blocks|key|1924415|text|字符串在Python中是不可变的。replace方法在替换后返回一个新字符串。尝试：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1924416|for+char+in+line:
++++if+char+in+"+?.!/;:":
++++++++line+=+line.replace(char,'')|code-block|syntax|javascript|1924417|这与您的原始代码完全相同，只是在循环中添加了对line的赋值。|1924418|请注意，string+replace()方法将替换字符串中出现的所有字符，因此，如果对要删除的每个字符使用replace()，而不是循环遍历字符串中的每个字符，则可以做得更好。|1924419|entityMap^0|H|7|0|0|N|4|0|B|9|1H|9|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|P|$]]

Strings are immutable in Python. The <code>replace</code> method returns a new string after the replacement. Try:
<pre><code>for char in line:
 if char in &quot; ?.!/;:&quot;:
 line = line.replace(char,'')
</code></pre>
This is identical to your original code, with the addition of an assignment to <code>line</code> inside the loop.
Note that the string <code>replace()</code> method replaces all of the occurrences of the character in the string, so you can do better by using <code>replace()</code> for each character you want to remove, instead of looping over each character in your string.

blocks|key|3325613|text|Python中的字符串是不可变的(不能更改)。正因为如此，line.replace(...)的作用只是创建一个新字符串，而不是更改旧字符串。您需要将其重新绑定(赋值)到line，以便使该变量获得新值，并删除这些字符。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3325614|而且，你做这件事的方式会相对缓慢。这也可能会让有经验的pythonators开发者感到困惑，他们会看到一个双重嵌套的结构，并认为有一些更复杂的东西正在发生。|3325615|从Python2.6和更新的Python2.x版本*开始，您可以改为使用str.translate，(参见下面的Python+3+answer+)：|BOLD|3325616|line+=+line.translate(None,+'!@#$')|code-block|syntax|javascript|3325617|或使用re.sub替换正则表达式|3325618|import+re
line+=+re.sub('[!@#$]',+'',+line)|3325619|括在括号中的字符构成了一个字符类。类中的line中的任何字符都将被sub的第二个参数替换:一个空字符串。|3325620|Python+3答案|3325621|在Python+3中，字符串是Unicode。你必须用不同的方式来翻译。凯夫派在其中一个答案的comment中提到了这一点，并在documentation+for+str.translate中提到了这一点。|3325622|在调用Unicode字符串的translate方法时，不能传递上面使用的第二个参数。您也不能将None作为第一个参数传递。相反，您需要传递一个转换表(通常是一个字典)作为唯一的参数。此表将字符的序数值(即，对字符调用ord的结果)映射到应替换它们的字符的序数值，或者-对我们有用-None，以指示应删除它们。|3325623|因此，要使用Unicode字符串执行上述操作，您可以调用如下代码|3325624|translation_table+=+dict.fromkeys(map(ord,+'!@#$'),+None)
unicode_line+=+unicode_line.translate(translation_table)|3325625|这里使用dict.fromkeys和map简洁地生成包含以下内容的字典|3325626|{ord('!'):+None,+ord('@'):+None,+...}|3325627|更简单的是，作为another+answer+puts+it，就地创建转换表：|3325628|unicode_line+=+unicode_line.translate({ord(c):+None+for+c+in+'!@#$'})|3325629|或者，按照Joseph+Lee的建议，使用str.maketrans创建相同的转换表|3325630|unicode_line+=+unicode_line.translate(str.maketrans('',+'',+'!@#$'))|3325631|3325632|*为了与早期的Python兼容，您可以创建一个"null“转换表来代替None|3325633|import+string
line+=+line.translate(string.maketrans('',+''),+'!@#$')|3325634|这里使用string.maketrans创建转换表，转换表只是一个字符串，其中包含序号值从0到255的字符。|3325635|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/2/library/stdtypes.html#str.translate|1|https://docs.python.org/2/library/re.html#re.sub|2|https://stackoverflow.com/questions/3939361/remove-specific-characters-from-a-string-in-python#comment-4205256|3|4|https://docs.python.org/2/library/functions.html#ord|5|https://docs.python.org/2/library/stdtypes.html#dict.fromkeys|6|https://docs.python.org/2/library/functions.html#map|7|https://stackoverflow.com/questions/23175809/typeerror-translate-takes-one-argument-2-given-python|8|https://stackoverflow.com/a/47030484/2932052|9|https://docs.python.org/3/library/stdtypes.html#str.maketrans|10|https://docs.python.org/2/library/string.html#string.maketrans^0|T|H|2C|4|0|0|10|D|1K|F|10|D|0|0|0|3|6|3|6|1|0|0|K|4|X|3|0|0|2A|D|1B|7|2|1S|V|3|0|E|9|1B|4|30|3|3W|4|30|3|4|0|0|0|4|D|I|3|4|D|5|I|3|6|0|0|8|M|7|0|0|L|D|5|A|8|L|D|9|0|0|0|Z|4|0|0|4|G|4|G|A|0^^$0|@$1|2|3|4|5|6|7|2F|8|@$9|2G|A|2H|B|C]|$9|2I|A|2J|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|2K|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|2L|8|@$9|2M|A|2N|B|C]|$9|2O|A|2P|B|J]]|D|@$9|2Q|A|2R|1|2S]]|E|$]]|$1|K|3|L|5|M|7|2T|8|@]|D|@]|E|$N|O]]|$1|P|3|Q|5|6|7|2U|8|@$9|2V|A|2W|B|C]]|D|@$9|2X|A|2Y|1|2Z]]|E|$]]|$1|R|3|S|5|M|7|30|8|@]|D|@]|E|$N|O]]|$1|T|3|U|5|6|7|31|8|@$9|32|A|33|B|C]|$9|34|A|35|B|C]]|D|@]|E|$]]|$1|V|3|W|5|6|7|36|8|@]|D|@]|E|$]]|$1|X|3|Y|5|6|7|37|8|@$9|38|A|39|B|C]]|D|@$9|3A|A|3B|1|3C]|$9|3D|A|3E|1|3F]]|E|$]]|$1|Z|3|10|5|6|7|3G|8|@$9|3H|A|3I|B|C]|$9|3J|A|3K|B|C]|$9|3L|A|3M|B|C]|$9|3N|A|3O|B|C]]|D|@$9|3P|A|3Q|1|3R]]|E|$]]|$1|11|3|12|5|6|7|3S|8|@]|D|@]|E|$]]|$1|13|3|14|5|M|7|3T|8|@]|D|@]|E|$N|O]]|$1|15|3|16|5|6|7|3U|8|@$9|3V|A|3W|B|C]|$9|3X|A|3Y|B|C]]|D|@$9|3Z|A|40|1|41]|$9|42|A|43|1|44]]|E|$]]|$1|17|3|18|5|M|7|45|8|@]|D|@]|E|$N|O]]|$1|19|3|1A|5|6|7|46|8|@]|D|@$9|47|A|48|1|49]]|E|$]]|$1|1B|3|1C|5|M|7|4A|8|@]|D|@]|E|$N|O]]|$1|1D|3|1E|5|6|7|4B|8|@$9|4C|A|4D|B|C]]|D|@$9|4E|A|4F|1|4G]|$9|4H|A|4I|1|4J]]|E|$]]|$1|1F|3|1G|5|M|7|4K|8|@]|D|@]|E|$N|O]]|$1|1H|3|-4|5|6|7|4L|8|@]|D|@]|E|$]]|$1|1I|3|1J|5|6|7|4M|8|@$9|4N|A|4O|B|C]]|D|@]|E|$]]|$1|1K|3|1L|5|M|7|4P|8|@]|D|@]|E|$N|O]]|$1|1M|3|1N|5|6|7|4Q|8|@$9|4R|A|4S|B|C]]|D|@$9|4T|A|4U|1|4V]]|E|$]]|$1|1O|3|-4|5|6|7|4W|8|@]|D|@]|E|$]]]|1P|$1Q|$5|1R|1S|1T|E|$1U|1V]]|1W|$5|1R|1S|1T|E|$1U|1X]]|1Y|$5|1R|1S|1T|E|$1U|1Z]]|20|$5|1R|1S|1T|E|$1U|1V]]|21|$5|1R|1S|1T|E|$1U|22]]|23|$5|1R|1S|1T|E|$1U|24]]|25|$5|1R|1S|1T|E|$1U|26]]|27|$5|1R|1S|1T|E|$1U|28]]|29|$5|1R|1S|1T|E|$1U|2A]]|2B|$5|1R|1S|1T|E|$1U|2C]]|2D|$5|1R|1S|1T|E|$1U|2E]]]]

Strings in Python are immutable (can't be changed). Because of this, the effect of <code>line.replace(...)</code> is just to create a new string, rather than changing the old one. You need to rebind (assign) it to <code>line</code> in order to have that variable take the new value, with those characters removed.
Also, the way you are doing it is going to be kind of slow, relatively. It's also likely to be a bit confusing to experienced pythonators, who will see a doubly-nested structure and think for a moment that something more complicated is going on.
Starting in Python 2.6 and newer Python 2.x versions *, you can instead use <a href="https://docs.python.org/2/library/stdtypes.html#str.translate" rel="noreferrer"><code>str.translate</code></a>, (see Python 3 answer below):
<pre><code>line = line.translate(None, '!@#$')
</code></pre>
or regular expression replacement with <a href="https://docs.python.org/2/library/re.html#re.sub" rel="noreferrer"><code>re.sub</code></a>
<pre><code>import re
line = re.sub('[!@#$]', '', line)
</code></pre>
The characters enclosed in brackets constitute a character class. Any characters in <code>line</code> which are in that class are replaced with the second parameter to <code>sub</code>: an empty string.
<h3>Python 3 answer</h3>
In Python 3, strings are Unicode. You'll have to translate a little differently. kevpie mentions this in a <a href="https://stackoverflow.com/questions/3939361/remove-specific-characters-from-a-string-in-python#comment-4205256">comment</a> on one of the answers, and it's noted in the <a href="https://docs.python.org/2/library/stdtypes.html#str.translate" rel="noreferrer">documentation for <code>str.translate</code></a>.
When calling the <code>translate</code> method of a Unicode string, you cannot pass the second parameter that we used above. You also can't pass <code>None</code> as the first parameter. Instead, you pass a translation table (usually a dictionary) as the only parameter. This table maps the ordinal values of characters (i.e. the result of calling <a href="https://docs.python.org/2/library/functions.html#ord" rel="noreferrer"><code>ord</code></a> on them) to the ordinal values of the characters which should replace them, or—usefully to us—<code>None</code> to indicate that they should be deleted.
So to do the above dance with a Unicode string you would call something like
<pre><code>translation_table = dict.fromkeys(map(ord, '!@#$'), None)
unicode_line = unicode_line.translate(translation_table)
</code></pre>
Here <a href="https://docs.python.org/2/library/stdtypes.html#dict.fromkeys" rel="noreferrer"><code>dict.fromkeys</code></a> and <a href="https://docs.python.org/2/library/functions.html#map" rel="noreferrer"><code>map</code></a> are used to succinctly generate a dictionary containing
<pre><code>{ord('!'): None, ord('@'): None, ...}
</code></pre>
Even simpler, as <a href="https://stackoverflow.com/questions/23175809/typeerror-translate-takes-one-argument-2-given-python">another answer puts it</a>, create the translation table in place:
<pre><code>unicode_line = unicode_line.translate({ord(c): None for c in '!@#$'})
</code></pre>
Or, as brought up by <a href="https://stackoverflow.com/a/47030484/2932052">Joseph Lee</a>, create the same translation table with <a href="https://docs.python.org/3/library/stdtypes.html#str.maketrans" rel="noreferrer"><code>str.maketrans</code></a>:
<pre><code>unicode_line = unicode_line.translate(str.maketrans('', '', '!@#$'))
</code></pre>
<hr />
* for compatibility with earlier Pythons, you can create a &quot;null&quot; translation table to pass in place of <code>None</code>:
<pre><code>import string
line = line.translate(string.maketrans('', ''), '!@#$')
</code></pre>
Here <a href="https://docs.python.org/2/library/string.html#string.maketrans" rel="noreferrer"><code>string.maketrans</code></a> is used to create a translation table, which is just a string containing the characters with ordinal values 0 to 255.

blocks|key|4736033|text|line+=+line.translate(None,+"+?.!/;:")|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4736034|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>line = line.translate(None, " ?.!/;:")
</code></pre>

blocks|key|4736036|text|>>>+line+=+"abc#@!?efg12;:?"
>>>+''.join(+c+for+c+in+line+if++c+not+in+'?:!/;'+)
'abc#@efg12'|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4736037|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|$]]

blocks|key|1929271|text|我是不是错过了这一点，或者仅仅是以下几点：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1929272|string+=+"ab1cd1ef"
string+=+string.replace("1",+"")+

print(string)
#+result:+"abcdef"|code-block|syntax|javascript|1929273|将其放入循环中：|1929274|a+=+"a!b@c#d$"
b+=+"!@#$"
for+char+in+b:
++++a+=+a.replace(char,+"")

print(a)
#+result:+"abcd"|1929275|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|$]]

Am I missing the point here, or is it just the following:
<pre class="lang-py prettyprint-override"><code>string = &quot;ab1cd1ef&quot;
string = string.replace(&quot;1&quot;, &quot;&quot;) 

print(string)
# result: &quot;abcdef&quot;
</code></pre>
Put it in a loop:
<pre class="lang-py prettyprint-override"><code>a = &quot;a!b@c#d$&quot;
b = &quot;!@#$&quot;
for char in b:
 a = a.replace(char, &quot;&quot;)

print(a)
# result: &quot;abcd&quot;
</code></pre>

blocks|key|3325737|text|发问者差一点就成功了。像Python中的大多数事情一样，答案比您想象的要简单。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3325738|>>>+line+=+"H+E?.LL!/;O::+"++
>>>+for+char+in+'+?.!/;:':++
...++line+=+line.replace(char,'')++
...
>>>+print+line
HELLO|code-block|syntax|javascript|3325739|您不必执行嵌套的if/for循环，但需要逐个检查每个字符。|3325740|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|$]]

The asker almost had it. Like most things in Python, the answer is simpler than you think.

<pre><code>&gt;&gt;&gt; line = "H E?.LL!/;O:: " 
&gt;&gt;&gt; for char in ' ?.!/;:': 
... line = line.replace(char,'') 
...
&gt;&gt;&gt; print line
HELLO
</code></pre>

You don't have to do the nested if/for loop thing, but you DO need to check each character individually.

blocks|key|4736192|text|对于only+allowing+某些字符+in+a+string的相反要求，您可以使用带有set+can运算符[%5EABCabc]的正则表达式。例如，要删除除ascii字母、数字和连字符之外的所有内容：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|CODE|entityRanges|data|4736193|>>>+import+string
>>>+import+re
>>>
>>>+phrase+=+'++There+were+"nine"+(9)+chick-peas+in+my+pocket!!!++++++'
>>>+allow+=+string.letters+%2B+string.digits+%2B+'-'
>>>+re.sub('[%5E%25s]'+%25+allow,+'',+phrase)

'Therewerenine9chick-peasinmypocket'|code-block|syntax|javascript|4736194|来自the+python+regular+expression+documentation|4736195|不在范围内的|4736196|字符可以通过对集合进行补充来匹配。如果集合的第一个字符是'%5E'，则将匹配不在集合中的所有字符。例如，[%5E5]将匹配除“5”之外的任何字符，而[%5E%5E]将匹配除'%5E'之外的任何字符。如果%5E不是集合中的第一个字符，那么它没有特殊意义。|blockquote|4736197|4736198|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/2/library/re.html^0|2|4|7|8|G|4|1J|9|0|0|2|17|0|0|0|S|3|1E|4|1Y|4|26|3|2J|1|0|0^^$0|@$1|2|3|4|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]|$9|16|A|17|B|C]|$9|18|A|19|B|D]]|E|@]|F|$]]|$1|G|3|H|5|I|7|1A|8|@]|E|@]|F|$J|K]]|$1|L|3|M|5|6|7|1B|8|@]|E|@$9|1C|A|1D|1|1E]]|F|$]]|$1|N|3|O|5|6|7|1F|8|@]|E|@]|F|$]]|$1|P|3|Q|5|R|7|1G|8|@$9|1H|A|1I|B|D]|$9|1J|A|1K|B|D]|$9|1L|A|1M|B|D]|$9|1N|A|1O|B|D]|$9|1P|A|1Q|B|D]]|E|@]|F|$]]|$1|S|3|-4|5|6|7|1R|8|@]|E|@]|F|$]]|$1|T|3|-4|5|6|7|1S|8|@]|E|@]|F|$]]]|U|$V|$5|W|X|Y|F|$Z|10]]]]

For the inverse requirement of only allowing certain characters in a string, you can use regular expressions with a set complement operator <code>[^ABCabc]</code>. For example, to remove everything except ascii letters, digits, and the hyphen:

<pre><code>&gt;&gt;&gt; import string
&gt;&gt;&gt; import re
&gt;&gt;&gt;
&gt;&gt;&gt; phrase = ' There were "nine" (9) chick-peas in my pocket!!! '
&gt;&gt;&gt; allow = string.letters + string.digits + '-'
&gt;&gt;&gt; re.sub('[^%s]' % allow, '', phrase)

'Therewerenine9chick-peasinmypocket'
</code></pre>

From <a href="http://docs.python.org/2/library/re.html">the python regular expression documentation</a>:

<blockquote>
 Characters that are not within a range can be matched by complementing
 the set. If the first character of the set is <code>'^'</code>, all the characters
 that are not in the set will be matched. For example, <code>[^5]</code> will match
 any character except '5', and <code>[^^]</code> will match any character except
 <code>'^'</code>. <code>^</code> has no special meaning if it’s not the first character in the
 set.
</blockquote>

blocks|key|1182012|text|#!/usr/bin/python
import+re

strs+=+"how%5E+much+for{}+the+maple+syrup?+$20.99?+That's[]+ricidulous!!!"
print+strs
nstr+=+re.sub(r'[?%7C$%7C.%7C!%7Ca%7Cb]',r'+',strs)#i+have+taken+special+character+to+remove+but+any+#character+can+be+added+here
print+nstr
nestr+=+re.sub(r'[%5Ea-zA-Z0-9+]',r'',nstr)#for+removing+special+character
print+nestr|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1182013|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>#!/usr/bin/python
import re

strs = "how^ much for{} the maple syrup? $20.99? That's[] ricidulous!!!"
print strs
nstr = re.sub(r'[?|$|.|!|a|b]',r' ',strs)#i have taken special character to remove but any #character can be added here
print nstr
nestr = re.sub(r'[^a-zA-Z0-9 ]',r'',nstr)#for removing special character
print nestr
</code></pre>

blocks|key|1924527|text|这样如何：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1924528|def+text_cleanup(text):
++++new+=+""
++++for+i+in+text:
++++++++if+i+not+in+"+?.!/;:":
++++++++++++new+%2B=+i
++++return+new|code-block|syntax|javascript|1924529|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|$]]

How about this:

<pre><code>def text_cleanup(text):
 new = ""
 for i in text:
 if i not in " ?.!/;:":
 new += i
 return new
</code></pre>

blocks|key|3325835|text|低于1.。不使用正则表达式概念..|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3325836|ipstring+="text+with+symbols!@#$%5E&*(+ends+here"
opstring=''
for+i+in+ipstring:
++++if+i.isalnum()==1+or+i=='+':
++++++++opstring%2B=i
++++pass
print+opstring|code-block|syntax|javascript|3325837|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|$]]

Below one.. with out using regular expression concept.. 

<pre><code>ipstring ="text with symbols!@#$^&amp;*( ends here"
opstring=''
for i in ipstring:
 if i.isalnum()==1 or i==' ':
 opstring+=i
 pass
print opstring
</code></pre>

blocks|key|1929358|text|您还可以使用一个函数来将不同种类的正则表达式或其他模式替换为使用列表。这样，你就可以混合使用正则表达式、字符类和真正的基本文本模式。当你需要替换很多像HTML这样的元素时，它真的很有用。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1929359|*注意:适用于Python+3.x|1929360|import+re++#+Regular+expression+library


def+string_cleanup(x,+notwanted):
++++for+item+in+notwanted:
++++++++x+=+re.sub(item,+'',+x)
++++return+x

line+=+"<title>My+example:+A+text+%25very%25+$clean!!</title>"
print("Uncleaned:+",+line)

#+Get+rid+of+html+elements
html_elements+=+["<title>",+"</title>",+"",+""]
line+=+string_cleanup(line,+html_elements)
print("1st+clean:+",+line)

#+Get+rid+of+special+characters
special_chars+=+["[!@#$]",+"%25"]
line+=+string_cleanup(line,+special_chars)
print("2nd+clean:+",+line)|code-block|syntax|javascript|1929361|在函数string_cleanup中，它将字符串x和列表作为参数。对于该元素或模式列表中的每一项，如果需要替换，就会这样做。|1929362|输出：|1929363|Uncleaned:++<title>My+example:+A+text+%25very%25+$clean!!</title>
1st+clean:++My+example:+A+text+%25very%25+$clean!!
2nd+clean:++My+example:+A+text+very+clean|1929364|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|V|8|@]|9|@]|A|$G|H]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

You can also use a function in order to substitute different kind of regular expression or other pattern with the use of a list. With that, you can mixed regular expression, character class, and really basic text pattern. It's really useful when you need to substitute a lot of elements like HTML ones.

*NB: works with Python 3.x

<pre><code>import re # Regular expression library


def string_cleanup(x, notwanted):
 for item in notwanted:
 x = re.sub(item, '', x)
 return x

line = "&lt;title&gt;My example: &lt;strong&gt;A text %very% $clean!!&lt;/strong&gt;&lt;/title&gt;"
print("Uncleaned: ", line)

# Get rid of html elements
html_elements = ["&lt;title&gt;", "&lt;/title&gt;", "&lt;strong&gt;", "&lt;/strong&gt;"]
line = string_cleanup(line, html_elements)
print("1st clean: ", line)

# Get rid of special characters
special_chars = ["[!@#$]", "%"]
line = string_cleanup(line, special_chars)
print("2nd clean: ", line)
</code></pre>

In the function string_cleanup, it takes your string x and your list notwanted as arguments. For each item in that list of elements or pattern, if a substitute is needed it will be done.

The output:

<pre><code>Uncleaned: &lt;title&gt;My example: &lt;strong&gt;A text %very% $clean!!&lt;/strong&gt;&lt;/title&gt;
1st clean: My example: A text %very% $clean!!
2nd clean: My example: A text very clean
</code></pre>

blocks|key|3325885|text|>>>+s+=+'a1b2c3'
>>>+''.join(c+for+c+in+s+if+c+not+in+'123')
'abc'|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3325886|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|$]]

blocks|key|4736334|text|我使用的方法可能不会那么有效，但它非常简单。我可以使用切片和格式化一次删除不同位置的多个字符。下面是一个例子：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4736335|words+=+"things"
removed+=+"%25s%25s"+%25+(words[:3],+words[-1:])|code-block|syntax|javascript|4736336|这将导致“删除”保留单词“This”。|4736337|格式化对于在打印字符串中途打印变量非常有用。它可以使用%25后跟变量的数据类型插入任何数据类型；所有数据类型都可以使用%25s，浮点数(又称小数)和整数可以使用%25d。|offset|length|style|BOLD|4736338|切片可用于对字符串的复杂控制。当我输入words:3时，它允许我选择字符串中从开头(冒号在数字之前，这意味着‘从开始到’)到第四个字符(包括第四个字符)的所有字符。3等于第4位的原因是因为Python从0开始。然后，当我输入word-1:时，它表示倒数第二个字符(冒号在数字后面)。放入-1将使Python从最后一个字符开始计数，而不是从第一个字符开始计数。同样，Python将从0开始。因此，word-1:的基本意思是‘从倒数第二个字符到字符串末尾。|4736339|因此，通过删除我想要删除的字符之前的字符和之后的字符，并将它们夹在一起，我可以删除不需要的字符。把它想像成一根香肠。在中间，它很脏，所以我想把它扔掉。我简单地剪掉我想要的两个末端，然后把它们放在一起，中间没有不想要的部分。|4736340|如果我想删除多个连续的字符，我只需在slicing+part中移动数字即可。或者，如果我想从不同的位置删除多个字符，我可以简单地一次将多个切片夹在一起。|4736341|示例：|4736342|+words+=+"control"
+removed+=+"%25s%25s"+%25+(words[:2],+words[-2:])|4736343|删除了等于'cool‘的。|4736344|words+=+"impacts"
removed+=+"%25s%25s%25s"+%25+(words[1],+words[3:5],+words[-1])|4736345|删除了等于‘mac’的。|4736346|在这种情况下，3:5表示位置3的字符到位置5的字符(不包括最终位置的字符)。|4736347|请记住，Python从0开始计数，因此您也需要从0开始计数。|4736348|entityMap|0|LINK|mutability|MUTABLE|url|slicing+part^0|0|0|0|0|3|R|1|1L|2|24|2|0|0|2|J|7|34|7|5H|7|0|1C|A|0|I|C|0|0|0|0|0|C|0|0|0|B|0|7|3|C|2|J|2|0|0^^$0|@$1|2|3|4|5|6|7|1G|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1H|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1I|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|1J|8|@$K|1K|L|1L|M|N]|$K|1M|L|1N|M|N]|$K|1O|L|1P|M|N]|$K|1Q|L|1R|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|1S|8|@$K|1T|L|1U|M|N]|$K|1V|L|1W|M|N]|$K|1X|L|1Y|M|N]|$K|1Z|L|20|M|N]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|21|8|@$K|22|L|23|M|N]]|9|@]|A|$]]|$1|S|3|T|5|6|7|24|8|@]|9|@$K|25|L|26|1|27]]|A|$]]|$1|U|3|V|5|6|7|28|8|@]|9|@]|A|$]]|$1|W|3|X|5|D|7|29|8|@]|9|@]|A|$E|F]]|$1|Y|3|Z|5|6|7|2A|8|@$K|2B|L|2C|M|N]]|9|@]|A|$]]|$1|10|3|11|5|D|7|2D|8|@]|9|@]|A|$E|F]]|$1|12|3|13|5|6|7|2E|8|@$K|2F|L|2G|M|N]]|9|@]|A|$]]|$1|14|3|15|5|6|7|2H|8|@$K|2I|L|2J|M|N]|$K|2K|L|2L|M|N]|$K|2M|L|2N|M|N]]|9|@]|A|$]]|$1|16|3|17|5|6|7|2O|8|@]|9|@]|A|$]]|$1|18|3|-4|5|6|7|2P|8|@]|9|@]|A|$]]]|19|$1A|$5|1B|1C|1D|A|$1E|1F]]]]

My method I'd use probably wouldn't work as efficiently, but it is massively simple. I can remove multiple characters at different positions all at once, using slicing and formatting.
Here's an example:

<pre><code>words = "things"
removed = "%s%s" % (words[:3], words[-1:])
</code></pre>

This will result in 'removed' holding the word 'this'.

Formatting can be very helpful for printing variables midway through a print string. It can insert any data type using a % followed by the variable's data type; all data types can use %s, and floats (aka decimals) and integers can use %d. 

Slicing can be used for intricate control over strings. When I put words[:3], it allows me to select all the characters in the string from the beginning (the colon is before the number, this will mean 'from the beginning to') to the 4th character (it includes the 4th character). The reason 3 equals till the 4th position is because Python starts at 0. Then, when I put word[-1:], it means the 2nd last character to the end (the colon is behind the number). Putting -1 will make Python count from the last character, rather than the first. Again, Python will start at 0. So, word[-1:] basically means 'from the second last character to the end of the string.

So, by cutting off the characters before the character I want to remove and the characters after and sandwiching them together, I can remove the unwanted character. Think of it like a sausage. In the middle it's dirty, so I want to get rid of it. I simply cut off the two ends I want then put them together without the unwanted part in the middle. 

If I want to remove multiple consecutive characters, I simply shift the numbers around in the [] (slicing part). Or if I want to remove multiple characters from different positions, I can simply sandwich together multiple slices at once.

Examples:

<pre><code> words = "control"
 removed = "%s%s" % (words[:2], words[-2:])
</code></pre>

removed equals 'cool'.

<pre><code>words = "impacts"
removed = "%s%s%s" % (words[1], words[3:5], words[-1])
</code></pre>

removed equals 'macs'.

In this case, [3:5] means character at position 3 through character at position 5 (excluding the character at the final position). 

Remember, Python starts counting at 0, so you will need to as well.

blocks|key|4736368|text|这是我的Python2/3兼容版本。因为转换api已更改。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4736369|def+remove(str_,+chars):
++++"""Removes+each+char+in+`chars`+from+`str_`.

++++Args:
++++++++str_:+String+to+remove+characters+from
++++++++chars:+String+of+to-be+removed+characters

++++Returns:
++++++++A+copy+of+str_+with+`chars`+removed

++++Example:
++++++++++++remove("What?!?:+darn;",+"+?.!:;")+=>+'Whatdarn'
++++"""
++++try:
++++++++#+Python2.x
++++++++return+str_.translate(None,+chars)
++++except+TypeError:
++++++++#+Python+3.x
++++++++table+=+{ord(char):+None+for+char+in+chars}
++++++++return+str_.translate(table)|code-block|syntax|javascript|4736370|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|$]]

Here's my Python 2/3 compatible version. Since the translate api has changed. 

<pre><code>def remove(str_, chars):
 """Removes each char in `chars` from `str_`.

 Args:
 str_: String to remove characters from
 chars: String of to-be removed characters

 Returns:
 A copy of str_ with `chars` removed

 Example:
 remove("What?!?: darn;", " ?.!:;") =&gt; 'Whatdarn'
 """
 try:
 # Python2.x
 return str_.translate(None, chars)
 except TypeError:
 # Python 3.x
 table = {ord(char): None for char in chars}
 return str_.translate(table)
</code></pre>

blocks|key|4736383|text|在Python+3.5中|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4736384|例如，|4736385|os.rename(file_name,+file_name.translate({ord(c):+None+for+c+in+'0123456789'}))|code-block|syntax|javascript|4736386|从字符串中删除所有数字|4736387|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|$]]

<h2>In Python 3.5</h2>

e.g.,

<pre><code>os.rename(file_name, file_name.translate({ord(c): None for c in '0123456789'}))
</code></pre>

To remove all the number from the string

blocks|key|4736405|text|即使是下面的方法也是可行的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4736406|line+=+"a,b,c,d,e"
alpha+=+list(line)
++++++++while+','+in+alpha:
++++++++++++alpha.remove(',')
finalString+=+''.join(alpha)
print(finalString)|code-block|syntax|javascript|4736407|输出：abcde|offset|length|style|CODE|4736408|entityMap^0|0|0|3|5|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|$]]

Even the below approach works

<pre><code>line = "a,b,c,d,e"
alpha = list(line)
 while ',' in alpha:
 alpha.remove(',')
finalString = ''.join(alpha)
print(finalString)
</code></pre>

output: <code>abcde</code>

blocks|key|1179737|text|使用re.sub正则表达式|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1179738|从Python3.5开始，可以使用正则表达式re.sub进行替换：|1179739|import+re
re.sub('\+%7C\?%7C\.%7C\!%7C\/%7C\;%7C\:',+'',+line)|code-block|syntax|javascript|1179740|示例|1179741|import+re
line+=+'Q:+Do+I+write+;/.???+No!!!'
re.sub('\+%7C\?%7C\.%7C\!%7C\/%7C\;%7C\:',+'',+line)

'QDoIwriteNo'|1179742|解释|1179743|在+(正则表达式)中，%7C是逻辑OR，\对空格和特殊字符进行转义，这些字符可能是实际的正则表达式命令。而sub代表替换，在本例中为，使用空字符串''。|BOLD|1179744|entityMap^0|2|6|0|M|6|0|0|0|0|0|B|1|I|1|1F|3|1Z|2|1K|8|0^^$0|@$1|2|3|4|5|6|7|X|8|@$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|13|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|14|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|15|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|16|8|@]|D|@]|E|$]]|$1|S|3|T|5|6|7|17|8|@$9|18|A|19|B|C]|$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]|$9|1E|A|1F|B|C]|$9|1G|A|1H|B|U]]|D|@]|E|$]]|$1|V|3|-4|5|6|7|1I|8|@]|D|@]|E|$]]]|W|$]]

<h1>With <code>re.sub</code> regular expression</h1>
Since Python 3.5, substitution using regular expressions <code>re.sub</code> became available:
<pre class="lang-py prettyprint-override"><code>import re
re.sub('\ |\?|\.|\!|\/|\;|\:', '', line)
</code></pre>
<h2>Example</h2>
<pre class="lang-py prettyprint-override"><code>import re
line = 'Q: Do I write ;/.??? No!!!'
re.sub('\ |\?|\.|\!|\/|\;|\:', '', line)

'QDoIwriteNo'
</code></pre>
<h2>Explanation</h2>
In <a href="https://en.wikipedia.org/wiki/Regular_expression" rel="noreferrer">regular expressions</a> (regex), <code>|</code> is a logical OR and <code>\</code> escapes spaces and special characters that might be actual regex commands. Whereas <code>sub</code> stands for substitution, in this case with the empty string <code>''</code>.

blocks|key|3326089|text|使用filter，您只需要一行代码|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3326090|line+=+filter(lambda+char:+char+not+in+"+?.!/;:",+line)|code-block|syntax|javascript|3326091|这会将字符串视为可迭代对象，并在lambda返回True时检查每个字符|3326092|有关模块__builtin__中内置函数filter的|3326093|help(filter)帮助:+filter(...)filter(function+or+None，sequence)+->列表、元组或字符串返回function(item)为true的序列项目。如果function为None，则返回为true的项。如果序列是元组或字符串，则返回相同类型，否则返回list.|blockquote|3326094|3326095|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/2/library/functions.html#filter^0|2|6|2|6|0|0|0|G|6|O|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]]|E|$]]|$1|F|3|G|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|17|8|@$9|18|A|19|B|C]|$9|1A|A|1B|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|1C|8|@]|D|@]|E|$]]|$1|O|3|P|5|Q|7|1D|8|@]|D|@]|E|$]]|$1|R|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]|$1|S|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|Z]]]]

Using <a href="https://docs.python.org/2/library/functions.html#filter" rel="noreferrer"><code>filter</code></a>, you'd just need one line


<pre><code>line = filter(lambda char: char not in " ?.!/;:", line)
</code></pre>

This treats the string as an iterable and checks every character if the <code>lambda</code> returns <code>True</code>:

<blockquote>
<pre><code>&gt;&gt;&gt; help(filter)
Help on built-in function filter in module __builtin__:

filter(...)
 filter(function or None, sequence) -&gt; list, tuple, or string

 Return those items of sequence for which function(item) is true. If
 function is None, return the items that are true. If sequence is a tuple
 or string, return the same type, else return a list.
</code></pre>
</blockquote>

blocks|key|1924654|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1924655|def+rm_char(original_str,+need2rm):
++++'''+Remove+charecters+in+"need2rm"+from+"original_str"+'''
++++return+original_str.translate(str.maketrans('','',need2rm))|code-block|syntax|javascript|1924656|此方法在Python+3中运行良好|1924657|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|$]]

Try this one:
<pre><code>def rm_char(original_str, need2rm):
 ''' Remove charecters in &quot;need2rm&quot; from &quot;original_str&quot; '''
 return original_str.translate(str.maketrans('','',need2rm))
</code></pre>
This method works well in Python 3

blocks|key|1929527|text|递归拆分:+s=string；要删除的chars=chars|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1929528|def+strip(s,chars):
if+len(s)==1:
++++return+""+if+s+in+chars+else+s
return+strip(s[0:int(len(s)/2)],chars)+%2B++strip(s[int(len(s)/2):len(s)],chars)|code-block|syntax|javascript|1929529|示例：|1929530|print(strip("Hello!","lo"))++++#He!|1929531|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|$]]

Recursive split:
s=string ; chars=chars to remove

<pre><code>def strip(s,chars):
if len(s)==1:
 return "" if s in chars else s
return strip(s[0:int(len(s)/2)],chars) + strip(s[int(len(s)/2):len(s)],chars)
</code></pre>

example: 

<pre><code>print(strip("Hello!","lo")) #He!
</code></pre>

blocks|key|4736613|text|以下是实现此任务的一些可能方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4736614|def+attempt1(string):
++++return+"".join([v+for+v+in+string+if+v+not+in+("a",+"e",+"i",+"o",+"u")])


def+attempt2(string):
++++for+v+in+("a",+"e",+"i",+"o",+"u"):
++++++++string+=+string.replace(v,+"")
++++return+string


def+attempt3(string):
++++import+re
++++for+v+in+("a",+"e",+"i",+"o",+"u"):
++++++++string+=+re.sub(v,+"",+string)
++++return+string


def+attempt4(string):
++++return+string.replace("a",+"").replace("e",+"").replace("i",+"").replace("o",+"").replace("u",+"")


for+attempt+in+[attempt1,+attempt2,+attempt3,+attempt4]:
++++print(attempt("murcielago"))|code-block|syntax|javascript|4736615|PS:而不是使用“?.!/;:”示例使用元音...是的，"murcielago“是西班牙语里的蝙蝠...有趣的单词，因为它包含所有元音:)|4736616|PS2:如果你对性能感兴趣，你可以用一个简单的代码来衡量这些尝试，比如：|4736617|import+timeit


K+=+1000000
for+i+in+range(1,5):
++++t+=+timeit.Timer(
++++++++f"attempt{i}('murcielago')",
++++++++setup=f"from+__main__+import+attempt{i}"
++++).repeat(1,+K)
++++print(f"attempt{i}",min(t))|4736618|在我的邮箱里，你会看到：|4736619|attempt1+2.2334518376057244
attempt2+1.8806643818474513
attempt3+7.214925774955572
attempt4+1.7271184513757465|4736620|因此，对于这种特定的输入，attempt4似乎是最快的。|4736621|entityMap^0|0|0|0|0|0|0|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]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

Here's some possible ways to achieve this task:

<pre><code>def attempt1(string):
 return "".join([v for v in string if v not in ("a", "e", "i", "o", "u")])


def attempt2(string):
 for v in ("a", "e", "i", "o", "u"):
 string = string.replace(v, "")
 return string


def attempt3(string):
 import re
 for v in ("a", "e", "i", "o", "u"):
 string = re.sub(v, "", string)
 return string


def attempt4(string):
 return string.replace("a", "").replace("e", "").replace("i", "").replace("o", "").replace("u", "")


for attempt in [attempt1, attempt2, attempt3, attempt4]:
 print(attempt("murcielago"))
</code></pre>

PS: Instead using " ?.!/;:" the examples use the vowels... and yeah, "murcielago" is the Spanish word to say bat... funny word as it contains all the vowels :)

PS2: If you're interested on performance you could measure these attempts with a simple code like:

<pre><code>import timeit


K = 1000000
for i in range(1,5):
 t = timeit.Timer(
 f"attempt{i}('murcielago')",
 setup=f"from __main__ import attempt{i}"
 ).repeat(1, K)
 print(f"attempt{i}",min(t))
</code></pre>

In my box you'd get:

<pre><code>attempt1 2.2334518376057244
attempt2 1.8806643818474513
attempt3 7.214925774955572
attempt4 1.7271184513757465
</code></pre>

So it seems attempt4 is the fastest one for this particular input.

blocks|key|4736650|text|您可以使用re模块的正则表达式替换。使用%5E表达式可以让您准确地从字符串中选取所需内容。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4736651|++++import+re
++++text+=+"This+is+absurd!"
++++text+=+re.sub("[%5Ea-zA-Z]","",text)+#+Keeps+only+Alphabets
++++print(text)|code-block|syntax|javascript|4736652|它的输出将是"Thisisabsurd“。只会显示%5E符号后指定的内容。|4736653|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 could use the re module's regular expression replacement. Using the ^ expression allows you to pick exactly what you want from your string.

<pre><code> import re
 text = "This is absurd!"
 text = re.sub("[^a-zA-Z]","",text) # Keeps only Alphabets
 print(text)
</code></pre>

Output to this would be "Thisisabsurd". Only things specified after the ^ symbol will appear.

blocks|key|3326257|text|filename#对于目录中的每个文件，重命名|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3326258|+++file_list+=+os.listdir+(r"D:\Dev\Python")

+++for+file_name+in+file_list:

+++++++os.rename(file_name,+re.sub(r'\d%2B','',file_name))|code-block|syntax|javascript|3326259|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|$]]

# for each file on a directory, rename filename

<pre><code> file_list = os.listdir (r"D:\Dev\Python")

 for file_name in file_list:

 os.rename(file_name, re.sub(r'\d+','',file_name))
</code></pre>

blocks|key|1929583|text|string方法replace不修改原始字符串。它保留原始文件不变，并返回修改后的副本。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1929584|您需要的内容类似于：line+=+line.replace(char,'')|1929585|def+replace_all(line,+)for+char+in+line:
++++if+char+in+"+?.!/;:":
++++++++line+=+line.replace(char,'')
++++return+line|code-block|syntax|javascript|1929586|但是，每次删除字符时都会创建一个新字符串，效率非常低。我建议使用以下方法：|1929587|def+replace_all(line,+baddies,+*):
++++"""
++++The+following+is+documentation+on+how+to+use+the+class,
++++without+reference+to+the+implementation+details:

++++For+implementation+notes,+please+see+comments+begining+with+`#`
++++in+the+source+file.

++++[*crickets+chirp*]

++++"""

++++is_bad+=+lambda+ch,+baddies=baddies:+return+ch+in+baddies
++++filter_baddies+=+lambda+ch,+*,+is_bad=is_bad:+""+if+is_bad(ch)+else+ch
++++mahp+=+replace_all.map(filter_baddies,+line)
++++return+replace_all.join('',+join(mahp))

++++#+-------------------------------------------------
++++#+WHY+`baddies=baddies`?!?
++++#+++++`is_bad=is_bad`
++++#+-------------------------------------------------
++++#+Default+arguments+to+a+lambda+function+are+evaluated
++++#+at+the+same+time+as+when+a+lambda+function+is
++++#+**defined**.
++++#
++++#+global+variables+of+a+lambda+function
++++#+are+evaluated+when+the+lambda+function+is
++++#+**called**
++++#
++++#+The+following+prints+"as+yellow+as+snow"
++++#
++++#+++++fleece_color+=+"white"
++++#+++++little_lamb+=+lambda+end:+return+"as+"+%2B+fleece_color+%2B+end
++++#
++++#+++++#+sometime+later...
++++#
++++#+++++fleece_color+=+"yellow"
++++#+++++print(little_lamb("+as+snow"))
++++#+--------------------------------------------------
replace_all.map+=+map
replace_all.join+=+str.join|1929588|entityMap^0|8|7|0|A|S|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|Z|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|R|$]]

The string method <code>replace</code> does not modify the original string. It leaves the original alone and returns a modified copy.

What you want is something like: <code>line = line.replace(char,'')</code>

<pre><code>def replace_all(line, )for char in line:
 if char in " ?.!/;:":
 line = line.replace(char,'')
 return line
</code></pre>

However, creating a new string each and every time that a character is removed is very inefficient. I recommend the following instead:

<pre><code>def replace_all(line, baddies, *):
 """
 The following is documentation on how to use the class,
 without reference to the implementation details:

 For implementation notes, please see comments begining with `#`
 in the source file.

 [*crickets chirp*]

 """

 is_bad = lambda ch, baddies=baddies: return ch in baddies
 filter_baddies = lambda ch, *, is_bad=is_bad: "" if is_bad(ch) else ch
 mahp = replace_all.map(filter_baddies, line)
 return replace_all.join('', join(mahp))

 # -------------------------------------------------
 # WHY `baddies=baddies`?!?
 # `is_bad=is_bad`
 # -------------------------------------------------
 # Default arguments to a lambda function are evaluated
 # at the same time as when a lambda function is
 # **defined**.
 #
 # global variables of a lambda function
 # are evaluated when the lambda function is
 # **called**
 #
 # The following prints "as yellow as snow"
 #
 # fleece_color = "white"
 # little_lamb = lambda end: return "as " + fleece_color + end
 #
 # # sometime later...
 #
 # fleece_color = "yellow"
 # print(little_lamb(" as snow"))
 # --------------------------------------------------
replace_all.map = map
replace_all.join = str.join
</code></pre>

blocks|key|1924752|text|如果您希望您的字符串只允许使用ASCII代码，您可以使用这段代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1924753|for+char+in+s:
++++if+ord(char)+<+96+or+ord(char)+>+123:
++++++++s+=+s.replace(char,+"")|code-block|syntax|javascript|1924754|它将删除a...z以外的所有字符，甚至大写字母。|1924755|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|$]]

If you want your string to be just allowed characters by using ASCII codes, you can use this piece of code:
<pre><code>for char in s:
 if ord(char) &lt; 96 or ord(char) &gt; 123:
 s = s.replace(char, &quot;&quot;)
</code></pre>
It will remove all the characters beyond a....z even upper cases.

I'm trying to remove specific characters from a string using Python. This is the code I'm using right now. Unfortunately it appears to do nothing to the string.

<pre><code>for char in line:
 if char in " ?.!/;:":
 line.replace(char,'')
</code></pre>

How do I do this properly?

Remove specific characters from 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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我正在尝试使用Python从字符串中删除特定字符。这就是我现在使用的代码。不幸的是，它似乎没有对字符串做任何事情。for char in line:    if char in " ?.!/;:":        line.replace(char,'')我该如何正确地做到这一点？

问在Python中移除字符串中的特定字符
EN

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