检查字符串是否包含数字

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.

我发现的大多数问题都是因为他们正在寻找他们的数字中的字母，而我正在寻找我希望成为无数字符串的数字。我需要输入一个字符串并检查它是否包含任何数字，如果它确实拒绝它。

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

如果所有字符都是数字，则函数isdigit（）仅返回true。我只是想看看用户是否输入了一个数字，如“我拥有1只狗”之类的句子。

Any ideas?

有任何想法吗？

10 个解决方案

#1

146

You can use any function, with the str.isdigit function, like this

你可以使用任何函数，使用str.isdigit函数，就像这样

>>> def hasNumbers(inputString):
...     return any(char.isdigit() for char in inputString)
... 
>>> hasNumbers("I own 1 dog")
True
>>> hasNumbers("I own no dog")
False

Alternatively you can use a Regular Expression, like this

或者，您可以使用正则表达式，如下所示

>>> import re
>>> def hasNumbers(inputString):
...     return bool(re.search(r'\d', inputString))
... 
>>> hasNumbers("I own 1 dog")
True
>>> hasNumbers("I own no dog")
False

#2

You can use a combination of any and str.isdigit:

您可以使用any和str.isdigit的组合：

def num_there(s):
    return any(i.isdigit() for i in s)

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

如果字符串中存在数字，则该函数将返回True，否则返回False。

Demo:

演示：

>>> king = 'I shall have 3 cakes'
>>> num_there(king)
True
>>> servant = 'I do not have any cakes'
>>> num_there(servant)
False

#3

https://docs.python.org/2/library/re.html

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

你应该更好地使用正则表达式。它要快得多。

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

#4

use

使用

str.isalpha()

Ref: https://docs.python.org/2/library/stdtypes.html#str.isalpha

参考：https：//docs.python.org/2/library/stdtypes.html#str.isalpha

Return true if all characters in the string are alphabetic and there is at least one character, false otherwise.

如果字符串中的所有字符都是字母并且至少有一个字符，则返回true，否则返回false。

#5

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

您可以对String中的每个字符应用函数isdigit（）。或者你可以使用正则表达式。

Also I found How do I find one number in a string in Python? with very suitable ways to return numbers. The solution below is from the answer in that question.

另外我发现如何在Python中的字符串中找到一个数字？以非常合适的方式返回数字。以下解决方案来自该问题的答案。

number = re.search(r'\d+', yourString).group()

Alternatively:

或者：

number = filter(str.isdigit, yourString)

For further Information take a look at the regex docu: http://docs.python.org/2/library/re.html

有关详细信息，请查看正则表达式文档：http：//docs.python.org/2/library/re.html

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

编辑：返回实际数字，而不是布尔值，因此上面的答案对于您的情况更正确

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.

第一种方法将返回第一个数字和后续连续数字。因此，1.56将返回1. 10,000将返回10. 0207-100-1000将返回0207。

The second method does not work.

第二种方法不起作用。

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

要提取所有数字，点和逗号，而不是丢失非连续数字，请使用：

re.sub('[^\d.,]' , '', yourString)

re.sub（'[^ \ d。，]'，''，yourString）

#6

What about this one?

这个如何？

import string

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

print containsNumber('234.12 a22') # returns True
print containsNumber('234.12L a22') # returns False
print containsNumber('234.12, a22') # returns True

#7

You can accomplish this as follows:

您可以按如下方式完成此操作：

if a_string.isdigit(): do_this() else: do_that()

如果a_string.isdigit（）：do_this（）else：do_that（）

https://docs.python.org/2/library/stdtypes.html#str.isdigit

Using .isdigit() also means not having to resort to exception handling (try/except) in cases where you need to use list comprehension (try/except is not possible inside a list comprehension).

使用.isdigit（）也意味着在需要使用列表理解的情况下不必诉诸异常处理（try / except）（在列表解析中不可能尝试/ except）。

#8

You can use NLTK method for it.

您可以使用NLTK方法。

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

这将在文本中找到'1'和'One'：

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.')

#9

You can use range with count to check how many times a number appears in the string by checking it against the range:

您可以使用带有计数的范围来检查数字在字符串中出现的次数，方法是在范围内进行检查：

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

#10

Simpler way to solve is as

更简单的解决方法是

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

#1

146