Checking if a string can be converted to float in Python

I would just use..

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

..it's simple, and it works. Note that it will still throw OverflowError if element is e.g. 1<<1024.

Another option would be a regular expression:

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

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

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()

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()

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.

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

You can use the try-except-else clause , this will catch any conversion/ value errors raised when the value passed cannot be converted to a float

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

a simple function that get you the type of number without try and except operation

def number_type(number):
    if number.isdigit():
        return int(number)
    elif number.replace(".","").isdigit():
        return float(number)
    else:
        return(type(number))

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'

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

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

It seems many regex given miss one thing or another. This has been working for me so far:

(?i)^\s*[+-]?(?:inf(inity)?|nan|(?:\d+\.?\d*|\.\d+)(?:e[+-]?\d+)?)\s*$

It allows for infinity (or inf) with sign, nan, no digit before the decimal, and leading/trailing spaces (if desired). The ^ and $ are needed to keep from partially matching something like 1.2f-2 as 1.2.

You could use [ed] instead of just e if you need to parse some files where D is used for double-precision scientific notation. You would want to replace it afterward or just replace them before checking since the float() function won't allow it.

I found a way that could also work. need to verify this. first time putting something here.

def isfloat(a_str):
    try:
        x=float(a_str)
        if x%1 == 0:
            return False
        elif x%1 != 0: #an else also do
            return True
    except Exception as error:
            return False

This works like a charm:

[dict([a,int(x) if isinstance(x, str)
 and x.isnumeric() else float(x) if isinstance(x, str)
 and x.replace('.', '', 1).isdigit() else x] for a, x in json_data.items())][0]

I've written my own functions. Instead of float(value), I use floatN() or floatZ(). which return None or 0.0 if the value can't be cast as a float. I keep them in a module I've called safeCasts.

def floatN(value):
    try:
        if value is not None:
            fvalue = float(value)
        else:
            fvalue = None
    except ValueError:
        fvalue = None

    return fvalue


def floatZ(value):
    try:
        if value is not None:
            fvalue = float(value)
        else:
            fvalue = 0.0
    except ValueError:
        fvalue = 0.0

    return fvalue

In other modules I import them

from safeCasts import floatN, floatZ

then use floatN(value) or floatZ(value) instead of float(). Obviously, you can use this technique for any cast function you need.

It's a simple, yet interesting question. Solution presented below works fine for me:

import re

val = "25,000.93$"

regex = r"\D"

splitted = re.split(regex, val)
splitted = list(filter(str.isdecimal, splitted))

if splitted:
    if len(splitted) > 1:
        splitted.insert(-1, ".")

    try:
        f = float("".join(splitted))
        print(f, "is float.")
        
    except ValueError:
        print("Not a float.")
        
else:
    print("Not a float.")

Important note: this solution is based on assumption that the last value in splitted contains decimal places.

You can create a function isfloat(), and use in place of isdigit() for both integers and floats, but not strings as you expect.

a = raw_input('How much is 1 share in that company? \n')

def isfloat(num):
    try:
        float(num)
        return True
    except:
        return False
       
while not isfloat(a):
    print("You need to write a number!\n")
    a = raw_input('How much is 1 share in that company? \n')

We can use regex as: import re if re.match('[0-9]*.?[0-9]+', <your_string>): print("Its a float/int") else: print("Its something alien") let me explain the regex in english,

• * -> 0 or more occurence
• + -> 1 or more occurence
• ? -> 0/1 occurence

now, lets convert

• '[0-9]* -> let there be 0 or more occurence of digits in between 0-9
• \.? -> followed by a 0 or one '.'(if you need to check if it can be int/float else we can also use instead of ?, use {1})
• [0-9]+ -> followed by 0 or more occurence of digits in between 0-9