blocks|key|4706499|text|您可以使用带有str.isdigit函数的any函数，如下所示|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4706500|>>>+def+has_numbers(inputString):
...+++++return+any(char.isdigit()+for+char+in+inputString)
...+
>>>+has_numbers("I+own+1+dog")
True
>>>+has_numbers("I+own+no+dog")
False|code-block|syntax|javascript|4706501|或者，您可以使用正则表达式，如下所示|4706502|>>>+import+re
>>>+def+has_numbers(inputString):
...+++++return+bool(re.search(r'\d',+inputString))
...+
>>>+has_numbers("I+own+1+dog")
True
>>>+has_numbers("I+own+no+dog")
False|4706503|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/stdtypes.html#str.isdigit|1|https://docs.python.org/3/library/functions.html#any^0|7|B|L|3|7|B|0|L|3|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]|$9|16|A|17|1|18]]|E|$]]|$1|F|3|G|5|H|7|19|8|@]|D|@]|E|$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]]|$1|O|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]|W|$5|R|S|T|E|$U|X]]]]

You can use <a href="https://docs.python.org/3/library/functions.html#any" rel="noreferrer"><code>any</code></a> function, with the <a href="https://docs.python.org/3/library/stdtypes.html#str.isdigit" rel="noreferrer"><code>str.isdigit</code></a> function, like this
<pre><code>&gt;&gt;&gt; def has_numbers(inputString):
... return any(char.isdigit() for char in inputString)
... 
&gt;&gt;&gt; has_numbers(&quot;I own 1 dog&quot;)
True
&gt;&gt;&gt; has_numbers(&quot;I own no dog&quot;)
False
</code></pre>
Alternatively you can use a Regular Expression, like this
<pre><code>&gt;&gt;&gt; import re
&gt;&gt;&gt; def has_numbers(inputString):
... return bool(re.search(r'\d', inputString))
... 
&gt;&gt;&gt; has_numbers(&quot;I own 1 dog&quot;)
True
&gt;&gt;&gt; has_numbers(&quot;I own no dog&quot;)
False
</code></pre>

blocks|key|200752|text|您可以组合使用any和str.isdigit|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|200753|def+num_there(s):
++++return+any(i.isdigit()+for+i+in+s)|code-block|syntax|javascript|200754|如果字符串中存在数字，该函数将返回True，否则返回False。|200755|演示：|200756|>>>+king+=+'I+shall+have+3+cakes'
>>>+num_there(king)
True
>>>+servant+=+'I+do+not+have+any+cakes'
>>>+num_there(servant)
False|200757|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/2/library/functions.html#any|1|https://docs.python.org/3/library/stdtypes.html#str.isdigit^0|7|3|B|B|7|3|0|B|B|1|0|0|H|4|Q|5|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@$9|15|A|16|1|17]|$9|18|A|19|1|1A]]|E|$]]|$1|F|3|G|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1C|8|@$9|1D|A|1E|B|C]|$9|1F|A|1G|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|1H|8|@]|D|@]|E|$]]|$1|O|3|P|5|H|7|1I|8|@]|D|@]|E|$I|J]]|$1|Q|3|-4|5|6|7|1J|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]|Y|$5|T|U|V|E|$W|Z]]]]

You can use a combination of <a href="http://docs.python.org/2/library/functions.html#any" rel="noreferrer"><code>any</code></a> and <a href="https://docs.python.org/3/library/stdtypes.html#str.isdigit" rel="noreferrer"><code>str.isdigit</code></a>:

<pre><code>def num_there(s):
 return any(i.isdigit() for i in s)
</code></pre>

The function will return <code>True</code> if a digit exists in the string, otherwise <code>False</code>.

Demo:

<pre><code>&gt;&gt;&gt; king = 'I shall have 3 cakes'
&gt;&gt;&gt; num_there(king)
True
&gt;&gt;&gt; servant = 'I do not have any cakes'
&gt;&gt;&gt; num_there(servant)
False
</code></pre>

blocks|key|4706605|text|您可以对字符串中的每个字符应用isdigit()函数。或者，您可以使用正则表达式。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4706606|我还发现How+do+I+find+one+number+in+a+string+in+Python?有非常合适的返回数字的方法。下面的解决方案来自该问题的答案。|offset|length|4706607|number+=+re.search(r'\d%2B',+yourString).group()|code-block|syntax|javascript|4706608|或者：|4706609|number+=+filter(str.isdigit,+yourString)|4706610|有关更多信息，请查看正则表达式文档：http://docs.python.org/2/library/re.html|4706611|编辑:这将返回实际数字，而不是布尔值，因此上面的答案更适合您的情况|4706612|第一个方法将返回第一个数字和随后的连续数字。因此，1.56将作为1返回。10,000将作为10.0返回。0207-100-1000将作为0207返回。|4706613|第二种方法不起作用。|4706614|要提取所有数字、点和逗号，并且不丢失不连续的数字，请使用：|4706615|re.sub('[%5E\d.,]'+,+'',+yourString)|4706616|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/8234641/how-do-i-find-one-number-in-a-string-in-python|1|http://docs.python.org/2/library/re.html^0|0|4|1B|0|0|0|0|0|I|14|1|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1A|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1B|8|@]|9|@$D|1C|E|1D|1|1E]]|A|$]]|$1|F|3|G|5|H|7|1F|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|1G|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|1H|8|@]|9|@]|A|$I|J]]|$1|O|3|P|5|6|7|1I|8|@]|9|@$D|1J|E|1K|1|1L]]|A|$]]|$1|Q|3|R|5|6|7|1M|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1N|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|1O|8|@]|9|@]|A|$]]|$1|W|3|X|5|6|7|1P|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|H|7|1Q|8|@]|9|@]|A|$I|J]]|$1|10|3|-4|5|6|7|1R|8|@]|9|@]|A|$]]]|11|$12|$5|13|14|15|A|$16|17]]|18|$5|13|14|15|A|$16|19]]]]

You could apply the function isdigit() on every character in the String. Or you could use regular expressions.

Also I found <a href="https://stackoverflow.com/questions/8234641/how-do-i-find-one-number-in-a-string-in-python">How do I find one number in a string in Python?</a> with very suitable ways to return numbers. The solution below is from the answer in that question.

<pre><code>number = re.search(r'\d+', yourString).group()
</code></pre>

Alternatively:

<pre><code>number = filter(str.isdigit, yourString)
</code></pre>

For further Information take a look at the regex docu: <a href="http://docs.python.org/2/library/re.html" rel="nofollow noreferrer">http://docs.python.org/2/library/re.html</a>

Edit: This Returns the actual numbers, not a boolean value, so the answers above are more correct for your case

The first method will return the first digit and subsequent consecutive digits. Thus 1.56 will be returned as 1. 10,000 will be returned as 10. 0207-100-1000 will be returned as 0207. 

The second method does not work.

To extract all digits, dots and commas, and not lose non-consecutive digits, use:

<pre><code>re.sub('[^\d.,]' , '', yourString)
</code></pre>

blocks|key|4706621|text|这件怎么样？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4706622|import+string

def+containsNumber(line):
++++res+=+False
++++try:
++++++++for+val+in+line.split():
++++++++++++if+(float(val.strip(string.punctuation))):
++++++++++++++++res+=+True
++++++++++++++++break
++++except+ValueError:
++++++++pass
++++return+res

containsNumber('234.12+a22')+#+returns+True
containsNumber('234.12L+a22')+#+returns+False
containsNumber('234.12,+a22')+#+returns+True|code-block|syntax|javascript|4706623|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|$]]

What about this one?

<pre><code>import string

def containsNumber(line):
 res = False
 try:
 for val in line.split():
 if (float(val.strip(string.punctuation))):
 res = True
 break
 except ValueError:
 pass
 return res

containsNumber('234.12 a22') # returns True
containsNumber('234.12L a22') # returns False
containsNumber('234.12, a22') # returns True
</code></pre>

blocks|key|4706645|text|https://docs.python.org/2/library/re.html|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4706646|你最好使用正则表达式。这要快得多。|4706647|import+re

def+f1(string):
++++return+any(i.isdigit()+for+i+in+string)


def+f2(string):
++++return+re.search('\d',+string)


#+if+you+compile+the+regex+string+first,+it's+even+faster
RE_D+=+re.compile('\d')
def+f3(string):
++++return+RE_D.search(string)

#+Output+from+iPython
#+In+[18]:+%25timeit++f1('assdfgag123')
#+1000000+loops,+best+of+3:+1.18+µs+per+loop

#+In+[19]:+%25timeit++f2('assdfgag123')
#+1000000+loops,+best+of+3:+923+ns+per+loop

#+In+[20]:+%25timeit++f3('assdfgag123')
#+1000000+loops,+best+of+3:+384+ns+per+loop|code-block|syntax|javascript|4706648|entityMap|0|LINK|mutability|MUTABLE|url^0|0|15|0|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@$A|S|B|T|1|U]]|C|$]]|$1|D|3|E|5|6|7|V|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|W|8|@]|9|@]|C|$I|J]]|$1|K|3|-4|5|6|7|X|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|4]]]]

<a href="https://docs.python.org/2/library/re.html">https://docs.python.org/2/library/re.html</a>

You should better use regular expression. It's much faster.

<pre><code>import re

def f1(string):
 return any(i.isdigit() for i in string)


def f2(string):
 return re.search('\d', string)


# if you compile the regex string first, it's even faster
RE_D = re.compile('\d')
def f3(string):
 return RE_D.search(string)

# Output from iPython
# In [18]: %timeit f1('assdfgag123')
# 1000000 loops, best of 3: 1.18 µs per loop

# In [19]: %timeit f2('assdfgag123')
# 1000000 loops, best of 3: 923 ns per loop

# In [20]: %timeit f3('assdfgag123')
# 1000000 loops, best of 3: 384 ns per loop
</code></pre>

blocks|key|4706677|text|您可以使用range+with+count来检查数字在字符串中出现的次数，方法是根据范围进行检查：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4706678|def+count_digit(a):
++++sum+=+0
++++for+i+in+range(10):
++++++++sum+%2B=+a.count(str(i))
++++return+sum

ans+=+count_digit("apple3rh5")
print(ans)

#This+print+2|code-block|syntax|javascript|4706679|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 use range with count to check how many times a number appears in the string by checking it against the range:

<pre><code>def count_digit(a):
 sum = 0
 for i in range(10):
 sum += a.count(str(i))
 return sum

ans = count_digit("apple3rh5")
print(ans)

#This print 2
</code></pre>

blocks|key|4706700|text|您可以使用NLTK方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4706701|这将在文本中同时找到'1‘和’1‘：|4706702|import+nltk+

def+existence_of_numeric_data(text):
++++text=nltk.word_tokenize(text)
++++pos+=+nltk.pos_tag(text)
++++count+=+0
++++for+i+in+range(len(pos)):
++++++++word+,+pos_tag+=+pos[i]
++++++++if+pos_tag+==+'CD':
++++++++++++return+True
++++return+False

existence_of_numeric_data('We+are+going+out.+Just+five+you+and+me.')|code-block|syntax|javascript|4706703|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|$]]

You can use NLTK method for it.

This will find both '1' and 'One' in the text:

<pre><code>import nltk 

def existence_of_numeric_data(text):
 text=nltk.word_tokenize(text)
 pos = nltk.pos_tag(text)
 count = 0
 for i in range(len(pos)):
 word , pos_tag = pos[i]
 if pos_tag == 'CD':
 return True
 return False

existence_of_numeric_data('We are going out. Just five you and me.')
</code></pre>

blocks|key|200967|text|更简单的解决方法是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|200968|s+=+'1dfss3sw235fsf7s'
count+=+0
temp+=+list(s)
for+item+in+temp:
++++if(item.isdigit()):
++++++++count+=+count+%2B+1
++++else:
++++++++pass
print+count|code-block|syntax|javascript|200969|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|$]]

Simpler way to solve is as

<pre><code>s = '1dfss3sw235fsf7s'
count = 0
temp = list(s)
for item in temp:
 if(item.isdigit()):
 count = count + 1
 else:
 pass
print count
</code></pre>

blocks|key|201027|text|我很惊讶竟然没有人提到any和map的这种组合|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|201028|def+contains_digit(s):
++++isdigit+=+str.isdigit
++++return+any(map(isdigit,s))|code-block|syntax|javascript|201029|在Python3中，它可能是最快的(除了正则表达式)，因为它不包含任何循环(函数的别名可以避免在str中查找它)。|201030|不要在Python2中使用它，因为map返回一个list，这会打破any短路|201031|entityMap^0|B|3|F|3|0|0|1C|3|0|H|3|O|4|X|3|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]|$9|14|A|15|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|P|$]]

I'm surprised that no-one mentionned this combination of <code>any</code> and <code>map</code>:

<pre><code>def contains_digit(s):
 isdigit = str.isdigit
 return any(map(isdigit,s))
</code></pre>

in python 3 it's probably the fastest there (except maybe for regexes) is because it doesn't contain any loop (and aliasing the function avoids looking it up in <code>str</code>). 

Don't use that in python 2 as <code>map</code> returns a <code>list</code>, which breaks <code>any</code> short-circuiting

blocks|key|4706786|text|import+string
import+random
n+=+10

p+=+''

while+(string.ascii_uppercase+not+in+p)+and+(string.ascii_lowercase+not+in+p)+and+(string.digits+not+in+p):
++++for+_+in+range(n):
++++++++state+=+random.randint(0,+2)
++++++++if+state+==+0:
++++++++++++p+=+p+%2B+chr(random.randint(97,+122))
++++++++elif+state+==+1:
++++++++++++p+=+p+%2B+chr(random.randint(65,+90))
++++++++else:
++++++++++++p+=+p+%2B+str(random.randint(0,+9))
++++break
print(p)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4706787|此代码生成一个大小为n的序列，该序列至少包含一个大写、小写和一个数字。通过使用while循环，我们保证了这个事件。|unstyled|4706788|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>import string
import random
n = 10

p = ''

while (string.ascii_uppercase not in p) and (string.ascii_lowercase not in p) and (string.digits not in p):
 for _ in range(n):
 state = random.randint(0, 2)
 if state == 0:
 p = p + chr(random.randint(97, 122))
 elif state == 1:
 p = p + chr(random.randint(65, 90))
 else:
 p = p + str(random.randint(0, 9))
 break
print(p)
</code></pre>

This code generates a sequence with size n which at least contain an uppercase, lowercase, and a digit. By using the while loop, we have guaranteed this event.

blocks|key|4706815|text|alp_num+=+[x+for+x+in+string.split()+if+x.isalnum()+and+re.search(r'\d',x)+and+
re.search(r'[a-z]',x)]

print(alp_num)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4706816|这将返回同时包含字母和数字的所有字符串。isalpha()返回全数字或全字符的字符串。|unstyled|4706817|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>alp_num = [x for x in string.split() if x.isalnum() and re.search(r'\d',x) and 
re.search(r'[a-z]',x)]

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

This returns all the string that has both alphabets and numbers in it. isalpha() returns the string with all digits or all characters.

blocks|key|4706840|text|我会让@zyxue的答案更明确一点：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4706841|RE_D+=+re.compile('\d')

def+has_digits(string):
++++res+=+RE_D.search(string)
++++return+res+is+not+None

has_digits('asdf1')
Out:+True

has_digits('asdf')
Out:+False|code-block|syntax|javascript|4706842|从@zyxue在答案中提出的解决方案中，哪个是具有最快基准的解决方案。|4706843|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|$]]

I'll make the @zyxue answer a bit more explicit:

<pre><code>RE_D = re.compile('\d')

def has_digits(string):
 res = RE_D.search(string)
 return res is not None

has_digits('asdf1')
Out: True

has_digits('asdf')
Out: False
</code></pre>

which is the solution with the fastest benchmark from the solutions that @zyxue proposed on the answer.

blocks|key|201192|text|这可能不是Python中最好的方法，但作为一个Haskeller，这种lambda/map方法对我来说非常有意义，而且非常简短：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|201193|anydigit+=+lambda+x:+any(map(str.isdigit,+x))|offset|length|style|CODE|201194|当然不需要命名。它的名字可以像anydigit("abc123")一样使用，这感觉就是我想要的！|201195|entityMap^0|0|0|19|0|F|I|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|P|8|@$D|Q|E|R|F|G]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|K|$]]

This probably isn't the best approach in Python, but as a Haskeller this lambda/map approach made perfect sense to me and is very short:

<code>anydigit = lambda x: any(map(str.isdigit, x))</code>

Doesn't need to be named of course. Named it could be used like <code>anydigit("abc123")</code>, which feels like what I was looking for!

blocks|key|201236|text|此外，您还可以使用regex+findall。这是一个更通用的解决方案，因为它增加了对数字长度的更多控制。在需要最小长度的数字的情况下，它可能会很有帮助。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|201237|s+=+'67389kjsdk'+
contains_digit+=+len(re.findall('\d%2B',+s))+>+0|code-block|syntax|javascript|201238|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|$]]

Also, you could use regex findall. It's a more general solution since it adds more control over the length of the number. It could be helpful in cases where you require a number with minimal length.
<pre><code>s = '67389kjsdk' 
contains_digit = len(re.findall('\d+', s)) &gt; 0
</code></pre>

blocks|key|4706970|text|这也是可行的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4706971|if+any(i.isdigit()+for+i+in+s):
++++print("True")|code-block|syntax|javascript|4706972|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 too will work.
<pre><code>if any(i.isdigit() for i in s):
 print(&quot;True&quot;)
</code></pre>

Most of the questions I've found are biased on the fact they're looking for letters in their numbers, whereas I'm looking for numbers in what I'd like to be a numberless string.
I need to enter a string and check to see if it contains any numbers and if it does reject it.

The function <code>isdigit()</code> only returns <code>True</code> if ALL of the characters are numbers. I just want to see if the user has entered a number so a sentence like <code>"I own 1 dog"</code> or something.

Any ideas?

Check if a string contains a number

Python

我发现的大多数问题都是基于这样一个事实，即他们在数字中寻找字母，而我在寻找数字中的数字，我希望它是一个无数的字符串。我需要输入一个字符串并检查它是否包含任何数字，以及它是否拒绝它。如果所有字符都是数字，则函数isdigit()仅返回True。我只想看看用户是否输入了一个数字，比如像"I own 1 dog"之类的句子。...

问检查字符串是否包含数字
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