blocks|key|1148141|text|>>>+s+=+"aaa+bbb+aaa+bbb"
>>>+s[::-1].replace('bbb','xxx',1)[::-1]
'aaa+bbb+aaa+xxx'|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1148142|对于您的第二个示例|unstyled|1148143|>>>+s+=+"123123"
>>>+s[::-1].replace('2','x',1)[::-1]
'1231x3'|1148144|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>&gt;&gt;&gt; s = "aaa bbb aaa bbb"
&gt;&gt;&gt; s[::-1].replace('bbb','xxx',1)[::-1]
'aaa bbb aaa xxx'
</code></pre>

For your second example

<pre><code>&gt;&gt;&gt; s = "123123"
&gt;&gt;&gt; s[::-1].replace('2','x',1)[::-1]
'1231x3'
</code></pre>

blocks|key|1148179|text|当需要的匹配项在字符串的末尾时，re.sub会提供帮助。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1148180|>>>+import+re
>>>+s+=+"aaa+bbb+aaa+bbb"
>>>+s
'aaa+bbb+aaa+bbb'
>>>+re.sub('bbb$',+'xxx',+s)
'aaa+bbb+aaa+xxx'
>>>+|code-block|syntax|javascript|1148181|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/re.html#re.sub^0|G|6|G|6|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

When the wanted match is at the end of string, <a href="http://docs.python.org/library/re.html#re.sub" rel="nofollow noreferrer"><code>re.sub</code></a> comes to the rescue.

<pre><code>&gt;&gt;&gt; import re
&gt;&gt;&gt; s = "aaa bbb aaa bbb"
&gt;&gt;&gt; s
'aaa bbb aaa bbb'
&gt;&gt;&gt; re.sub('bbb$', 'xxx', s)
'aaa bbb aaa xxx'
&gt;&gt;&gt; 
</code></pre>

blocks|key|933813|text|使用正则表达式函数re.sub替换字符串末尾的单词|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|933814|import+re
s+=+"123123"
s+=+re.sub('23$',+'penguins',+s)
print+s|code-block|syntax|javascript|933815|打印：|933816|1231penguins|933817|或|933818|import+re
s+=+"123123"
s+=+re.sub('%5E12',+'penguins',+s)
print+s|933819|933820|penguins3123|933821|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/re.html#re.sub^0|9|6|9|6|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@$9|16|A|17|1|18]]|E|$]]|$1|F|3|G|5|H|7|19|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1A|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1C|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1D|8|@]|D|@]|E|$I|J]]|$1|S|3|L|5|6|7|1E|8|@]|D|@]|E|$]]|$1|T|3|U|5|H|7|1F|8|@]|D|@]|E|$I|J]]|$1|V|3|-4|5|6|7|1G|8|@]|D|@]|E|$]]]|W|$X|$5|Y|Z|10|E|$11|12]]]]

<h2>Using regular expression function <a href="https://docs.python.org/3/library/re.html#re.sub" rel="noreferrer"><code>re.sub</code></a> to replace words at end of string</h2>

<pre><code>import re
s = "123123"
s = re.sub('23$', 'penguins', s)
print s
</code></pre>

Prints:

<pre><code>1231penguins
</code></pre>

or 

<pre><code>import re
s = "123123"
s = re.sub('^12', 'penguins', s)
print s
</code></pre>

Prints:

<pre><code>penguins3123
</code></pre>

blocks|key|933856|text|这是为数不多的没有左右版本的字符串函数之一，但是我们可以使用一些有左右版本的字符串函数来模拟这种行为。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|933857|>>>+s+=+'123123'
>>>+t+=+s.rsplit('2',+1)
>>>+u+=+'x'.join(t)
>>>+u
'1231x3'|code-block|syntax|javascript|933858|或|933859|>>>+'x'.join('123123'.rsplit('2',+1))
'1231x3'|933860|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|$]]

This is one of the few string functions that doesn't have a left and right version, but we can mimic the behaviour using some of the string functions that do.

<pre><code>&gt;&gt;&gt; s = '123123'
&gt;&gt;&gt; t = s.rsplit('2', 1)
&gt;&gt;&gt; u = 'x'.join(t)
&gt;&gt;&gt; u
'1231x3'
</code></pre>

or

<pre><code>&gt;&gt;&gt; 'x'.join('123123'.rsplit('2', 1))
'1231x3'
</code></pre>

blocks|key|933915|text|我得到了一个棘手的答案，但效率不够高。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|933916|++++>>>+fname+=+'12345.png.pngasdfg.png'
++++>>>+suffix+=+'.png'
++++>>>+fname_rev+=+fname[::-1]
++++>>>+suffix_rev+=+suffix[::-1]
++++>>>+fullname_rev+=+fname_rev.replace(suffix_rev,+'',+1)
++++>>>+fullname+=+fullname_rev[::-1]
++++>>>+fullname+'12345.png.pngasdfg'|code-block|syntax|javascript|933917|内置函数replace()接受三个参数str.replace(old,+new,+max_time)，因此您可以从原始字符串中删除最后处理的字符串|offset|length|style|CODE|933918|entityMap^0|0|0|J|V|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I got a tricky answer, but it is not efficient enough

<pre><code>
 &gt;&gt;&gt; fname = '12345.png.pngasdfg.png'
 &gt;&gt;&gt; suffix = '.png'
 &gt;&gt;&gt; fname_rev = fname[::-1]
 &gt;&gt;&gt; suffix_rev = suffix[::-1]
 &gt;&gt;&gt; fullname_rev = fname_rev.replace(suffix_rev, '', 1)
 &gt;&gt;&gt; fullname = fullname_rev[::-1]
 &gt;&gt;&gt; fullname '12345.png.pngasdfg'

</code></pre>

Built-in function replace() takes three arguments
<code>str.replace(old, new, max_time)</code>
So you can delete the last mached string from the origional string

blocks|key|933980|text|有一种方法可以处理函数replace，使其反向工作。其思想是读取字符串、要替换的文本以及从头到尾的新文本。逻辑保持不变。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|933981|这是基于使用[::-1]向后切换字符串的方式。|933982|例如：|933983|>>>+a+=+"123+456+123+456+123+456"
>>>+print(a)
123+456+123+456+123+456|code-block|syntax|javascript|933984|并且我们希望用"abc“替换最后一个(或两个或n个)出现的"123”。|933985|>>>+res+=+a[::-1].replace("123"[::-1],+"abc"[::-1],+1)[::-1]
>>>+print(res)
123+456+123+456+abc+456|933986|这可以作为一个函数，其行为与replace完全相同。|933987|def+replace_right(text,+old,+new,+count=-1):
++++"""
++++Return+a+copy+with+all+occurrences+of+substring+old+replaced+by+new+starting+from+right.
++++"""
++++return+text[::-1].replace(old[::-1],+new[::-1],+count)[::-1]|933988|执行：|933989|>>>+replace_right(a,+"123",+"abc",+1)
'123+456+123+456+abc+456'

>>>+replace_right(a,+"123",+"abc",+2)
'123+456+abc+456+abc+456'|933990|entityMap^0|B|7|0|6|6|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|18|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|19|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|1A|8|@]|D|@]|E|$]]|$1|Q|3|R|5|L|7|1B|8|@]|D|@]|E|$M|N]]|$1|S|3|T|5|6|7|1C|8|@]|D|@]|E|$]]|$1|U|3|V|5|L|7|1D|8|@]|D|@]|E|$M|N]]|$1|W|3|X|5|6|7|1E|8|@]|D|@]|E|$]]|$1|Y|3|Z|5|L|7|1F|8|@]|D|@]|E|$M|N]]|$1|10|3|-4|5|6|7|1G|8|@]|D|@]|E|$]]]|11|$]]

There is a way to handle function <code>replace</code> to make is work backwards.
The idea is to read the string, text to replace and new text from end to start. The logic stays the same.
This is based on the way to switch a string backwards using <code>[::-1]</code>.
For example:
<pre><code>&gt;&gt;&gt; a = &quot;123 456 123 456 123 456&quot;
&gt;&gt;&gt; print(a)
123 456 123 456 123 456
</code></pre>
and we want to replace the last one, (or two, or n) occurrences of &quot;123&quot; with &quot;abc&quot;
<pre><code>&gt;&gt;&gt; res = a[::-1].replace(&quot;123&quot;[::-1], &quot;abc&quot;[::-1], 1)[::-1]
&gt;&gt;&gt; print(res)
123 456 123 456 abc 456
</code></pre>
This could be put as a function that behaves exactly like replace.
<pre><code>def replace_right(text, old, new, count=-1):
 &quot;&quot;&quot;
 Return a copy with all occurrences of substring old replaced by new starting from right.
 &quot;&quot;&quot;
 return text[::-1].replace(old[::-1], new[::-1], count)[::-1]
</code></pre>
Execution:
<pre><code>&gt;&gt;&gt; replace_right(a, &quot;123&quot;, &quot;abc&quot;, 1)
'123 456 123 456 abc 456'

&gt;&gt;&gt; replace_right(a, &quot;123&quot;, &quot;abc&quot;, 2)
'123 456 abc 456 abc 456'
</code></pre>

blocks|key|934007|text|txt+=+"123123"
txt+=+txt[::-1]
txt+=+txt.replace("2",+"*",+1)
txt+=+txt[::-1]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|934008|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>txt = &quot;123123&quot;
txt = txt[::-1]
txt = txt.replace(&quot;2&quot;, &quot;*&quot;, 1)
txt = txt[::-1]
</code></pre>

I want to replace characters at the end of a python string. I have this string:

<pre><code> s = "123123"
</code></pre>

I want to replace the last <code>2</code> with <code>x</code>. Suppose there is a method called <code>replace_last</code>:

<pre><code> r = replace_last(s, '2', 'x')
 print r
 1231x3
</code></pre>

Is there any built-in or easy method to do this?

How to replace the some characters from the end of a string?

我想替换python字符串末尾的字符。我有这个字符串： s = "123123"我想用x替换最后一个2。假设有一个名为replace_last的方法 r = replace_last(s, '2', 'x') print r 1231x3有没有什么内置的或者简单的方法可以做到这一点呢？

问如何替换字符串末尾的一些字符？
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