blocks|key|1482949|text|不确定这是否是最有效的方法，但是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1482950|>>>+''.join(c+for+c+in+"abc123def456"+if+c.isdigit())
'123456'|code-block|syntax|javascript|1482951|''.join部分意味着将所有结果字符组合在一起，中间没有任何字符。然后它的其余部分是列表理解，其中(正如您可能猜到的)我们只获取字符串中与条件isdigit匹配的部分。|offset|length|style|CODE|1482952|entityMap^0|0|0|0|7|20|7|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]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

Not sure if this is the most efficient way, but:

<pre><code>&gt;&gt;&gt; ''.join(c for c in "abc123def456" if c.isdigit())
'123456'
</code></pre>

The <code>''.join</code> part means to combine all the resulting characters together without any characters in between. Then the rest of it is a list comprehension, where (as you can probably guess) we only take the parts of the string that match the condition <code>isdigit</code>.

blocks|key|1482956|text|>>>+import+re
>>>+re.sub("[%5E0-9]",+"",+"sdkjh987978asd098as0980a98sd")
'987978098098098'|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1482957|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>&gt;&gt;&gt; import re
&gt;&gt;&gt; re.sub("[^0-9]", "", "sdkjh987978asd098as0980a98sd")
'987978098098098'
</code></pre>

blocks|key|1268489|text|最快的方法是，如果你需要执行超过一两个这样的删除操作(或者甚至只是一个，但是对一个非常长的字符串！-)，则依赖于字符串的translate方法，尽管它确实需要一些准备：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1268490|>>>+import+string
>>>+allchars+=+''.join(chr(i)+for+i+in+xrange(256))
>>>+identity+=+string.maketrans('',+'')
>>>+nondigits+=+allchars.translate(identity,+string.digits)
>>>+s+=+'abc123def456'
>>>+s.translate(identity,+nondigits)
'123456'|code-block|syntax|javascript|1268491|在Unicode字符串上的translate方法与在字节字符串上的方法不同，而且使用起来可能更简单一些。|1268492|>>>+unondig+=+dict.fromkeys(xrange(65536))
>>>+for+x+in+string.digits:+del+unondig[ord(x)]
...+
>>>+s+=+u'abc123def456'
>>>+s.translate(unondig)
u'123456'|1268493|您可能希望使用映射类而不是实际的dict，尤其是当Unicode字符串可能包含具有非常高的ord值的字符时(这会使dict过大；-)。例如：|1268494|>>>+class+keeponly(object):
...+++def+__init__(self,+keep):+
...+++++self.keep+=+set(ord(c)+for+c+in+keep)
...+++def+__getitem__(self,+key):
...+++++if+key+in+self.keep:
...+++++++return+key
...+++++return+None
...+
>>>+s.translate(keeponly(string.digits))
u'123456'
>>>+|1268495|entityMap^0|1O|9|0|0|D|9|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|12|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|T|$]]

Fastest approach, if you need to perform more than just one or two such removal operations (or even just one, but on a very long string!-), is to rely on the <code>translate</code> method of strings, even though it does need some prep:

<pre><code>&gt;&gt;&gt; import string
&gt;&gt;&gt; allchars = ''.join(chr(i) for i in xrange(256))
&gt;&gt;&gt; identity = string.maketrans('', '')
&gt;&gt;&gt; nondigits = allchars.translate(identity, string.digits)
&gt;&gt;&gt; s = 'abc123def456'
&gt;&gt;&gt; s.translate(identity, nondigits)
'123456'
</code></pre>

The <code>translate</code> method is different, and maybe a tad simpler simpler to use, on Unicode strings than it is on byte strings, btw:

<pre><code>&gt;&gt;&gt; unondig = dict.fromkeys(xrange(65536))
&gt;&gt;&gt; for x in string.digits: del unondig[ord(x)]
... 
&gt;&gt;&gt; s = u'abc123def456'
&gt;&gt;&gt; s.translate(unondig)
u'123456'
</code></pre>

You might want to use a mapping class rather than an actual dict, especially if your Unicode string may potentially contain characters with very high ord values (that would make the dict excessively large;-). For example:

<pre><code>&gt;&gt;&gt; class keeponly(object):
... def __init__(self, keep): 
... self.keep = set(ord(c) for c in keep)
... def __getitem__(self, key):
... if key in self.keep:
... return key
... return None
... 
&gt;&gt;&gt; s.translate(keeponly(string.digits))
u'123456'
&gt;&gt;&gt; 
</code></pre>

blocks|key|1483057|text|这应该适用于Python2中的字符串和unicode对象，以及Python3中的字符串和字节：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1483058|#+python+<3.0
def+only_numerics(seq):
++++return+filter(type(seq).isdigit,+seq)

#+python+≥3.0
def+only_numerics(seq):
++++seq_type=+type(seq)
++++return+seq_type().join(filter(seq_type.isdigit,+seq))|code-block|syntax|javascript|1483059|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|$]]

This should work for both strings and unicode objects in Python2, and both strings and bytes in Python3:
<pre><code># python &lt;3.0
def only_numerics(seq):
 return filter(type(seq).isdigit, seq)

# python ≥3.0
def only_numerics(seq):
 seq_type= type(seq)
 return seq_type().join(filter(seq_type.isdigit, seq))
</code></pre>

blocks|key|1268593|text|@Ned+Batchelder和@newacct提供了正确的答案，但是...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1268594|以防万一您有逗号(，)小数(.)在您的字符串中：|1268595|import+re
re.sub("[%5E\d\.]",+"",+"$1,999,888.77")
'1999888.77'|code-block|syntax|javascript|1268596|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

@Ned Batchelder and @newacct provided the right answer, but ...

Just in case if you have comma(,) decimal(.) in your string:

<pre><code>import re
re.sub("[^\d\.]", "", "$1,999,888.77")
'1999888.77'
</code></pre>

blocks|key|1268649|text|许多正确的答案，但如果你想要它的浮点数，直接，而不使用正则表达式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1268650|x=+'$123.45M'

float(''.join(c+for+c+in+x+if+(c.isdigit()+or+c+=='.'))|code-block|syntax|javascript|1268651|123.45|1268652|您可以根据需要更改逗号的位置。|1268653|如果您知道您的数字是一个整数，请为此进行更改|1268654|x='$1123'++++
int(''.join(c+for+c+in+x+if+c.isdigit())|1268655|1123|1268656|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Many right answers but in case you want it in a float, directly, without using regex:

<pre><code>x= '$123.45M'

float(''.join(c for c in x if (c.isdigit() or c =='.'))
</code></pre>

123.45

You can change the point for a comma depending on your needs.

change for this if you know your number is an integer

<pre><code>x='$1123' 
int(''.join(c for c in x if c.isdigit())
</code></pre>

1123

How do we remove all non-numeric characters from a string in Python?

Removing all non-numeric characters from string in Python

Python

如何在Python中删除字符串中的所有非数字字符？

问在Python中删除字符串中的所有非数字字符
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问在Python中删除字符串中的所有非数字字符EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问在Python中删除字符串中的所有非数字字符
EN