blocks|key|1449162|text|使用re.sub，如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1449163|>>>+import+re
>>>+re.sub('\D',+'',+'aas30dsa20')
'3020'|code-block|syntax|javascript|1449164|\D匹配任何非数字字符，因此，上面的代码实质上是用空字符串替换每个非数字字符。|1449165|或者您可以使用filter，如下所示(在Python2中)：|1449166|>>>+filter(str.isdigit,+'aas30dsa20')
'3020'|1449167|由于在Python3中，filter返回的是迭代器而不是list，因此您可以使用以下代码：|1449168|>>>+''.join(filter(str.isdigit,+'aas30dsa20'))
'3020'|1449169|entityMap^0|2|6|0|0|0|2|0|7|6|0|0|C|6|S|4|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@]|E|$]]|$1|O|3|P|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|17|8|@$9|18|A|19|B|C]|$9|1A|A|1B|B|C]]|D|@]|E|$]]|$1|S|3|T|5|H|7|1C|8|@]|D|@]|E|$I|J]]|$1|U|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]]|V|$]]

Use <code>re.sub</code>, like so:

<pre class="lang-py prettyprint-override"><code>&gt;&gt;&gt; import re
&gt;&gt;&gt; re.sub('\D', '', 'aas30dsa20')
'3020'
</code></pre>

<code>\D</code> matches any non-digit character so, the code above, is essentially replacing every non-digit character for the empty string.

Or you can use <code>filter</code>, like so (in Python 2):

<pre class="lang-py prettyprint-override"><code>&gt;&gt;&gt; filter(str.isdigit, 'aas30dsa20')
'3020'
</code></pre>

Since in Python 3, <code>filter</code> returns an iterator instead of a <code>list</code>, you can use the following instead:

<pre class="lang-py prettyprint-override"><code>&gt;&gt;&gt; ''.join(filter(str.isdigit, 'aas30dsa20'))
'3020'
</code></pre>

blocks|key|1234527|text|使用生成器表达式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1234528|>>>+s+=+"foo200bar"
>>>+new_s+=+"".join(i+for+i+in+s+if+i+in+"0123456789")|code-block|syntax|javascript|1234529|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|$]]

Use a generator expression:

<pre><code>&gt;&gt;&gt; s = "foo200bar"
&gt;&gt;&gt; new_s = "".join(i for i in s if i in "0123456789")
</code></pre>

blocks|key|1234583|text|丑陋但有效：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1234584|>>>+s
'aaa12333bb445bb54b5b52'
>>>+a+=+''.join(filter(lambda+x+:+x.isdigit(),+s))
>>>+a
'1233344554552'
>>>|code-block|syntax|javascript|1234585|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|$]]

Ugly but works:

<pre><code>&gt;&gt;&gt; s
'aaa12333bb445bb54b5b52'
&gt;&gt;&gt; a = ''.join(filter(lambda x : x.isdigit(), s))
&gt;&gt;&gt; a
'1233344554552'
&gt;&gt;&gt;
</code></pre>

blocks|key|1234609|text|按照拜耳的回答：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1234610|''.join(i+for+i+in+s+if+i.isdigit())|code-block|syntax|javascript|1234611|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|$]]

along the lines of bayer's answer:

<pre><code>''.join(i for i in s if i.isdigit())
</code></pre>

blocks|key|1449393|text|您可以使用filter：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1449394|filter(lambda+x:+x.isdigit(),+"dasdasd2313dsa")|code-block|syntax|javascript|1449395|在python3.0上，你必须加入这个(有点难看:(+))|1449396|''.join(filter(lambda+x:+x.isdigit(),+"dasdasd2313dsa"))|1449397|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|$]]

You can use filter:

<pre><code>filter(lambda x: x.isdigit(), "dasdasd2313dsa")
</code></pre>

On python3.0 you have to join this (kinda ugly :( )

<pre><code>''.join(filter(lambda x: x.isdigit(), "dasdasd2313dsa"))
</code></pre>

blocks|key|1449427|text|s=''.join(i+for+i+in+s+if+i.isdigit())|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1449428|另一个生成器变体。|unstyled|1449429|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>s=''.join(i for i in s if i.isdigit())
</code></pre>

Another generator variant.

blocks|key|1449461|text|操作员在评论中提到，他希望保留小数位。这可以使用re.sub方法(根据第二个和IMHO最佳答案)通过显式列出要保留的字符来完成。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1449462|>>>+re.sub("[%5E0123456789\.]","","poo123.4and5fish")
'123.45'|code-block|syntax|javascript|1449463|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|$]]

The op mentions in the comments that he wants to keep the decimal place. This can be done with the re.sub method (as per the second and IMHO best answer) by explicitly listing the characters to keep e.g.

<pre><code>&gt;&gt;&gt; re.sub("[^0123456789\.]","","poo123.4and5fish")
'123.45'
</code></pre>

blocks|key|1449518|text|x.translate(None,+string.digits)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1449519|将删除字符串中的所有数字。要删除字母并保留数字，请执行以下操作：|unstyled|1449520|x.translate(None,+string.letters)|1449521|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>x.translate(None, string.digits)
</code></pre>

will delete all digits from string. To delete letters and keep the digits, do this:

<pre><code>x.translate(None, string.letters)
</code></pre>

blocks|key|1449610|text|Python+3的快速版本：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1449611|#+xx3.py
from+collections+import+defaultdict
import+string
_NoneType+=+type(None)

def+keeper(keep):
++++table+=+defaultdict(_NoneType)
++++table.update({ord(c):+c+for+c+in+keep})
++++return+table

digit_keeper+=+keeper(string.digits)|code-block|syntax|javascript|1449612|下面是与正则表达式的性能比较：|1449613|$+python3.3+-mtimeit+-s'import+xx3;+x="aaa12333bb445bb54b5b52"'+'x.translate(xx3.digit_keeper)'
1000000+loops,+best+of+3:+1.02+usec+per+loop
$+python3.3+-mtimeit+-s'import+re;+r+=+re.compile(r"\D");+x="aaa12333bb445bb54b5b52"'+'r.sub("",+x)'
100000+loops,+best+of+3:+3.43+usec+per+loop|1449614|所以对我来说，它比regex快3倍多一点。它也比上面的Python语言更快，因为defaultdict在C中完成所有的查找，而不是(慢的)+class+Del。这里是我相同系统上的版本，以供比较。|offset|length|style|CODE|1449615|$+python3.3+-mtimeit+-s'import+xx;+x="aaa12333bb445bb54b5b52"'+'x.translate(xx.DD)'
100000+loops,+best+of+3:+13.6+usec+per+loop|1449616|entityMap^0|0|0|0|0|14|B|1Y|9|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]|$M|11|N|12|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|T|$]]

A fast version for Python 3:

<pre><code># xx3.py
from collections import defaultdict
import string
_NoneType = type(None)

def keeper(keep):
 table = defaultdict(_NoneType)
 table.update({ord(c): c for c in keep})
 return table

digit_keeper = keeper(string.digits)
</code></pre>

Here's a performance comparison vs. regex:

<pre><code>$ python3.3 -mtimeit -s'import xx3; x="aaa12333bb445bb54b5b52"' 'x.translate(xx3.digit_keeper)'
1000000 loops, best of 3: 1.02 usec per loop
$ python3.3 -mtimeit -s'import re; r = re.compile(r"\D"); x="aaa12333bb445bb54b5b52"' 'r.sub("", x)'
100000 loops, best of 3: 3.43 usec per loop
</code></pre>

So it's a little bit more than 3 times faster than regex, for me. It's also faster than <code>class Del</code> above, because <code>defaultdict</code> does all its lookups in C, rather than (slow) Python. Here's that version on my same system, for comparison.

<pre><code>$ python3.3 -mtimeit -s'import xx; x="aaa12333bb445bb54b5b52"' 'x.translate(xx.DD)'
100000 loops, best of 3: 13.6 usec per loop
</code></pre>

blocks|key|1449649|text|您可以使用正则表达式轻松完成此操作|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1449650|>>>+import+re
>>>+re.sub("\D","","£70,000")
70000|code-block|syntax|javascript|1449651|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 can easily do it using Regex

<pre><code>&gt;&gt;&gt; import re
&gt;&gt;&gt; re.sub("\D","","£70,000")
70000
</code></pre>

blocks|key|1449678|text|不是单行代码，而是非常简单：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1449679|buffer+=+""
some_str+=+"aas30dsa20"

for+char+in+some_str:
++++if+not+char.isdigit():
++++++++buffer+%2B=+char

print(+buffer+)|code-block|syntax|javascript|1449680|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|$]]

Not a one liner but very simple:

<pre><code>buffer = ""
some_str = "aas30dsa20"

for char in some_str:
 if not char.isdigit():
 buffer += char

print( buffer )
</code></pre>

blocks|key|1235164|text|你可以读懂每个字符。如果是数字，则将其包括在答案中。str.isdigit()+method是一种知道字符是否为数字的方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1235165|your_input+=+'12kjkh2nnk34l34'
your_output+=+''.join(c+for+c+in+your_input+if+c.isdigit())
print(your_output)+#+'1223434'|code-block|syntax|javascript|1235166|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/stdtypes.html?highlight=isdigit#str.isdigit^0|Q|D|14|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]]]]

You can read each character. If it is digit, then include it in the answer. The <code>str.isdigit()</code> <a href="https://docs.python.org/3/library/stdtypes.html?highlight=isdigit#str.isdigit" rel="nofollow noreferrer">method</a> is a way to know if a character is digit.

<pre class="lang-py prettyprint-override"><code>your_input = '12kjkh2nnk34l34'
your_output = ''.join(c for c in your_input if c.isdigit())
print(your_output) # '1223434'
</code></pre>

blocks|key|1235247|text|我用了这个。'letters'应该包含您想要删除的所有字母：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1235248|Output+=+Input.translate({ord(i):+None+for+i+in+'letters'}))|1235249|示例：|1235250|Input+=+"I+would+like+20+dollars+for+that+suit"+Output+=+Input.translate({ord(i):+None+for+i+in+'abcdefghijklmnopqrstuvwxzy'}))+print(Output)|1235251|输出：20|1235252|entityMap^0|6|9|0|0|1O|0|0|0|3X|0|3|2|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|V|8|@]|D|@]|E|$]]|$1|J|3|K|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|L|3|M|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|N|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|O|$]]

I used this. <code>'letters'</code> should contain all the letters that you want to get rid of:

<code>Output = Input.translate({ord(i): None for i in 'letters'}))</code>

Example:

<code>Input = "I would like 20 dollars for that suit"
Output = Input.translate({ord(i): None for i in 'abcdefghijklmnopqrstuvwxzy'}))
print(Output)</code>

Output:
<code>20</code>

blocks|key|1450094|text|尝试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1450095|import+re

string+=+'1abcd2XYZ3'
string_without_letters+=+re.sub(r'[a-z]',+'',+string.lower())|code-block|syntax|javascript|1450096|这应该会给出：|1450097|123|1450098|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|$]]

Try:
<pre><code>import re

string = '1abcd2XYZ3'
string_without_letters = re.sub(r'[a-z]', '', string.lower())
</code></pre>
this should give:
<pre><code>123
</code></pre>

How can I remove all characters except numbers from string?

Remove characters except digits from string using Python?

Python

如何从字符串中删除除数字以外的所有字符？

问是否使用Python从字符串中删除除数字以外的字符？
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