blocks|key|864968|text|我会使用regexp：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|864969|>>>+import+re
>>>+re.findall(r'\d%2B',+'hello+42+I\'m+a+32+string+30')
['42',+'32',+'30']|code-block|syntax|javascript|864970|这也将匹配来自bla42bla的42。如果您只想使用单词边界(空格、句点、逗号)分隔数字，则可以使用\b：|offset|length|style|CODE|864971|>>>+re.findall(r'\b\d%2B\b',+'he33llo+42+I\'m+a+32+string+30')
['42',+'32',+'30']|864972|要以数字列表而不是字符串列表结束：|864973|>>>+[int(s)+for+s+in+re.findall(r'\b\d%2B\b',+'he33llo+42+I\'m+a+32+string+30')]
[42,+32,+30]|864974|entityMap^0|0|0|7|8|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]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

I'd use a regexp :

<pre><code>&gt;&gt;&gt; import re
&gt;&gt;&gt; re.findall(r'\d+', 'hello 42 I\'m a 32 string 30')
['42', '32', '30']
</code></pre>

This would also match 42 from <code>bla42bla</code>. If you only want numbers delimited by word boundaries (space, period, comma), you can use \b :

<pre><code>&gt;&gt;&gt; re.findall(r'\b\d+\b', 'he33llo 42 I\'m a 32 string 30')
['42', '32', '30']
</code></pre>

To end up with a list of numbers instead of a list of strings:

<pre><code>&gt;&gt;&gt; [int(s) for s in re.findall(r'\b\d+\b', 'he33llo 42 I\'m a 32 string 30')]
[42, 32, 30]
</code></pre>

blocks|key|1079634|text|我假设你想要浮点数，而不仅仅是整数，所以我会这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1079635|l+=+[]
for+t+in+s.split():
++++try:
++++++++l.append(float(t))
++++except+ValueError:
++++++++pass|code-block|syntax|javascript|1079636|请注意，这里发布的一些其他解决方案不适用于负数：|1079637|>>>+re.findall(r'\b\d%2B\b',+'he33llo+42+I\'m+a+32+string+-30')
['42',+'32',+'30']

>>>+'-3'.isdigit()
False|1079638|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|$]]

I'm assuming you want floats not just integers so I'd do something like this:

<pre><code>l = []
for t in s.split():
 try:
 l.append(float(t))
 except ValueError:
 pass
</code></pre>

Note that some of the other solutions posted here don't work with negative numbers:

<pre><code>&gt;&gt;&gt; re.findall(r'\b\d+\b', 'he33llo 42 I\'m a 32 string -30')
['42', '32', '30']

&gt;&gt;&gt; '-3'.isdigit()
False
</code></pre>

blocks|key|1079681|text|如果只想提取正整数，请尝试以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1079682|>>>+txt+=+"h3110+23+cat+444.4+rabbit+11+2+dog"
>>>+[int(s)+for+s+in+txt.split()+if+s.isdigit()]
[23,+11,+2]|code-block|syntax|javascript|1079683|我认为这比正则表达式示例更好，因为您不需要另一个模块，而且它更具可读性，因为您不需要解析(和学习)+regex+mini-language。|offset|length|1079684|这将不识别浮点数、负整数或十六进制格式的整数。如果你不能接受这些限制，jmnas's+answer+below可以解决这个问题。|1079685|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/re.html#regular-expression-syntax|1|https://stackoverflow.com/questions/4289331/python-extract-numbers-of-a-string/4289415#4289415^0|0|0|1E|J|0|0|Z|K|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@$I|Z|J|10|1|11]]|A|$]]|$1|K|3|L|5|6|7|12|8|@]|9|@$I|13|J|14|1|15]]|A|$]]|$1|M|3|-4|5|6|7|16|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]|U|$5|P|Q|R|A|$S|V]]]]

If you only want to extract only positive integers, try the following:
<pre><code>&gt;&gt;&gt; txt = &quot;h3110 23 cat 444.4 rabbit 11 2 dog&quot;
&gt;&gt;&gt; [int(s) for s in txt.split() if s.isdigit()]
[23, 11, 2]
</code></pre>
I would argue that this is better than the regex example because you don't need another module and it's more readable because you don't need to parse (and learn) the <a href="https://docs.python.org/3/library/re.html#regular-expression-syntax" rel="noreferrer">regex mini-language</a>.
This will not recognize floats, negative integers, or integers in hexadecimal format. If you can't accept these limitations, <a href="https://stackoverflow.com/questions/4289331/python-extract-numbers-of-a-string/4289415#4289415">jmnas's answer below</a> will do the trick.

blocks|key|1079835|text|@jmnas，我喜欢你的回答，但它没有找到浮动。我正在编写一个脚本来解析送到CNC工厂的代码，并且需要找到可以是整数或浮点数的X和Y维，所以我将您的代码调整为以下内容。这将查找int，float，其中有正值和负值。仍然找不到十六进制格式值，但您可以将"x“和"A”到"F“添加到num_char元组中，我认为它可以解析类似”0x23AC“的内容。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1079836|s+=+'hello+X42+I\'m+a+Y-32.35+string+Z30'
xy+=+("X",+"Y")
num_char+=+(".",+"%2B",+"-")

l+=+[]

tokens+=+s.split()
for+token+in+tokens:

++++if+token.startswith(xy):
++++++++num+=+""
++++++++for+char+in+token:
++++++++++++#+print(char)
++++++++++++if+char.isdigit()+or+(char+in+num_char):
++++++++++++++++num+=+num+%2B+char

++++++++try:
++++++++++++l.append(float(num))
++++++++except+ValueError:
++++++++++++pass

print(l)|code-block|syntax|javascript|1079837|entityMap^0|3V|8|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

@jmnas, I liked your answer, but it didn't find floats. I'm working on a script to parse code going to a CNC mill and needed to find both X and Y dimensions that can be integers or floats, so I adapted your code to the following. This finds int, float with positive and negative vals. Still doesn't find hex formatted values but you could add "x" and "A" through "F" to the <code>num_char</code> tuple and I think it would parse things like '0x23AC'.

<pre><code>s = 'hello X42 I\'m a Y-32.35 string Z30'
xy = ("X", "Y")
num_char = (".", "+", "-")

l = []

tokens = s.split()
for token in tokens:

 if token.startswith(xy):
 num = ""
 for char in token:
 # print(char)
 if char.isdigit() or (char in num_char):
 num = num + char

 try:
 l.append(float(num))
 except ValueError:
 pass

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

blocks|key|1079875|text|这有点晚了，但是您也可以扩展正则表达式来说明科学记数法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1079876|import+re

#+Format+is+[(<string>,+<expected+output>),+...]
ss+=+[("apple-12.34+ba33na+fanc-14.23e-2yapple%2B45e5%2B67.56E%2B3",
+++++++['-12.34',+'33',+'-14.23e-2',+'%2B45e5',+'%2B67.56E%2B3']),
++++++('hello+X42+I\'m+a+Y-32.35+string+Z30',
+++++++['42',+'-32.35',+'30']),
++++++('he33llo+42+I\'m+a+32+string+-30',+
+++++++['33',+'42',+'32',+'-30']),
++++++('h3110+23+cat+444.4+rabbit+11+2+dog',+
+++++++['3110',+'23',+'444.4',+'11',+'2']),
++++++('hello+12+hi+89',+
+++++++['12',+'89']),
++++++('4',+
+++++++['4']),
++++++('I+like+74,600+commas+not,500',+
+++++++['74,600',+'500']),
++++++('I+like+bad+math+1%2B2=.001',+
+++++++['1',+'%2B2',+'.001'])]

for+s,+r+in+ss:
++++rr+=+re.findall("[-%2B]?[.]?[\d]%2B(?:,\d\d\d)*[\.]?\d*(?:[eE][-%2B]?\d%2B)?",+s)
++++if+rr+==+r:
++++++++print('GOOD')
++++else:
++++++++print('WRONG',+rr,+'should+be',+r)|code-block|syntax|javascript|1079877|给所有的好！|1079878|此外，您还可以查看AWS+Glue+built-in+regex|offset|length|1079879|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.aws.amazon.com/glue/latest/dg/custom-classifier.html#classifier-builtin-patterns?icmpid=docs_glue_console^0|0|0|0|9|N|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]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|X|8|@]|9|@$K|Y|L|Z|1|10]]|A|$]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

This is more than a bit late, but you can extend the regex expression to account for scientific notation too.

<pre><code>import re

# Format is [(&lt;string&gt;, &lt;expected output&gt;), ...]
ss = [("apple-12.34 ba33na fanc-14.23e-2yapple+45e5+67.56E+3",
 ['-12.34', '33', '-14.23e-2', '+45e5', '+67.56E+3']),
 ('hello X42 I\'m a Y-32.35 string Z30',
 ['42', '-32.35', '30']),
 ('he33llo 42 I\'m a 32 string -30', 
 ['33', '42', '32', '-30']),
 ('h3110 23 cat 444.4 rabbit 11 2 dog', 
 ['3110', '23', '444.4', '11', '2']),
 ('hello 12 hi 89', 
 ['12', '89']),
 ('4', 
 ['4']),
 ('I like 74,600 commas not,500', 
 ['74,600', '500']),
 ('I like bad math 1+2=.001', 
 ['1', '+2', '.001'])]

for s, r in ss:
 rr = re.findall("[-+]?[.]?[\d]+(?:,\d\d\d)*[\.]?\d*(?:[eE][-+]?\d+)?", s)
 if rr == r:
 print('GOOD')
 else:
 print('WRONG', rr, 'should be', r)
</code></pre>

Gives all good!

Additionally, you can look at the <a href="http://docs.aws.amazon.com/glue/latest/dg/custom-classifier.html#classifier-builtin-patterns?icmpid=docs_glue_console" rel="noreferrer">AWS Glue built-in regex</a>

blocks|key|1079944|text|下面是我找到的最好的选择。它将提取一个数字，并可以消除任何类型的字符。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1079945|def+extract_nbr(input_str):
++++if+input_str+is+None+or+input_str+==+'':
++++++++return+0

++++out_number+=+''
++++for+ele+in+input_str:
++++++++if+ele.isdigit():
++++++++++++out_number+%2B=+ele
++++return+float(out_number)++++|code-block|syntax|javascript|1079946|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 best option I found is below. It will extract a number and can eliminate any type of char.

<pre><code>def extract_nbr(input_str):
 if input_str is None or input_str == '':
 return 0

 out_number = ''
 for ele in input_str:
 if ele.isdigit():
 out_number += ele
 return float(out_number) 
</code></pre>

blocks|key|1080032|text|此答案还包含数字在字符串中为浮点型时的情况|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1080033|def+get_first_nbr_from_str(input_str):
++++'''
++++:param+input_str:+strings+that+contains+digit+and+words
++++:return:+the+number+extracted+from+the+input_str
++++demo:
++++'ab324.23.123xyz':+324.23
++++'.5abc44':+0.5
++++'''
++++if+not+input_str+and+not+isinstance(input_str,+str):
++++++++return+0
++++out_number+=+''
++++for+ele+in+input_str:
++++++++if+(ele+==+'.'+and+'.'+not+in+out_number)+or+ele.isdigit():
++++++++++++out_number+%2B=+ele
++++++++elif+out_number:
++++++++++++break
++++return+float(out_number)|code-block|syntax|javascript|1080034|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 answer also contains the case when the number is float in the string

<pre><code>def get_first_nbr_from_str(input_str):
 '''
 :param input_str: strings that contains digit and words
 :return: the number extracted from the input_str
 demo:
 'ab324.23.123xyz': 324.23
 '.5abc44': 0.5
 '''
 if not input_str and not isinstance(input_str, str):
 return 0
 out_number = ''
 for ele in input_str:
 if (ele == '.' and '.' not in out_number) or ele.isdigit():
 out_number += ele
 elif out_number:
 break
 return float(out_number)
</code></pre>

blocks|key|1080097|text|如果你知道字符串中只有一个数字，也就是'hello+12+hi'，你可以试试filter。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1080098|例如：|1080099|In+[1]:+int(''.join(filter(str.isdigit,+'200+grams')))
Out[1]:+200
In+[2]:+int(''.join(filter(str.isdigit,+'Counters:+55')))
Out[2]:+55
In+[3]:+int(''.join(filter(str.isdigit,+'more+than+23+times')))
Out[3]:+23|code-block|syntax|javascript|1080100|但要小心！：|1080101|In+[4]:+int(''.join(filter(str.isdigit,+'200+grams+5')))
Out[4]:+2005|1080102|entityMap^0|J|D|12|6|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|Y|8|@]|D|@]|E|$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|$]]

If you know it will be only one number in the string, i.e <code>'hello 12 hi'</code>, you can try <code>filter</code>.
For example:
<pre><code>In [1]: int(''.join(filter(str.isdigit, '200 grams')))
Out[1]: 200
In [2]: int(''.join(filter(str.isdigit, 'Counters: 55')))
Out[2]: 55
In [3]: int(''.join(filter(str.isdigit, 'more than 23 times')))
Out[3]: 23
</code></pre>
But be carefull !!! :
<pre><code>In [4]: int(''.join(filter(str.isdigit, '200 grams 5')))
Out[4]: 2005
</code></pre>

blocks|key|1080194|text|由于这些文档都不涉及我需要查找的excel和word文档中的真实金融数据，因此我的变体如下。它处理整数、浮点数、负数、货币数(因为它不会在拆分时回复)，并且可以选择去掉小数部分，只返回整数，或者返回所有内容。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1080195|它还处理印度Laks数字系统，其中逗号出现不规则，而不是每隔3个数字。|1080196|它不处理科学记数法或在预算中放入括号内的负数--将显示为正数。|1080197|它也不提取日期。在字符串中查找日期有更好的方法。|1080198|import+re
def+find_numbers(string,+ints=True):++++++++++++
++++numexp+=+re.compile(r'[-]?\d[\d,]*[\.]?[\d{2}]*')+#optional+-+in+front
++++numbers+=+numexp.findall(string)++++
++++numbers+=+[x.replace(',','')+for+x+in+numbers]
++++if+ints+is+True:
++++++++return+[int(x.replace(',','').split('.')[0])+for+x+in+numbers]++++++++++++
++++else:
++++++++return+numbers|code-block|syntax|javascript|1080199|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Q|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|R|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|S|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Since none of these dealt with real world financial numbers in excel and word docs that I needed to find, here is my variation. It handles ints, floats, negative numbers, currency numbers (because it doesn't reply on split), and has the option to drop the decimal part and just return ints, or return everything. 

It also handles Indian Laks number system where commas appear irregularly, not every 3 numbers apart.

It does not handle scientific notation or negative numbers put inside parentheses in budgets -- will appear positive. 

It also does not extract dates. There are better ways for finding dates in strings.

<pre><code>import re
def find_numbers(string, ints=True): 
 numexp = re.compile(r'[-]?\d[\d,]*[\.]?[\d{2}]*') #optional - in front
 numbers = numexp.findall(string) 
 numbers = [x.replace(',','') for x in numbers]
 if ints is True:
 return [int(x.replace(',','').split('.')[0]) for x in numbers] 
 else:
 return numbers
</code></pre>

blocks|key|865616|text|#+extract+numbers+from+garbage+string:
s+=+'12//n,_@#$%253.14kjlw0xdadfackvj1.6e-19&*ghn334'
newstr+=+''.join((ch+if+ch+in+'0123456789.-e'+else+'+')+for+ch+in+s)
listOfNumbers+=+[float(i)+for+i+in+newstr.split()]
print(listOfNumbers)
[12.0,+3.14,+0.0,+1.6e-19,+334.0]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|865617|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># extract numbers from garbage string:
s = '12//n,_@#$%3.14kjlw0xdadfackvj1.6e-19&amp;*ghn334'
newstr = ''.join((ch if ch in '0123456789.-e' else ' ') for ch in s)
listOfNumbers = [float(i) for i in newstr.split()]
print(listOfNumbers)
[12.0, 3.14, 0.0, 1.6e-19, 334.0]
</code></pre>

blocks|key|1080273|text|我正在寻找一个解决方案来删除字符串的面具，特别是从巴西的电话号码，这篇文章没有回应，但启发了我。这是我的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1080274|>>>+phone_number+=+'%2B55(11)8715-9877'
>>>+''.join([n+for+n+in+phone_number+if+n.isdigit()])
'551187159877'|code-block|syntax|javascript|1080275|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|$]]

I was looking for a solution to remove strings' masks, specifically from Brazilian phones numbers, this post not answered but inspired me. This is my solution:

<pre><code>&gt;&gt;&gt; phone_number = '+55(11)8715-9877'
&gt;&gt;&gt; ''.join([n for n in phone_number if n.isdigit()])
'551187159877'
</code></pre>

blocks|key|865738|text|使用下面的正则表达式就是方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|865739|lines+=+"hello+12+hi+89"
import+re
output+=+[]
#repl_str+=+re.compile('\d%2B.?\d*')
repl_str+=+re.compile('%5E\d%2B$')
#t+=+r'\d%2B.?\d*'
line+=+lines.split()
for+word+in+line:
++++++++match+=+re.search(repl_str,+word)
++++++++if+match:
++++++++++++output.append(float(match.group()))
print+(output)|code-block|syntax|javascript|865740|使用findall+re.findall(r'\d%2B',+"hello+12+hi+89")|offset|length|style|CODE|865741|['12',+'89']|865742|re.findall(r'\b\d%2B\b',+"hello+12+hi+89+33F+AC+777")|865743|['12',+'89',+'777']|865744|entityMap^0|0|0|A|10|0|0|0|1F|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]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|10|8|@$I|11|J|12|K|L]]|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|$]]

Using Regex below is the way
<pre><code>lines = &quot;hello 12 hi 89&quot;
import re
output = []
#repl_str = re.compile('\d+.?\d*')
repl_str = re.compile('^\d+$')
#t = r'\d+.?\d*'
line = lines.split()
for word in line:
 match = re.search(repl_str, word)
 if match:
 output.append(float(match.group()))
print (output)
</code></pre>
with findall
<code>re.findall(r'\d+', &quot;hello 12 hi 89&quot;)</code>
<pre><code>['12', '89']
</code></pre>
<code>re.findall(r'\b\d+\b', &quot;hello 12 hi 89 33F AC 777&quot;)</code>
<pre><code>['12', '89', '777']
</code></pre>

blocks|key|865790|text|我之所以添加这个答案，是因为没有人添加使用异常处理的答案，而且这也适用于浮点数|type|unstyled|depth|inlineStyleRanges|entityRanges|data|865791|a+=+[]
line+=+"abcd+1234+efgh+56.78+ij"
for+word+in+line.split():
++++try:
++++++++a.append(float(word))
++++except+ValueError:
++++++++pass
print(a)|code-block|syntax|javascript|865792|输出：|865793|[1234.0,+56.78]|865794|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|$]]

I am just adding this answer because no one added one using Exception handling and because this also works for floats

<pre><code>a = []
line = "abcd 1234 efgh 56.78 ij"
for word in line.split():
 try:
 a.append(float(word))
 except ValueError:
 pass
print(a)
</code></pre>

Output :

<pre><code>[1234.0, 56.78]
</code></pre>

blocks|key|865849|text|line2+=+"hello+12+hi+89"++#+this+is+the+given+string+
temp1+=+re.findall(r'\d%2B',+line2)+#+find+number+of+digits+through+regular+expression
res2+=+list(map(int,+temp1))
print(res2)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|865850|嗨，|unstyled|865851|您可以使用findall表达式通过数字查找字符串中的所有整数。|865852|在第二步中，创建一个列表res2，并将字符串中的数字添加到该列表中|865853|希望这能有所帮助|865854|致敬，Diwakar+Sharma|865855|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|R|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|F|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|F|7|W|8|@]|9|@]|A|$]]]|P|$]]

<pre><code>line2 = &quot;hello 12 hi 89&quot; # this is the given string 
temp1 = re.findall(r'\d+', line2) # find number of digits through regular expression
res2 = list(map(int, temp1))
print(res2)
</code></pre>
Hi ,
you can search all the integers in the string through digit by using findall expression .
In the second step create a list res2 and add the digits found in string to this list
hope this helps
Regards,
Diwakar Sharma

blocks|key|1080538|text|我发现的最干净的方法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1080539|>>>+data+=+'hs122+125+&55,58,+25'
>>>+new_data+=+''.join((ch+if+ch+in+'0123456789.-e'+else+'+')+for+ch+in+data)
>>>+numbers+=+[i+for+i+in+new_data.split()]
>>>+print(numbers)
['122',+'125',+'55',+'58',+'25']|code-block|syntax|javascript|1080540|或者这样：|1080541|>>>+import+re
>>>+data+=+'hs122+125+&55,58,+25'
>>>+numbers+=+re.findall(r'\d%2B',+data)
>>>+print(numbers)
['122',+'125',+'55',+'58',+'25']|1080542|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|$]]

The cleanest way i found:
<pre><code>&gt;&gt;&gt; data = 'hs122 125 &amp;55,58, 25'
&gt;&gt;&gt; new_data = ''.join((ch if ch in '0123456789.-e' else ' ') for ch in data)
&gt;&gt;&gt; numbers = [i for i in new_data.split()]
&gt;&gt;&gt; print(numbers)
['122', '125', '55', '58', '25']
</code></pre>
or this:
<pre><code>&gt;&gt;&gt; import re
&gt;&gt;&gt; data = 'hs122 125 &amp;55,58, 25'
&gt;&gt;&gt; numbers = re.findall(r'\d+', data)
&gt;&gt;&gt; print(numbers)
['122', '125', '55', '58', '25']
</code></pre>

I would like to extract all the numbers contained in a string. Which is better suited for the purpose, regular expressions or the <code>isdigit()</code> method?
Example:
<pre><code>line = &quot;hello 12 hi 89&quot;
</code></pre>
Result:
<pre><code>[12, 89]
</code></pre>

How to extract numbers from a string in Python?

我想提取字符串中包含的所有数字。正则表达式和isdigit()方法，哪个更适合这个目的？示例：line = "hello 12 hi 89"结果：[12, 89]

问如何在Python中从字符串中提取数字？
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