def num(s):
    """num(s)
    num(3),num(3.7)-->3
    num('3')-->3, num('3.7')-->3.7
    num('3,700')-->ValueError
    num('3a'),num('a3'),-->ValueError
    num('3e4') --> 30000.0
    """
    try:
        return int(s)
    except ValueError:
        try:
            return float(s)
        except ValueError:
            raise ValueError('argument is not a string of number')

这是对你的问题的另一种解释（提示：它很模糊）。你可能正在寻找这样的东西：

def parseIntOrFloat( aString ):
    return eval( aString )

它是这样工作的。。。

>>> parseIntOrFloat("545.2222")
545.22220000000004
>>> parseIntOrFloat("545")
545

理论上，存在注入漏洞。例如，字符串可以是“import os；os.art（）”。然而，如果没有任何关于字符串来源的背景信息，理论上可能是推测。因为这个问题很模糊，所以根本不清楚这个漏洞是否真的存在。

处理十六进制、八进制、二进制、十进制和浮点

这个解决方案将处理数字的所有字符串约定（我所知道的）。

def to_number(n):
    ''' Convert any number representation to a number
    This covers: float, decimal, hex, and octal numbers.
    '''

    try:
        return int(str(n), 0)
    except:
        try:
            # Python 3 doesn't accept "010" as a valid octal.  You must use the
            # '0o' prefix
            return int('0o' + n, 0)
        except:
            return float(n)

这个测试用例输出说明了我所说的内容。

======================== CAPTURED OUTPUT =========================
to_number(3735928559)   = 3735928559 == 3735928559
to_number("0xFEEDFACE") = 4277009102 == 4277009102
to_number("0x0")        =          0 ==          0
to_number(100)          =        100 ==        100
to_number("42")         =         42 ==         42
to_number(8)            =          8 ==          8
to_number("0o20")       =         16 ==         16
to_number("020")        =         16 ==         16
to_number(3.14)         =       3.14 ==       3.14
to_number("2.72")       =       2.72 ==       2.72
to_number("1e3")        =     1000.0 ==       1000
to_number(0.001)        =      0.001 ==      0.001
to_number("0xA")        =         10 ==         10
to_number("012")        =         10 ==         10
to_number("0o12")       =         10 ==         10
to_number("0b01010")    =         10 ==         10
to_number("10")         =         10 ==         10
to_number("10.0")       =       10.0 ==         10
to_number("1e1")        =       10.0 ==         10

下面是测试：

class test_to_number(unittest.TestCase):

    def test_hex(self):
        # All of the following should be converted to an integer
        #
        values = [

                 #          HEX
                 # ----------------------
                 # Input     |   Expected
                 # ----------------------
                (0xDEADBEEF  , 3735928559), # Hex
                ("0xFEEDFACE", 4277009102), # Hex
                ("0x0"       ,          0), # Hex

                 #        Decimals
                 # ----------------------
                 # Input     |   Expected
                 # ----------------------
                (100         ,        100), # Decimal
                ("42"        ,         42), # Decimal
            ]



        values += [
                 #        Octals
                 # ----------------------
                 # Input     |   Expected
                 # ----------------------
                (0o10        ,          8), # Octal
                ("0o20"      ,         16), # Octal
                ("020"       ,         16), # Octal
            ]


        values += [
                 #        Floats
                 # ----------------------
                 # Input     |   Expected
                 # ----------------------
                (3.14        ,       3.14), # Float
                ("2.72"      ,       2.72), # Float
                ("1e3"       ,       1000), # Float
                (1e-3        ,      0.001), # Float
            ]

        values += [
                 #        All ints
                 # ----------------------
                 # Input     |   Expected
                 # ----------------------
                ("0xA"       ,         10),
                ("012"       ,         10),
                ("0o12"      ,         10),
                ("0b01010"   ,         10),
                ("10"        ,         10),
                ("10.0"      ,         10),
                ("1e1"       ,         10),
            ]

        for _input, expected in values:
            value = to_number(_input)

            if isinstance(_input, str):
                cmd = 'to_number("{}")'.format(_input)
            else:
                cmd = 'to_number({})'.format(_input)

            print("{:23} = {:10} == {:10}".format(cmd, value, expected))
            self.assertEqual(value, expected)

如果您不反对第三方模块，可以查看fastnumbers模块。它提供了一个名为fast_real的函数，它完全符合这个问题的要求，并且比纯Python实现更快：

>>> from fastnumbers import fast_real
>>> fast_real("545.2222")
545.2222
>>> type(fast_real("545.2222"))
float
>>> fast_real("31")
31
>>> type(fast_real("31"))
int

>>> a = "545.2222"
>>> float(a)
545.22220000000004
>>> int(float(a))
545

这个问题似乎有点过时了。但让我建议一个函数parseStr，它有类似的功能，即返回整数或浮点数，如果给定的ASCII字符串不能转换为任何一个，它将不受影响地返回。当然，代码可能会被调整为只执行您想要的操作：

   >>> import string
   >>> parseStr = lambda x: x.isalpha() and x or x.isdigit() and \
   ...                      int(x) or x.isalnum() and x or \
   ...                      len(set(string.punctuation).intersection(x)) == 1 and \
   ...                      x.count('.') == 1 and float(x) or x
   >>> parseStr('123')
   123
   >>> parseStr('123.3')
   123.3
   >>> parseStr('3HC1')
   '3HC1'
   >>> parseStr('12.e5')
   1200000.0
   >>> parseStr('12$5')
   '12$5'
   >>> parseStr('12.2.2')
   '12.2.2'