在Python中循环列表并修改它 [英] Loop through a list in Python and modify it

问题描述

这段代码来自Python的文档。

  words = ['cat'，'window'，'defenestrate'] 
对于w而言[：]：
 if len（w）> 6：
 words.insert（0，w）
 print（words）
  

以下是我首先想到的：

  words = ['cat'，'window'，'defenestrate'] 
 for w in words：
 if len（w）> 6：
 words.insert（0，w）
 print（words）
  

为什么这段代码创建了一个无限循环，而第一个没有？

解决方案

这是一个难题！的python，可以逃脱初学者。

字[：] 是魔术酱。

观察：

 >>> words = ['cat'，'window'，'defenestrate'] 
>>> words2 = words [：] 
>>> words2.insert（0，'hello'）
>>> words2 
 ['hello'，'cat'，'window'，'defenestrate'] 
>>>单词
 ['cat'，'window'，'defenestrate'] 
  

[：] ：

 >>> words = ['cat'，'window'，'defenestrate'] 
>>> words2 = words 
>>> words2.insert（0，'hello'）
>>> words2 
 ['hello'，'cat'，'window'，'defenestrate'] 
>>>单词
 ['hello'，'cat'，'window'，'defenestrate'] 
  

这里需要注意的是 words [：] 返回现有列表的 copy ，所以你遍历未修改的副本。

您可以使用 id（）来检查您是否引用相同的列表：

在第一种情况下：

 >>> words2 = words [：] 
>>> id（words2）
 4360026736 
>>> id（words）
 4360188992 
>>> words2是单词
 False 
  

在第二种情况下：

 >>> id（words2）
 4360188992 
>>> id（words）
 4360188992 
>>> words2 is words 
 True 
  

值得注意的是 [i ：j] 被称为切片操作符，它所做的是返回从索引 i 给你

 >>> [0：2] 
 ['hello'，'cat'] 
  

省略起始索引意味着它默认为 0 ，而省略最后一个索引意味着它默认为 len（words），最终的结果就是你收到了整个 列表的副本。

---

如果你想让你的代码更具可读性，我建议使用 copy 模块。

 从副本进口副本
 
 words = ['cat'，'window'，'defenestrate'] 
 for copy in copy（words）：
 if len（w）> ; 6：
 words.insert（0，w）
 print（words）
  

这和你的第一个代码片段基本上是一样的，并且更具可读性。

或者（如注释中的DSM所述）和python> = 3 ，你也可以使用 words.copy（）来做同样的事情。

This code is from Python's Documentation. I'm a little confused.

words = ['cat', 'window', 'defenestrate']
for w in words[:]:
    if len(w) > 6:
        words.insert(0, w)
print(words)

And the following is what I thought at first:

words = ['cat', 'window', 'defenestrate']
for w in words:
    if len(w) > 6:
        words.insert(0, w)
print(words)

Why does this code create a infinite loop and the first one doesn't?

解决方案

This is one of the gotchas! of python, that can escape beginners.

The words[:] is the magic sauce here.

Observe:

>>> words =  ['cat', 'window', 'defenestrate']
>>> words2 = words[:]
>>> words2.insert(0, 'hello')
>>> words2
['hello', 'cat', 'window', 'defenestrate']
>>> words
['cat', 'window', 'defenestrate']

And now without the [:]:

>>> words =  ['cat', 'window', 'defenestrate']
>>> words2 = words
>>> words2.insert(0, 'hello')
>>> words2
['hello', 'cat', 'window', 'defenestrate']
>>> words
['hello', 'cat', 'window', 'defenestrate']

The main thing to note here is that words[:] returns a copy of the existing list, so you are iterating over a copy, which is not modified.

You can check whether you are referring to the same lists using id():

In the first case:

>>> words2 = words[:]
>>> id(words2)
4360026736
>>> id(words)
4360188992
>>> words2 is words
False

In the second case:

>>> id(words2)
4360188992
>>> id(words)
4360188992
>>> words2 is words
True

It is worth noting that [i:j] is called the slicing operator, and what it does is it returns a fresh copy of the list starting from index i, upto (but not including) index j.

So, words[0:2] gives you

>>> words[0:2]
['hello', 'cat']

Omitting the starting index means it defaults to 0, while omitting the last index means it defaults to len(words), and the end result is that you receive a copy of the entire list.

---

If you want to make your code a little more readable, I recommend the copy module.

from copy import copy 

words = ['cat', 'window', 'defenestrate']
for w in copy(words):
    if len(w) > 6:
        words.insert(0, w)
print(words)

This basically does the same thing as your first code snippet, and is much more readable.

Alternatively (as mentioned by DSM in the comments) and on python >=3, you may also use words.copy() which does the same thing.

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