2的交互式会话产生了完全不同的结果：

 
 
 Python 3.2.3（默认，2013年2月20日， 17:02:41）
 [gcc 4.7.2] on linux2 
输入help，copyright，credits或license以获取更多信息。 
 >>> import re 
 >>> KEY =REC_PAPER
 >>> VALIDVALUES = filter（lambda x：re.search（r'^'+ KEY +'\ = '，x），[
 ...REC_METAL = | YES | NO |，
 ...REC_PAPER = | YES | NO |，
 ...REC_GLASS = | YES | NO |，
 ...REC_PLAST = | YES | NO |，
 ...DEBUG_FLAG = | 0 | 1 |
 ...]）＃结束一般清单。 
 >>> print（VALIDVALUES）
 0xb734268c处的<filter对象> 
 >>> 
 
 

我曾经在几个地方看过（包括stackoverflow），Python对Perl的grep和列表的等价物是过滤列表。这似乎在Python 2中工作。但是，假设在Python 3中的上述行为是正确的，似乎不再是这种情况。

第一个问题：是第二个问题：假设它是一个特性，我如何得到Python 2给出的输出结果？由于我不想进入的原因，我想远离定义一个函数或子程序，并像目前的代码一样内联。

我是否缺少显而易见的东西（对于新手来说很可能）？提前致谢。

文档， filter 返回一个迭代器比 2.x版中的列表要多。这比预先生成整个列表更有记忆效率。如果你想返回列表，你可以将迭代器包装在列表（）调用中：

  VALIDVALUES = list（filter（...））
  

Okay, I am a complete newbie to Python - and stackoverflow. I am coming from a ksh and Perl background.

The following in an interactive session with Python 2.7:

    Python 2.7.3 (default, Jan  2 2013, 16:53:07) 
    [GCC 4.7.2] on linux2
    Type "help", "copyright", "credits" or "license" for more information.
    >>> import re
    >>> KEY="REC_PAPER"
    >>> VALIDVALUES=filter(lambda x:re.search(r'^' + KEY + '\=', x), [
    ... "REC_METAL=|YES|NO|",
    ... "REC_PAPER=|YES|NO|",
    ... "REC_GLASS=|YES|NO|",
    ... "REC_PLAST=|YES|NO|",
    ... "DEBUG_FLAG=|0|1|"
    ... ])  #End general list.
    >>> print(VALIDVALUES)
    ['REC_PAPER=|YES|NO|']
    >>> 

Which is what I would expect VALIDVALUES to return. However, Python 3.2's interactive session yields completely different results:

    Python 3.2.3 (default, Feb 20 2013, 17:02:41) 
    [GCC 4.7.2] on linux2
    Type "help", "copyright", "credits" or "license" for more information.
    >>> import re
    >>> KEY="REC_PAPER"
    >>> VALIDVALUES=filter(lambda x:re.search(r'^' + KEY + '\=', x), [
        ... "REC_METAL=|YES|NO|",
        ... "REC_PAPER=|YES|NO|",
        ... "REC_GLASS=|YES|NO|",
        ... "REC_PLAST=|YES|NO|",
        ... "DEBUG_FLAG=|0|1|"
        ... ])  #End general list.
    >>> print(VALIDVALUES)
    <filter object at 0xb734268c>
    >>> 

I have seen in several places (including stackoverflow) where Python's equivalent of Perl's grep against a list is to filter the list. That appeared to work in Python 2. However, assuming the above behaviour in Python 3 is "correct," that no longer seems to be the case.

First question: Is the above beahviour a bug or feature in Python 3?

Second question: Assuming it is a feature, how do I get the output that Python 2 was giving? For reasons I won't go into, I want to stay away from defining a function or subroutine, and do it "inline" like the current code.

Am I missing something obvious (quite possible for a newbie)? Thanks in advance.

解决方案

As per the documentation, filter in Python 3.x returns an iterator, rather than a list as in version 2.x. This is more memory-efficient than generating the whole list up-front. If you want the list back, you can wrap the iterator in a list() call:

VALIDVALUES = list(filter(...))

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