blocks|key|1102108|text|它被认为是糟糕的形式。如果您需要保留对列表的现有引用，请使用列表理解，并使用切片分配。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1102109|a+=+[1,+3,+5]
b+=+a
a[:]+=+[x+%2B+2+for+x+in+a]
print(b)|code-block|syntax|javascript|1102110|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

It's considered poor form. Use a list comprehension instead, with slice assignment if you need to retain existing references to the list.

<pre><code>a = [1, 3, 5]
b = a
a[:] = [x + 2 for x in a]
print(b)
</code></pre>

blocks|key|887358|text|不，你不会改变列表的“内容”，如果你可以这样改变字符串的话。但在Python中，它们不是可变的。任何字符串操作都会返回一个新字符串。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|887359|如果你有一个你知道是可变的对象列表，你可以这样做，只要你不改变列表的实际内容。|887360|因此，您将需要进行某种映射。如果你使用生成器表达式，那么操作将在你迭代时完成，并且可以节省内存。|887361|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

No you wouldn't alter the "content" of the list, if you could mutate strings that way. But in Python they are not mutable. Any string operation returns a new string.

If you had a list of objects you knew were mutable, you could do this as long as you don't change the actual contents of the list.

Thus you will need to do a map of some sort. If you use a generator expression it [the operation] will be done as you iterate and you will save memory.

blocks|key|887379|text|多了一个for循环变体，在我看来比使用枚举()的循环更干净：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|887380|for+idx+in+range(len(list)):
++++list[idx]=...+#+set+a+new+value
++++#+some+other+code+which+doesn't+let+you+use+a+list+comprehension|code-block|syntax|javascript|887381|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

One more for loop variant, looks cleaner to me than one with enumerate():

<pre><code>for idx in range(len(list)):
 list[idx]=... # set a new value
 # some other code which doesn't let you use a list comprehension
</code></pre>

blocks|key|1102228|text|从您的问题中不清楚决定删除哪些字符串的标准是什么，但如果您有或可以列出要删除的字符串，则可以执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1102229|my_strings+=+['a','b','c','d','e']
undesirable_strings+=+['b','d']
for+undesirable_string+in+undesirable_strings:
++++for+i+in+range(my_strings.count(undesirable_string)):
++++++++my_strings.remove(undesirable_string)|code-block|syntax|javascript|1102230|将my_strings更改为'a'，'c'，'e‘|1102231|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

It is not clear from your question what the criteria for deciding what strings to remove is, but if you have or can make a list of the strings that you want to remove , you could do the following:

<pre><code>my_strings = ['a','b','c','d','e']
undesirable_strings = ['b','d']
for undesirable_string in undesirable_strings:
 for i in range(my_strings.count(undesirable_string)):
 my_strings.remove(undesirable_string)
</code></pre>

which changes my_strings to ['a', 'c', 'e']

blocks|key|887459|text|只要不改变向列表添加/删除元素，在迭代列表时修改每个元素都是可以的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|887460|您可以使用列表理解：|887461|l+=+['a',+'+list',+'of+',+'+string+']
l+=+[item.strip()+for+item+in+l]|code-block|syntax|javascript|887462|或者只执行C-style+for循环：|offset|length|style|CODE|887463|for+index,+item+in+enumerate(l):
++++l[index]+=+item.strip()|887464|entityMap^0|0|0|0|5|7|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|V|8|@$K|W|L|X|M|N]]|9|@]|A|$]]|$1|O|3|P|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Modifying each element while iterating a list is fine, as long as you do not change add/remove elements to list. 

You can use list comprehension:

<pre><code>l = ['a', ' list', 'of ', ' string ']
l = [item.strip() for item in l]
</code></pre>

or just do the <code>C-style</code> for loop:

<pre><code>for index, item in enumerate(l):
 l[index] = item.strip()
</code></pre>

blocks|key|887498|text|你可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|887499|a+=+[1,2,3,4,5]
b+=+[i**2+for+i+in+a]|code-block|syntax|javascript|887500|它被称为列表理解，以使您更容易在列表中循环。|887501|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You can do something like this:

<pre><code>a = [1,2,3,4,5]
b = [i**2 for i in a]
</code></pre>

It's called a list comprehension, to make it easier for you to loop inside a list.

blocks|key|887556|text|简而言之，在迭代相同的列表的同时对列表进行修改。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|887557|list[:]+=+["Modify+the+list"+for+each_element+in+list+"Condition+Check"]|code-block|syntax|javascript|887558|示例：|887559|list[:]+=+[list.remove(each_element)+for+each_element+in+list+if+each_element+in+["data1",+"data2"]]|887560|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

In short, to do modification on the list while iterating the same list.

<pre><code>list[:] = ["Modify the list" for each_element in list "Condition Check"]
</code></pre>

example:

<pre><code>list[:] = [list.remove(each_element) for each_element in list if each_element in ["data1", "data2"]]
</code></pre>

Now I know that it is not safe to modify the list during an iterative looping. However, suppose I have a list of strings, and I want to strip the strings themselves. Does replacement of mutable values count as modification?

How to modify list entries during for loop?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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现在我知道在迭代循环期间修改列表是不安全的。但是，假设我有一个字符串列表，并且我想剥离字符串本身。可变值的替换是否算作修改？

问如何在for循环过程中修改列表条目？
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问如何在for循环过程中修改列表条目？
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