检查字符串是否可以在Python中转换为float

问题:检查字符串是否可以在Python中转换为float

我有一些运行在字符串列表中的Python代码,并在可能的情况下将它们转换为整数或浮点数。对整数执行此操作非常简单

if element.isdigit():
  newelement = int(element)

浮点数比较困难。现在,我正在使用partition('.')分割字符串并检查以确保一侧或两侧都是数字。

partition = element.partition('.')
if (partition[0].isdigit() and partition[1] == '.' and partition[2].isdigit()) 
    or (partition[0] == '' and partition[1] == '.' and partition[2].isdigit()) 
    or (partition[0].isdigit() and partition[1] == '.' and partition[2] == ''):
  newelement = float(element)

这是可行的,但是显然,如果使用if语句有点让人头疼。我考虑的另一种解决方案是将转换仅包装在try / catch块中,然后查看转换是否成功,如本问题所述

还有其他想法吗?关于分区和尝试/捕获方法的相对优点的看法?

I’ve got some Python code that runs through a list of strings and converts them to integers or floating point numbers if possible. Doing this for integers is pretty easy

if element.isdigit():
  newelement = int(element)

Floating point numbers are more difficult. Right now I’m using partition('.') to split the string and checking to make sure that one or both sides are digits.

partition = element.partition('.')
if (partition[0].isdigit() and partition[1] == '.' and partition[2].isdigit()) 
    or (partition[0] == '' and partition[1] == '.' and partition[2].isdigit()) 
    or (partition[0].isdigit() and partition[1] == '.' and partition[2] == ''):
  newelement = float(element)

This works, but obviously the if statement for that is a bit of a bear. The other solution I considered is to just wrap the conversion in a try/catch block and see if it succeeds, as described in this question.

Anyone have any other ideas? Opinions on the relative merits of the partition and try/catch approaches?


回答 0

我会用..

try:
    float(element)
except ValueError:
    print "Not a float"

..它很简单,并且可以正常工作

另一个选择是正则表达式:

import re
if re.match(r'^-?\d+(?:\.\d+)?$', element) is None:
    print "Not float"

I would just use..

try:
    float(element)
except ValueError:
    print "Not a float"

..it’s simple, and it works

Another option would be a regular expression:

import re
if re.match(r'^-?\d+(?:\.\d+)?$', element) is None:
    print "Not float"

回答 1

检查浮点数的Python方法:

def isfloat(value):
  try:
    float(value)
    return True
  except ValueError:
    return False

不要被隐藏在浮船上的妖精所咬!做单元测试!

什么是浮动货币,哪些不是浮动货币,可能会让您感到惊讶:

Command to parse                        Is it a float?  Comment
--------------------------------------  --------------- ------------
print(isfloat(""))                      False
print(isfloat("1234567"))               True 
print(isfloat("NaN"))                   True            nan is also float
print(isfloat("NaNananana BATMAN"))     False
print(isfloat("123.456"))               True
print(isfloat("123.E4"))                True
print(isfloat(".1"))                    True
print(isfloat("1,234"))                 False
print(isfloat("NULL"))                  False           case insensitive
print(isfloat(",1"))                    False           
print(isfloat("123.EE4"))               False           
print(isfloat("6.523537535629999e-07")) True
print(isfloat("6e777777"))              True            This is same as Inf
print(isfloat("-iNF"))                  True
print(isfloat("1.797693e+308"))         True
print(isfloat("infinity"))              True
print(isfloat("infinity and BEYOND"))   False
print(isfloat("12.34.56"))              False           Two dots not allowed.
print(isfloat("#56"))                   False
print(isfloat("56%"))                   False
print(isfloat("0E0"))                   True
print(isfloat("x86E0"))                 False
print(isfloat("86-5"))                  False
print(isfloat("True"))                  False           Boolean is not a float.   
print(isfloat(True))                    True            Boolean is a float
print(isfloat("+1e1^5"))                False
print(isfloat("+1e1"))                  True
print(isfloat("+1e1.3"))                False
print(isfloat("+1.3P1"))                False
print(isfloat("-+1"))                   False
print(isfloat("(1)"))                   False           brackets not interpreted

Python method to check for float:

def isfloat(value):
  try:
    float(value)
    return True
  except ValueError:
    return False

Don’t get bit by the goblins hiding in the float boat! DO UNIT TESTING!

What is, and is not a float may surprise you:

Command to parse                        Is it a float?  Comment
--------------------------------------  --------------- ------------
print(isfloat(""))                      False
print(isfloat("1234567"))               True 
print(isfloat("NaN"))                   True            nan is also float
print(isfloat("NaNananana BATMAN"))     False
print(isfloat("123.456"))               True
print(isfloat("123.E4"))                True
print(isfloat(".1"))                    True
print(isfloat("1,234"))                 False
print(isfloat("NULL"))                  False           case insensitive
print(isfloat(",1"))                    False           
print(isfloat("123.EE4"))               False           
print(isfloat("6.523537535629999e-07")) True
print(isfloat("6e777777"))              True            This is same as Inf
print(isfloat("-iNF"))                  True
print(isfloat("1.797693e+308"))         True
print(isfloat("infinity"))              True
print(isfloat("infinity and BEYOND"))   False
print(isfloat("12.34.56"))              False           Two dots not allowed.
print(isfloat("#56"))                   False
print(isfloat("56%"))                   False
print(isfloat("0E0"))                   True
print(isfloat("x86E0"))                 False
print(isfloat("86-5"))                  False
print(isfloat("True"))                  False           Boolean is not a float.   
print(isfloat(True))                    True            Boolean is a float
print(isfloat("+1e1^5"))                False
print(isfloat("+1e1"))                  True
print(isfloat("+1e1.3"))                False
print(isfloat("+1.3P1"))                False
print(isfloat("-+1"))                   False
print(isfloat("(1)"))                   False           brackets not interpreted

回答 2

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

true仅当存在一个或没有“。”时返回。在数字字符串中。

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

将返回 false

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

将返回 false

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

which will return true only if there is one or no ‘.’ in the string of digits.

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

will return false

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

will return false


回答 3

TL; DR

  • 如果您输入的大部分内容都是可以转换为浮点数的字符串,try: except:方法是最好的本机Python方法。
  • 如果您的输入大部分是不能输入的字符串转换为浮点数的,则正则表达式或partition方法会更好。
  • 如果您1)不确定您的输入或需要更快的速度,并且2)不介意并可以安装第三方C扩展名,则fastnumbers效果很好。

通过第三方模块可以使用另一种称为fastnumbers的方法(公开,我是作者)。它提供了一个称为isfloat的功能。我在这个答案中采用了Jacob Gabrielson概述的单元测试示例,但是添加了该fastnumbers.isfloat方法。我还应该注意,Jacob的示例对正则表达式选项不公道,因为该示例中的大多数时间都因为点运算符而花费在全局查找中try: except:


def is_float_try(str):
    try:
        float(str)
        return True
    except ValueError:
        return False

import re
_float_regexp = re.compile(r"^[-+]?(?:\b[0-9]+(?:\.[0-9]*)?|\.[0-9]+\b)(?:[eE][-+]?[0-9]+\b)?$").match
def is_float_re(str):
    return True if _float_regexp(str) else False

def is_float_partition(element):
    partition=element.partition('.')
    if (partition[0].isdigit() and partition[1]=='.' and partition[2].isdigit()) or (partition[0]=='' and partition[1]=='.' and partition[2].isdigit()) or (partition[0].isdigit() and partition[1]=='.' and partition[2]==''):
        return True
    else:
        return False

from fastnumbers import isfloat


if __name__ == '__main__':
    import unittest
    import timeit

    class ConvertTests(unittest.TestCase):

        def test_re_perf(self):
            print
            print 're sad:', timeit.Timer('ttest.is_float_re("12.2x")', "import ttest").timeit()
            print 're happy:', timeit.Timer('ttest.is_float_re("12.2")', "import ttest").timeit()

        def test_try_perf(self):
            print
            print 'try sad:', timeit.Timer('ttest.is_float_try("12.2x")', "import ttest").timeit()
            print 'try happy:', timeit.Timer('ttest.is_float_try("12.2")', "import ttest").timeit()

        def test_fn_perf(self):
            print
            print 'fn sad:', timeit.Timer('ttest.isfloat("12.2x")', "import ttest").timeit()
            print 'fn happy:', timeit.Timer('ttest.isfloat("12.2")', "import ttest").timeit()


        def test_part_perf(self):
            print
            print 'part sad:', timeit.Timer('ttest.is_float_partition("12.2x")', "import ttest").timeit()
            print 'part happy:', timeit.Timer('ttest.is_float_partition("12.2")', "import ttest").timeit()

    unittest.main()

在我的机器上,输出为:

fn sad: 0.220988988876
fn happy: 0.212214946747
.
part sad: 1.2219619751
part happy: 0.754667043686
.
re sad: 1.50515985489
re happy: 1.01107215881
.
try sad: 2.40243887901
try happy: 0.425730228424
.
----------------------------------------------------------------------
Ran 4 tests in 7.761s

OK

如您所见,regex实际上并不像最初看起来的那样糟糕,并且如果您确实对速度有需求,则此fastnumbers方法相当不错。

TL;DR:

  • If your input is mostly strings that can be converted to floats, the try: except: method is the best native Python method.
  • If your input is mostly strings that cannot be converted to floats, regular expressions or the partition method will be better.
  • If you are 1) unsure of your input or need more speed and 2) don’t mind and can install a third-party C-extension, fastnumbers works very well.

There is another method available via a third-party module called fastnumbers (disclosure, I am the author); it provides a function called isfloat. I have taken the unittest example outlined by Jacob Gabrielson in this answer, but added the fastnumbers.isfloat method. I should also note that Jacob’s example did not do justice to the regex option because most of the time in that example was spent in global lookups because of the dot operator… I have modified that function to give a fairer comparison to try: except:.


def is_float_try(str):
    try:
        float(str)
        return True
    except ValueError:
        return False

import re
_float_regexp = re.compile(r"^[-+]?(?:\b[0-9]+(?:\.[0-9]*)?|\.[0-9]+\b)(?:[eE][-+]?[0-9]+\b)?$").match
def is_float_re(str):
    return True if _float_regexp(str) else False

def is_float_partition(element):
    partition=element.partition('.')
    if (partition[0].isdigit() and partition[1]=='.' and partition[2].isdigit()) or (partition[0]=='' and partition[1]=='.' and partition[2].isdigit()) or (partition[0].isdigit() and partition[1]=='.' and partition[2]==''):
        return True
    else:
        return False

from fastnumbers import isfloat


if __name__ == '__main__':
    import unittest
    import timeit

    class ConvertTests(unittest.TestCase):

        def test_re_perf(self):
            print
            print 're sad:', timeit.Timer('ttest.is_float_re("12.2x")', "import ttest").timeit()
            print 're happy:', timeit.Timer('ttest.is_float_re("12.2")', "import ttest").timeit()

        def test_try_perf(self):
            print
            print 'try sad:', timeit.Timer('ttest.is_float_try("12.2x")', "import ttest").timeit()
            print 'try happy:', timeit.Timer('ttest.is_float_try("12.2")', "import ttest").timeit()

        def test_fn_perf(self):
            print
            print 'fn sad:', timeit.Timer('ttest.isfloat("12.2x")', "import ttest").timeit()
            print 'fn happy:', timeit.Timer('ttest.isfloat("12.2")', "import ttest").timeit()


        def test_part_perf(self):
            print
            print 'part sad:', timeit.Timer('ttest.is_float_partition("12.2x")', "import ttest").timeit()
            print 'part happy:', timeit.Timer('ttest.is_float_partition("12.2")', "import ttest").timeit()

    unittest.main()

On my machine, the output is:

fn sad: 0.220988988876
fn happy: 0.212214946747
.
part sad: 1.2219619751
part happy: 0.754667043686
.
re sad: 1.50515985489
re happy: 1.01107215881
.
try sad: 2.40243887901
try happy: 0.425730228424
.
----------------------------------------------------------------------
Ran 4 tests in 7.761s

OK

As you can see, regex is actually not as bad as it originally seemed, and if you have a real need for speed, the fastnumbers method is quite good.


回答 4

如果您关心性能(我不建议您这样做),则只要您不期望太多,基于尝试的方法就是明显的赢家(与基于分区的方法或regexp方法相比)无效的字符串,在这种情况下,它可能会变慢(可能是由于异常处理的开销)。

再一次,我不建议您关心性能,只是给您数据,以防您每秒进行100亿次这样的操作。同样,基于分区的代码不能处理至少一个有效的字符串。

$ ./floatstr.py
F..
分区悲伤:3.1102449894
分区快乐:2.09208488464
..
难过:7.76906108856
重新开心:7.09421992302
..
尝试悲伤:12.1525540352
尝试快乐:1.44165301323
。
================================================== ====================
失败:test_partition(__main __。ConvertTests)
-------------------------------------------------- --------------------
追溯(最近一次通话):
  在test_partition中,文件“ ./floatstr.py”,第48行
    self.failUnless(is_float_partition(“ 20e2”))
断言错误

-------------------------------------------------- --------------------
在33.670秒内进行了8次测试

失败(失败= 1)

以下是代码(Python 2.6,来自John Gietzen的答案的正则表达式):

def is_float_try(str):
    try:
        float(str)
        return True
    except ValueError:
        return False

import re
_float_regexp = re.compile(r"^[-+]?(?:\b[0-9]+(?:\.[0-9]*)?|\.[0-9]+\b)(?:[eE][-+]?[0-9]+\b)?$")
def is_float_re(str):
    return re.match(_float_regexp, str)


def is_float_partition(element):
    partition=element.partition('.')
    if (partition[0].isdigit() and partition[1]=='.' and partition[2].isdigit()) or (partition[0]=='' and partition[1]=='.' and pa\
rtition[2].isdigit()) or (partition[0].isdigit() and partition[1]=='.' and partition[2]==''):
        return True

if __name__ == '__main__':
    import unittest
    import timeit

    class ConvertTests(unittest.TestCase):
        def test_re(self):
            self.failUnless(is_float_re("20e2"))

        def test_try(self):
            self.failUnless(is_float_try("20e2"))

        def test_re_perf(self):
            print
            print 're sad:', timeit.Timer('floatstr.is_float_re("12.2x")', "import floatstr").timeit()
            print 're happy:', timeit.Timer('floatstr.is_float_re("12.2")', "import floatstr").timeit()

        def test_try_perf(self):
            print
            print 'try sad:', timeit.Timer('floatstr.is_float_try("12.2x")', "import floatstr").timeit()
            print 'try happy:', timeit.Timer('floatstr.is_float_try("12.2")', "import floatstr").timeit()

        def test_partition_perf(self):
            print
            print 'partition sad:', timeit.Timer('floatstr.is_float_partition("12.2x")', "import floatstr").timeit()
            print 'partition happy:', timeit.Timer('floatstr.is_float_partition("12.2")', "import floatstr").timeit()

        def test_partition(self):
            self.failUnless(is_float_partition("20e2"))

        def test_partition2(self):
            self.failUnless(is_float_partition(".2"))

        def test_partition3(self):
            self.failIf(is_float_partition("1234x.2"))

    unittest.main()

If you cared about performance (and I’m not suggesting you should), the try-based approach is the clear winner (compared with your partition-based approach or the regexp approach), as long as you don’t expect a lot of invalid strings, in which case it’s potentially slower (presumably due to the cost of exception handling).

Again, I’m not suggesting you care about performance, just giving you the data in case you’re doing this 10 billion times a second, or something. Also, the partition-based code doesn’t handle at least one valid string.

$ ./floatstr.py
F..
partition sad: 3.1102449894
partition happy: 2.09208488464
..
re sad: 7.76906108856
re happy: 7.09421992302
..
try sad: 12.1525540352
try happy: 1.44165301323
.
======================================================================
FAIL: test_partition (__main__.ConvertTests)
----------------------------------------------------------------------
Traceback (most recent call last):
  File "./floatstr.py", line 48, in test_partition
    self.failUnless(is_float_partition("20e2"))
AssertionError

----------------------------------------------------------------------
Ran 8 tests in 33.670s

FAILED (failures=1)

Here’s the code (Python 2.6, regexp taken from John Gietzen’s answer):

def is_float_try(str):
    try:
        float(str)
        return True
    except ValueError:
        return False

import re
_float_regexp = re.compile(r"^[-+]?(?:\b[0-9]+(?:\.[0-9]*)?|\.[0-9]+\b)(?:[eE][-+]?[0-9]+\b)?$")
def is_float_re(str):
    return re.match(_float_regexp, str)


def is_float_partition(element):
    partition=element.partition('.')
    if (partition[0].isdigit() and partition[1]=='.' and partition[2].isdigit()) or (partition[0]=='' and partition[1]=='.' and pa\
rtition[2].isdigit()) or (partition[0].isdigit() and partition[1]=='.' and partition[2]==''):
        return True

if __name__ == '__main__':
    import unittest
    import timeit

    class ConvertTests(unittest.TestCase):
        def test_re(self):
            self.failUnless(is_float_re("20e2"))

        def test_try(self):
            self.failUnless(is_float_try("20e2"))

        def test_re_perf(self):
            print
            print 're sad:', timeit.Timer('floatstr.is_float_re("12.2x")', "import floatstr").timeit()
            print 're happy:', timeit.Timer('floatstr.is_float_re("12.2")', "import floatstr").timeit()

        def test_try_perf(self):
            print
            print 'try sad:', timeit.Timer('floatstr.is_float_try("12.2x")', "import floatstr").timeit()
            print 'try happy:', timeit.Timer('floatstr.is_float_try("12.2")', "import floatstr").timeit()

        def test_partition_perf(self):
            print
            print 'partition sad:', timeit.Timer('floatstr.is_float_partition("12.2x")', "import floatstr").timeit()
            print 'partition happy:', timeit.Timer('floatstr.is_float_partition("12.2")', "import floatstr").timeit()

        def test_partition(self):
            self.failUnless(is_float_partition("20e2"))

        def test_partition2(self):
            self.failUnless(is_float_partition(".2"))

        def test_partition3(self):
            self.failIf(is_float_partition("1234x.2"))

    unittest.main()

回答 5

仅出于多样性,这是另一种方法。

>>> all([i.isnumeric() for i in '1.2'.split('.',1)])
True
>>> all([i.isnumeric() for i in '2'.split('.',1)])
True
>>> all([i.isnumeric() for i in '2.f'.split('.',1)])
False

编辑:我确信它不会容纳所有的float情况,尽管特别是在有指数的时候。为了解决它看起来像这样。这将返回True,只有val是浮点数,而对于int则返回False,但性能可能不如regex。

>>> def isfloat(val):
...     return all([ [any([i.isnumeric(), i in ['.','e']]) for i in val],  len(val.split('.')) == 2] )
...
>>> isfloat('1')
False
>>> isfloat('1.2')
True
>>> isfloat('1.2e3')
True
>>> isfloat('12e3')
False

Just for variety here is another method to do it.

>>> all([i.isnumeric() for i in '1.2'.split('.',1)])
True
>>> all([i.isnumeric() for i in '2'.split('.',1)])
True
>>> all([i.isnumeric() for i in '2.f'.split('.',1)])
False

Edit: Im sure it will not hold up to all cases of float though especially when there is an exponent. To solve that it looks like this. This will return True only val is a float and False for int but is probably less performant than regex.

>>> def isfloat(val):
...     return all([ [any([i.isnumeric(), i in ['.','e']]) for i in val],  len(val.split('.')) == 2] )
...
>>> isfloat('1')
False
>>> isfloat('1.2')
True
>>> isfloat('1.2e3')
True
>>> isfloat('12e3')
False

回答 6

此正则表达式将检查科学的浮点数:

^[-+]?(?:\b[0-9]+(?:\.[0-9]*)?|\.[0-9]+\b)(?:[eE][-+]?[0-9]+\b)?$

但是,我相信您最好的选择是尝试使用解析器。

This regex will check for scientific floating point numbers:

^[-+]?(?:\b[0-9]+(?:\.[0-9]*)?|\.[0-9]+\b)(?:[eE][-+]?[0-9]+\b)?$

However, I believe that your best bet is to use the parser in a try.


回答 7

如果您不必担心数字的科学表达式或其他表达式,而只使用可能是带或不带句点的数字的字符串,则:

功能

def is_float(s):
    result = False
    if s.count(".") == 1:
        if s.replace(".", "").isdigit():
            result = True
    return result

Lambda版本

is_float = lambda x: x.replace('.','',1).isdigit() and "." in x

if is_float(some_string):
    some_string = float(some_string)
elif some_string.isdigit():
    some_string = int(some_string)
else:
    print "Does not convert to int or float."

这样,您就不会意外将应为int的内容转换为float。

If you don’t need to worry about scientific or other expressions of numbers and are only working with strings that could be numbers with or without a period:

Function

def is_float(s):
    result = False
    if s.count(".") == 1:
        if s.replace(".", "").isdigit():
            result = True
    return result

Lambda version

is_float = lambda x: x.replace('.','',1).isdigit() and "." in x

Example

if is_float(some_string):
    some_string = float(some_string)
elif some_string.isdigit():
    some_string = int(some_string)
else:
    print "Does not convert to int or float."

This way you aren’t accidentally converting what should be an int, into a float.


回答 8

函数的简化版本, is_digit(str)在大多数情况下就足够了(不考虑指数符号“ NaN”值):

def is_digit(str):
    return str.lstrip('-').replace('.', '').isdigit()

Simplified version of the function is_digit(str), which suffices in most cases (doesn’t consider exponential notation and “NaN” value):

def is_digit(str):
    return str.lstrip('-').replace('.', '').isdigit()

回答 9

我使用了已经提到的函数,但是很快我注意到,字符串“ Nan”,“ Inf”及其变体被视为数字。因此,我建议您对函数进行改进,使其在这些输入类型上返回false,并且不会使“ 1e3”变体失败:

def is_float(text):
    # check for nan/infinity etc.
    if text.isalpha():
        return False
    try:
        float(text)
        return True
    except ValueError:
        return False

I used the function already mentioned, but soon I notice that strings as “Nan”, “Inf” and it’s variation are considered as number. So I propose you improved version of the function, that will return false on those type of input and will not fail “1e3” variants:

def is_float(text):
    # check for nan/infinity etc.
    if text.isalpha():
        return False
    try:
        float(text)
        return True
    except ValueError:
        return False

回答 10

尝试转换为浮点数。如果有错误,请打印ValueError异常。

try:
    x = float('1.23')
    print('val=',x)
    y = float('abc')
    print('val=',y)
except ValueError as err:
    print('floatErr;',err)

输出:

val= 1.23
floatErr: could not convert string to float: 'abc'

Try to convert to float. If there is an error, print the ValueError exception.

try:
    x = float('1.23')
    print('val=',x)
    y = float('abc')
    print('val=',y)
except ValueError as err:
    print('floatErr;',err)

Output:

val= 1.23
floatErr: could not convert string to float: 'abc'

回答 11

将字典作为参数传递时,它将转换可以转换为float的字符串,并留下其他字符串

def covertDict_float(data):
        for i in data:
            if data[i].split(".")[0].isdigit():
                try:
                    data[i] = float(data[i])
                except:
                    continue
        return data

Passing dictionary as argument it will convert strings which can be converted to float and will leave others

def covertDict_float(data):
        for i in data:
            if data[i].split(".")[0].isdigit():
                try:
                    data[i] = float(data[i])
                except:
                    continue
        return data

回答 12

我一直在寻找一些类似的代码,但是看起来使用try / excepts是最好的方法。这是我正在使用的代码。如果输入无效,则包括重试功能。我需要检查输入是否大于0,是否将其转换为浮点型。

def cleanInput(question,retry=False): 
    inputValue = input("\n\nOnly positive numbers can be entered, please re-enter the value.\n\n{}".format(question)) if retry else input(question)
    try:
        if float(inputValue) <= 0 : raise ValueError()
        else : return(float(inputValue))
    except ValueError : return(cleanInput(question,retry=True))


willbefloat = cleanInput("Give me the number: ")

I was looking for some similar code, but it looks like using try/excepts is the best way. Here is the code I’m using. It includes a retry function if the input is invalid. I needed to check if the input was greater than 0 and if so convert it to a float.

def cleanInput(question,retry=False): 
    inputValue = input("\n\nOnly positive numbers can be entered, please re-enter the value.\n\n{}".format(question)) if retry else input(question)
    try:
        if float(inputValue) <= 0 : raise ValueError()
        else : return(float(inputValue))
    except ValueError : return(cleanInput(question,retry=True))


willbefloat = cleanInput("Give me the number: ")

回答 13

def try_parse_float(item):
  result = None
  try:
    float(item)
  except:
    pass
  else:
    result = float(item)
  return result
def try_parse_float(item):
  result = None
  try:
    float(item)
  except:
    pass
  else:
    result = float(item)
  return result

回答 14

我尝试了上述一些简单的选项,并使用了一个围绕转换为浮点数的try测试,发现大多数答复中都存在问题。

简单测试(沿以上答案行):

entry = ttk.Entry(self, validate='key')
entry['validatecommand'] = (entry.register(_test_num), '%P')

def _test_num(P):
    try: 
        float(P)
        return True
    except ValueError:
        return False

问题出现在以下情况:

  • 输入“-”以开始一个负数:

然后float('-'),您正在尝试失败

  • 您输入一个数字,但然后尝试删除所有数字

然后float(''),您正在尝试同样失败的尝试

我的快速解决方案是:

def _test_num(P):
    if P == '' or P == '-': return True
    try: 
        float(P)
        return True
    except ValueError:
        return False

I tried some of the above simple options, using a try test around converting to a float, and found that there is a problem in most of the replies.

Simple test (along the lines of above answers):

entry = ttk.Entry(self, validate='key')
entry['validatecommand'] = (entry.register(_test_num), '%P')

def _test_num(P):
    try: 
        float(P)
        return True
    except ValueError:
        return False

The problem comes when:

  • You enter ‘-‘ to start a negative number:

You are then trying float('-') which fails

  • You enter a number, but then try to delete all the digits

You are then trying float('') which likewise also fails

The quick solution I had is:

def _test_num(P):
    if P == '' or P == '-': return True
    try: 
        float(P)
        return True
    except ValueError:
        return False

回答 15

str(strval).isdigit()

似乎很简单。

处理以字符串或int或float形式存储的值

str(strval).isdigit()

seems to be simple.

Handles values stored in as a string or int or float