blocks|key|930298|text|TypeError:+'str'+object+does+not+support+item+assignment|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|930299|这是意料之中的--+python字符串是不可变的。|unstyled|930300|一种方法是做一些切片和切丁。如下所示：|930301|>>>+aa+=+'abcdefghijkl'
>>>+changed++=+aa[0:-2]+%2B+'A'+%2B+aa[-1]
>>>+print+changed
abcdefghijAl|930302|连接的结果，changed将是另一个不可变的字符串。请注意，这不是适用于所有替换场景的通用解决方案。|offset|length|style|CODE|930303|entityMap^0|0|0|0|0|6|7|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|T|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$B|C]]|$1|K|3|L|5|F|7|W|8|@$M|X|N|Y|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|F|7|Z|8|@]|9|@]|A|$]]]|R|$]]

<pre><code>TypeError: 'str' object does not support item assignment
</code></pre>

This is to be expected - python strings are immutable. 

One way is to do some slicing and dicing. Like this:

<pre><code>&gt;&gt;&gt; aa = 'abcdefghijkl'
&gt;&gt;&gt; changed = aa[0:-2] + 'A' + aa[-1]
&gt;&gt;&gt; print changed
abcdefghijAl
</code></pre>

The result of the concatenation, <code>changed</code> will be another immutable string. Mind you, this is not a generic solution that fits all substitution scenarios.

blocks|key|4230960|text|如果你要替换的总是同一个位置，你可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4230961|>>>+s+=+s[0:-2]+%2B+"A"+%2B+s[-1:]
>>>+print+s
abcdefghijAl|code-block|syntax|javascript|4230962|在一般情况下，您可以这样做：|4230963|>>>+rindex+=+-2+#Second+to+the+last+letter
>>>+s+=+s[0:rindex]+%2B+"A"+%2B+s[rindex%2B1:]
>>>+print+s
abcdefghijAl|4230964|编辑：很的一般情况下，如果你只想在替换中重复一个字母：|offset|length|style|BOLD|4230965|>>>+s+=+"abcdefghijkl"
>>>+repl_str+=+"A"
>>>+rindex+=+-4+#Start+at+4th+character+from+the+end
>>>+repl+=+3+#Replace+3+characters
>>>+s+=+s[0:rindex]+%2B+(repl_str+*+repl)+%2B+s[rindex%2Brepl:]
>>>+print+s
abcdefghAAAl|4230966|entityMap^0|0|0|0|0|3|1|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|Y|8|@$M|Z|N|10|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

If it's always the same position you're replacing, you could do something like:

<pre><code>&gt;&gt;&gt; s = s[0:-2] + "A" + s[-1:]
&gt;&gt;&gt; print s
abcdefghijAl
</code></pre>

In the general case, you could do:

<pre><code>&gt;&gt;&gt; rindex = -2 #Second to the last letter
&gt;&gt;&gt; s = s[0:rindex] + "A" + s[rindex+1:]
&gt;&gt;&gt; print s
abcdefghijAl
</code></pre>

Edit: The very general case, if you just want to repeat a single letter in the replacement:

<pre><code>&gt;&gt;&gt; s = "abcdefghijkl"
&gt;&gt;&gt; repl_str = "A"
&gt;&gt;&gt; rindex = -4 #Start at 4th character from the end
&gt;&gt;&gt; repl = 3 #Replace 3 characters
&gt;&gt;&gt; s = s[0:rindex] + (repl_str * repl) + s[rindex+repl:]
&gt;&gt;&gt; print s
abcdefghAAAl
</code></pre>

blocks|key|930339|text|你必须将它分割并实例化一个新的字符串，因为字符串是不可变的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|930340|aa+=+aa[:-2]+%2B+'A'+%2B+aa[-1:]|code-block|syntax|javascript|930341|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|$]]

You have to slice it up and instantiate a new string since strings are immutable:

<pre><code>aa = aa[:-2] + 'A' + aa[-1:]
</code></pre>

blocks|key|2822126|text|另一种方法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2822127|s+=+"abcdefghijkl"
l+=+list(+s+)
l[-2]+=+'A'
s+=+"".join(+l+)|code-block|syntax|javascript|2822128|Python中的字符串是不可变的序列-非常类似于元组。您可以从字符串或元组创建一个可变的列表，更改相关索引，然后使用join将列表转换回字符串，或者使用元组构造函数将列表转换回tuple。|offset|length|style|CODE|2822129|编辑：如果是字符串，可以使用bytearray+(docs)类型而不是list，这使得使用str构造函数转换为字符串成为可能。|BOLD|2822130|此外，您还可以实现自己的类或使用标准库中的这个类：http://docs.python.org/library/userdict.html#UserString.MutableString，如下所示：|2822131|>>>+from+UserString+import+MutableString
>>>+s+=+MutableString(+"zdfgzbdr+"+)
>>>+s
'zdfgzbdr+'
>>>+s[1:5]
'dfgz'
>>>+s[1:5]+=+"xxxx"
>>>+s
'zxxxxbdr+'|2822132|遗憾的是-+MutableString已经被弃用，并且在Py3k中不可用(我认为你仍然可以编写自己的类)。|2822133|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/py3k/library/functions.html#bytearray|1|http://docs.python.org/library/userdict.html#UserString.MutableString^0|0|0|1M|4|2G|5|0|0|3|Z|4|19|3|P|4|0|0|P|1X|1|0|0|6|D|0^^$0|@$1|2|3|4|5|6|7|15|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|16|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|17|8|@$I|18|J|19|K|L]|$I|1A|J|1B|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|1C|8|@$I|1D|J|1E|K|O]|$I|1F|J|1G|K|L]|$I|1H|J|1I|K|L]]|9|@$I|1J|J|1K|1|1L]]|A|$]]|$1|P|3|Q|5|6|7|1M|8|@]|9|@$I|1N|J|1O|1|1P]]|A|$]]|$1|R|3|S|5|D|7|1Q|8|@]|9|@]|A|$E|F]]|$1|T|3|U|5|6|7|1R|8|@$I|1S|J|1T|K|L]]|9|@]|A|$]]|$1|V|3|-4|5|6|7|1U|8|@]|9|@]|A|$]]]|W|$X|$5|Y|Z|10|A|$11|12]]|13|$5|Y|Z|10|A|$11|14]]]]

Another way to do this:

<pre><code>s = "abcdefghijkl"
l = list( s )
l[-2] = 'A'
s = "".join( l )
</code></pre>

Strings in Python are immutable sequences - very much like tuples. You can make a list, which is mutable, from string or tuple, change relevant indexes and transform the list back into string with <code>join</code> or into <code>tuple</code> with tuple constructor.

EDIT: in case of strings you can use bytearray (<a href="http://docs.python.org/py3k/library/functions.html#bytearray" rel="nofollow noreferrer">docs</a>) type instead of <code>list</code>, which makes transforming into string possible with <code>str</code> constructor.

Also, you could implement your own or use this class from standard library: <a href="http://docs.python.org/library/userdict.html#UserString.MutableString" rel="nofollow noreferrer">http://docs.python.org/library/userdict.html#UserString.MutableString</a> like so:

<pre><code>&gt;&gt;&gt; from UserString import MutableString
&gt;&gt;&gt; s = MutableString( "zdfgzbdr " )
&gt;&gt;&gt; s
'zdfgzbdr '
&gt;&gt;&gt; s[1:5]
'dfgz'
&gt;&gt;&gt; s[1:5] = "xxxx"
&gt;&gt;&gt; s
'zxxxxbdr '
</code></pre>

Sadly - <code>MutableString</code> is deprecated and not available in Py3k (you can still write your own class, I think).

blocks|key|1672566|text|回复其他帖子的评论：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1672567|要直接使用字符串替换三个字符(另请参阅bytearray帖子)，您可以使用序列进行分区，特别是如果您不知道位置而不进行搜索：|1672568|aa+=+'abcdefghijkl'
replace+=+'def'
withstring+=+'QRS'
newstr,found,endpart+=+aa.partition(replace)

if+found:
++++newstr%2B=withstring%2Bendpart
++++print+newstr
else:
++++print+"%25r+string+is+not+in+%25r"+%25+(replace,aa)|code-block|syntax|javascript|1672569|在这种情况下，替换的长度不需要与原始长度匹配。|1672570|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

To reply comment of other post:

To make three character substitution directly with strings (see also the bytearray post) you can do partition with the sequence, especially if you do not know the position without search:

<pre><code>aa = 'abcdefghijkl'
replace = 'def'
withstring = 'QRS'
newstr,found,endpart = aa.partition(replace)

if found:
 newstr+=withstring+endpart
 print newstr
else:
 print "%r string is not in %r" % (replace,aa)
</code></pre>

The length of replacement does not need to match the original in this case.

How to replace a set of characters inside another string in Python?

Here is what I'm trying to do: let's say I have a string 'abcdefghijkl' and want to replace the 2-d from the end symbol (k) with A. I'm getting an error:

<pre><code>&gt;&gt;&gt; aa = 'abcdefghijkl'
&gt;&gt;&gt; print aa[-2]
k
&gt;&gt;&gt; aa[-2]='A'

Traceback (most recent call last):
 File "&lt;pyshell#2&gt;", line 1, in &lt;module&gt;
 aa[-2]='A'
TypeError: 'str' object does not support item assignment
</code></pre>

So, the question: is there an elegant way to replace (substitute) with a string symbols inside another string starting from specified position? Something like:

<pre><code># subst(whole_string,symbols_to_substiture_with,starting_position)

&gt;&gt;&gt; print aa
abcdefghijkl
&gt;&gt;&gt; aa = subst(aa,'A',-2)
&gt;&gt;&gt; print aa
abcdefghijAl
</code></pre>

What would be a not-brute-force code for the subst?

how to substitute part of a string in python?

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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如何在Python中替换另一个字符串中的一组字符？下面是我正在尝试做的事情:假设我有一个字符串'abcdefghijkl‘，并想用A替换结束符号(k)中的2-d。我收到一个错误：>>> aa = 'abcdefghijkl'>>> print aa[-2]k>>> aa[-2]='A'Traceback (most r...

问如何在python中替换字符串的一部分？
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