如何在Python中检查字符串是否表示数值?

def is_number(s):
    try:
        float(s)
        return True
    except ValueError:
        return False

上述方法可行,但似乎很笨拙。


如果您正在测试的内容来自用户输入,那么即使它表示int或float,它仍然是一个字符串。请参阅如何将输入读取为数字?用于转换输入,并询问用户输入,直到他们给出有效响应以确保输入在继续之前表示int或float(或其他要求)。


当前回答

我想看看哪种方法最快。总的来说,check_replace函数给出了最佳和最一致的结果。check_exception函数给出了最快的结果,但前提是没有触发异常——这意味着它的代码是最有效的,但抛出异常的开销非常大。

请注意,检查成功的强制转换是唯一准确的方法,例如,这与check_exception一起工作,但其他两个测试函数将为有效的float返回False:

huge_number = float('1e+100')

以下是基准代码:

import time, re, random, string

ITERATIONS = 10000000

class Timer:    
    def __enter__(self):
        self.start = time.clock()
        return self
    def __exit__(self, *args):
        self.end = time.clock()
        self.interval = self.end - self.start

def check_regexp(x):
    return re.compile("^\d*\.?\d*$").match(x) is not None

def check_replace(x):
    return x.replace('.','',1).isdigit()

def check_exception(s):
    try:
        float(s)
        return True
    except ValueError:
        return False

to_check = [check_regexp, check_replace, check_exception]

print('preparing data...')
good_numbers = [
    str(random.random() / random.random()) 
    for x in range(ITERATIONS)]

bad_numbers = ['.' + x for x in good_numbers]

strings = [
    ''.join(random.choice(string.ascii_uppercase + string.digits) for _ in range(random.randint(1,10)))
    for x in range(ITERATIONS)]

print('running test...')
for func in to_check:
    with Timer() as t:
        for x in good_numbers:
            res = func(x)
    print('%s with good floats: %s' % (func.__name__, t.interval))
    with Timer() as t:
        for x in bad_numbers:
            res = func(x)
    print('%s with bad floats: %s' % (func.__name__, t.interval))
    with Timer() as t:
        for x in strings:
            res = func(x)
    print('%s with strings: %s' % (func.__name__, t.interval))

以下是2017年MacBook Pro 13上Python 2.7.10的结果:

check_regexp with good floats: 12.688639
check_regexp with bad floats: 11.624862
check_regexp with strings: 11.349414
check_replace with good floats: 4.419841
check_replace with bad floats: 4.294909
check_replace with strings: 4.086358
check_exception with good floats: 3.276668
check_exception with bad floats: 13.843092
check_exception with strings: 15.786169

以下是2017年MacBook Pro 13上Python 3.6.5的结果:

check_regexp with good floats: 13.472906000000009
check_regexp with bad floats: 12.977665000000016
check_regexp with strings: 12.417542999999995
check_replace with good floats: 6.011045999999993
check_replace with bad floats: 4.849356
check_replace with strings: 4.282754000000011
check_exception with good floats: 6.039081999999979
check_exception with bad floats: 9.322753000000006
check_exception with strings: 9.952595000000002

以下是2017年MacBook Pro 13上PyPy 2.7.13的结果:

check_regexp with good floats: 2.693217
check_regexp with bad floats: 2.744819
check_regexp with strings: 2.532414
check_replace with good floats: 0.604367
check_replace with bad floats: 0.538169
check_replace with strings: 0.598664
check_exception with good floats: 1.944103
check_exception with bad floats: 2.449182
check_exception with strings: 2.200056

其他回答

使用以下方法处理所有情况:-

import re
a=re.match('((\d+[\.]\d*$)|(\.)\d+$)' ,  '2.3') 
a=re.match('((\d+[\.]\d*$)|(\.)\d+$)' ,  '2.')
a=re.match('((\d+[\.]\d*$)|(\.)\d+$)' ,  '.3')
a=re.match('((\d+[\.]\d*$)|(\.)\d+$)' ,  '2.3sd')
a=re.match('((\d+[\.]\d*$)|(\.)\d+$)' ,  '2.3')

这个怎么样:

'3.14'.replace('.','',1).isdigit()

只有当有一个或没有“”时,它才会返回真在数字串中。

'3.14.5'.replace('.','',1).isdigit()

将返回false

编辑:刚刚看到另一条评论。。。可以为其他情况添加.replace(badstuff,“”,maxnum_badstuff)。如果你传递的是盐而不是任意的调味品(参考:xkcd#974),这会很好:P

因此,将所有这些放在一起,检查Nan、无穷大和复数(看起来它们是用j指定的,而不是i,即1+2j),结果是:

def is_number(s):
    try:
        n=str(float(s))
        if n == "nan" or n=="inf" or n=="-inf" : return False
    except ValueError:
        try:
            complex(s) # for complex
        except ValueError:
            return False
    return True

您可能需要考虑一个例外:字符串“NaN”

如果您希望is_number为“NaN”返回FALSE,则此代码将无法工作,因为Python将其转换为非数字的数字表示(请讨论身份问题):

>>> float('NaN')
nan

否则,我应该感谢您现在广泛使用的这段代码

G.

在浮点数的最常见情况下,我们希望处理整数和小数。让我们以字符串“1.1”为例。

我会尝试以下方法之一:

1.>isnumeric()

word = "1.1"

"".join(word.split(".")).isnumeric()
>>> True

2.>isdigit()

word = "1.1"

"".join(word.split(".")).isdigit()
>>> True

3.>isdecimal()

word = "1.1"

"".join(word.split(".")).isdecimal()
>>> True

速度:

► 所有上述方法具有相似的速度。

%timeit "".join(word.split(".")).isnumeric()
>>> 257 ns ± 12 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

%timeit "".join(word.split(".")).isdigit()
>>> 252 ns ± 11 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

%timeit "".join(word.split(".")).isdecimal()
>>> 244 ns ± 7.17 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)