检查后将值转换为字符串为整数，然后检查字符串第一个字符值为-或+，其余字符串为数字。最后检查isdigit。 Test = ['1'， '12015'， '1..]2 ', ' a2kk78”、“1.5”,2,1.24,“-8.5”,“+”、“1”、“88751.71 + 7)

检查

for k,v in enumerate(test): 
    print(k, v, 'test: ', True if isinstance(v, int) is not False else True if str(v)[0] in ['-', '+'] and str(v)[1:].isdigit() else str(v).isdigit())

结果

0 1 test:  True
1 12015 test:  True
2 1..2 test:  False
3 a2kk78 test:  False
4 1.5 test:  False
5 2 test:  True
6 1.24 test:  False
7 -8.5 test:  False
8 +88751.71 test:  False
9 -1 test:  True
10 +7 test:  True

对于正整数，可以使用.isdigit:

>>> '16'.isdigit()
True

但它对负整数不起作用。假设您可以尝试以下方法:

>>> s = '-17'
>>> s.startswith('-') and s[1:].isdigit()
True

它不适用于'16.0'格式，在这个意义上，'16.0'格式类似于int类型强制转换。

编辑:

def check_int(s):
    if s[0] in ('-', '+'):
        return s[1:].isdigit()
    return s.isdigit()

可以用下面的方法进行检查。

def check_if_string_is_int(string1):
    for character in string1:
        if not character.isdigit():
            return "Not a number"
    else:
        return "Is a number"

正确的RegEx解决方案应该结合Greg Hewgill和Nowell的思想，但不使用全局变量。可以通过将属性附加到方法来实现这一点。另外，我知道在方法中导入是不受欢迎的，但我想要的是像http://peak.telecommunity.com/DevCenter/Importing#lazy-imports这样的“惰性模块”效果

edit:到目前为止，我最喜欢的技术是使用String对象的独占方法。

#!/usr/bin/env python

# Uses exclusively methods of the String object
def isInteger(i):
    i = str(i)
    return i=='0' or (i if i.find('..') > -1 else i.lstrip('-+').rstrip('0').rstrip('.')).isdigit()

# Uses re module for regex
def isIntegre(i):
    import re
    if not hasattr(isIntegre, '_re'):
        print("I compile only once. Remove this line when you are confident in that.")
        isIntegre._re = re.compile(r"[-+]?\d+(\.0*)?$")
    return isIntegre._re.match(str(i)) is not None

# When executed directly run Unit Tests
if __name__ == '__main__':
    for obj in [
                # integers
                0, 1, -1, 1.0, -1.0,
                '0', '0.','0.0', '1', '-1', '+1', '1.0', '-1.0', '+1.0',
                # non-integers
                1.1, -1.1, '1.1', '-1.1', '+1.1',
                '1.1.1', '1.1.0', '1.0.1', '1.0.0',
                '1.0.', '1..0', '1..',
                '0.0.', '0..0', '0..',
                'one', object(), (1,2,3), [1,2,3], {'one':'two'}
            ]:
        # Notice the integre uses 're' (intended to be humorous)
        integer = ('an integer' if isInteger(obj) else 'NOT an integer')
        integre = ('an integre' if isIntegre(obj) else 'NOT an integre')
        # Make strings look like strings in the output
        if isinstance(obj, str):
            obj = ("'%s'" % (obj,))
        print("%30s is %14s is %14s" % (obj, integer, integre))

对于那些不太喜欢冒险的同学，输出如下:

I compile only once. Remove this line when you are confident in that.
                             0 is     an integer is     an integre
                             1 is     an integer is     an integre
                            -1 is     an integer is     an integre
                           1.0 is     an integer is     an integre
                          -1.0 is     an integer is     an integre
                           '0' is     an integer is     an integre
                          '0.' is     an integer is     an integre
                         '0.0' is     an integer is     an integre
                           '1' is     an integer is     an integre
                          '-1' is     an integer is     an integre
                          '+1' is     an integer is     an integre
                         '1.0' is     an integer is     an integre
                        '-1.0' is     an integer is     an integre
                        '+1.0' is     an integer is     an integre
                           1.1 is NOT an integer is NOT an integre
                          -1.1 is NOT an integer is NOT an integre
                         '1.1' is NOT an integer is NOT an integre
                        '-1.1' is NOT an integer is NOT an integre
                        '+1.1' is NOT an integer is NOT an integre
                       '1.1.1' is NOT an integer is NOT an integre
                       '1.1.0' is NOT an integer is NOT an integre
                       '1.0.1' is NOT an integer is NOT an integre
                       '1.0.0' is NOT an integer is NOT an integre
                        '1.0.' is NOT an integer is NOT an integre
                        '1..0' is NOT an integer is NOT an integre
                         '1..' is NOT an integer is NOT an integre
                        '0.0.' is NOT an integer is NOT an integre
                        '0..0' is NOT an integer is NOT an integre
                         '0..' is NOT an integer is NOT an integre
                         'one' is NOT an integer is NOT an integre
<object object at 0x103b7d0a0> is NOT an integer is NOT an integre
                     (1, 2, 3) is NOT an integer is NOT an integre
                     [1, 2, 3] is NOT an integer is NOT an integre
                {'one': 'two'} is NOT an integer is NOT an integre

我猜这个问题与速度有关，因为try/except有一个时间惩罚:

测试数据

首先，我创建了一个包含200个字符串、100个失败字符串和100个数字字符串的列表。

from random import shuffle
numbers = [u'+1'] * 100
nonumbers = [u'1abc'] * 100
testlist = numbers + nonumbers
shuffle(testlist)
testlist = np.array(testlist)

numpy解决方案(仅适用于数组和unicode)

Np.core.defchararray.isnumeric也可以用于unicode字符串，但它返回一个数组。所以，如果你必须做成千上万的转换，并且有丢失的数据或非数值数据，这是一个很好的解决方案。

import numpy as np
%timeit np.core.defchararray.isnumeric(testlist)
10000 loops, best of 3: 27.9 µs per loop # 200 numbers per loop

试/除了

def check_num(s):
  try:
    int(s)
    return True
  except:
    return False

def check_list(l):
  return [check_num(e) for e in l]

%timeit check_list(testlist)
1000 loops, best of 3: 217 µs per loop # 200 numbers per loop

numpy解决方案似乎更快。

如果你想只接受低ascii码的数字，这里有这样做的测试:

Python 3.7+:(u.isdecimal() and u.isascii()))

Python <= 3.6:(u.isdecimal() and u == str(int(u)))

其他答案建议使用.isdigit()或.isdecimal()，但它们都包含一些上unicode字符，如'缌' (u'\u0662'):

u = u'\u0662'     # '٢'
u.isdigit()       # True
u.isdecimal()     # True
u.isascii()       # False (Python 3.7+ only)
u == str(int(u))  # False