How do I check if a string is a number (float)?

Question

What is the best possible way to check if a string can be represented as a number in Python?

The function I currently have right now is:

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

Which, not only is ugly and slow, seems clunky. However I haven't found a better method because calling float in the main function is even worse.

Answer 1

In case you are looking for parsing (positive, unsigned) integers instead of floats, you can use the isdigit() function for string objects.

>>> a = "03523"
>>> a.isdigit()
True
>>> b = "963spam"
>>> b.isdigit()
False

String Methods - isdigit(): Python2, Python3

There's also something on Unicode strings, which I'm not too familiar with Unicode - Is decimal/decimal

Answer 2

Which, not only is ugly and slow

I'd dispute both.

A regex or other string parsing method would be uglier and slower.

I'm not sure that anything much could be faster than the above. It calls the function and returns. Try/Catch doesn't introduce much overhead because the most common exception is caught without an extensive search of stack frames.

The issue is that any numeric conversion function has two kinds of results

• A number, if the number is valid
• A status code (e.g., via errno) or exception to show that no valid number could be parsed.

C (as an example) hacks around this a number of ways. Python lays it out clearly and explicitly.

I think your code for doing this is perfect.

Answer 3

TL;DR The best solution is s.replace('.','',1).isdigit()

I did some benchmarks comparing the different approaches

def is_number_tryexcept(s):
    """ Returns True is string is a number. """
    try:
        float(s)
        return True
    except ValueError:
        return False

import re    
def is_number_regex(s):
    """ Returns True is string is a number. """
    if re.match("^\d+?\.\d+?$", s) is None:
        return s.isdigit()
    return True


def is_number_repl_isdigit(s):
    """ Returns True is string is a number. """
    return s.replace('.','',1).isdigit()

If the string is not a number, the except-block is quite slow. But more importantly, the try-except method is the only approach that handles scientific notations correctly.

funcs = [
          is_number_tryexcept, 
          is_number_regex,
          is_number_repl_isdigit
          ]

a_float = '.1234'

print('Float notation ".1234" is not supported by:')
for f in funcs:
    if not f(a_float):
        print('\t -', f.__name__)

Float notation ".1234" is not supported by:
- is_number_regex

scientific1 = '1.000000e+50'
scientific2 = '1e50'


print('Scientific notation "1.000000e+50" is not supported by:')
for f in funcs:
    if not f(scientific1):
        print('\t -', f.__name__)




print('Scientific notation "1e50" is not supported by:')
for f in funcs:
    if not f(scientific2):
        print('\t -', f.__name__)

Scientific notation "1.000000e+50" is not supported by:
- is_number_regex
- is_number_repl_isdigit
Scientific notation "1e50" is not supported by:
- is_number_regex
- is_number_repl_isdigit

EDIT: The benchmark results

import timeit

test_cases = ['1.12345', '1.12.345', 'abc12345', '12345']
times_n = {f.__name__:[] for f in funcs}

for t in test_cases:
    for f in funcs:
        f = f.__name__
        times_n[f].append(min(timeit.Timer('%s(t)' %f, 
                      'from __main__ import %s, t' %f)
                              .repeat(repeat=3, number=1000000)))

where the following functions were tested

from re import match as re_match
from re import compile as re_compile

def is_number_tryexcept(s):
    """ Returns True is string is a number. """
    try:
        float(s)
        return True
    except ValueError:
        return False

def is_number_regex(s):
    """ Returns True is string is a number. """
    if re_match("^\d+?\.\d+?$", s) is None:
        return s.isdigit()
    return True


comp = re_compile("^\d+?\.\d+?$")    

def compiled_regex(s):
    """ Returns True is string is a number. """
    if comp.match(s) is None:
        return s.isdigit()
    return True


def is_number_repl_isdigit(s):
    """ Returns True is string is a number. """
    return s.replace('.','',1).isdigit()

Answer 4

There is one exception that you may want to take into account: the string 'NaN'

If you want is_number to return FALSE for 'NaN' this code will not work as Python converts it to its representation of a number that is not a number (talk about identity issues):

>>> float('NaN')
nan

Otherwise, I should actually thank you for the piece of code I now use extensively. :)

G.

Answer 5

how about this:

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

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

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

will return false

edit: just saw another comment ... adding a .replace(badstuff,'',maxnum_badstuff) for other cases can be done. if you are passing salt and not arbitrary condiments (ref:xkcd#974) this will do fine :P

Answer 6

Updated after Alfe pointed out you don't need to check for float separately as complex handles both:

def is_number(s):
    try:
        complex(s) # for int, long, float and complex
    except ValueError:
        return False

    return True

---

Previously said: Is some rare cases you might also need to check for complex numbers (e.g. 1+2i), which can not be represented by a float:

def is_number(s):
    try:
        float(s) # for int, long and float
    except ValueError:
        try:
            complex(s) # for complex
        except ValueError:
            return False

    return True

Answer 7

Which, not only is ugly and slow, seems clunky.

It may take some getting used to, but this is the pythonic way of doing it. As has been already pointed out, the alternatives are worse. But there is one other advantage of doing things this way: polymorphism.

The central idea behind duck typing is that "if it walks and talks like a duck, then it's a duck." What if you decide that you need to subclass string so that you can change how you determine if something can be converted into a float? Or what if you decide to test some other object entirely? You can do these things without having to change the above code.

Other languages solve these problems by using interfaces. I'll save the analysis of which solution is better for another thread. The point, though, is that python is decidedly on the duck typing side of the equation, and you're probably going to have to get used to syntax like this if you plan on doing much programming in Python (but that doesn't mean you have to like it of course).

One other thing you might want to take into consideration: Python is pretty fast in throwing and catching exceptions compared to a lot of other languages (30x faster than .Net for instance). Heck, the language itself even throws exceptions to communicate non-exceptional, normal program conditions (every time you use a for loop). Thus, I wouldn't worry too much about the performance aspects of this code until you notice a significant problem.

Answer 8

For int use this:

>>> "1221323".isdigit()
True

But for float we need some tricks ;-). Every float number has one point...

>>> "12.34".isdigit()
False
>>> "12.34".replace('.','',1).isdigit()
True
>>> "12.3.4".replace('.','',1).isdigit()
False

Also for negative numbers just add lstrip():

>>> '-12'.lstrip('-')
'12'

And now we get a universal way:

>>> '-12.34'.lstrip('-').replace('.','',1).isdigit()
True
>>> '.-234'.lstrip('-').replace('.','',1).isdigit()
False

Answer 9

For strings of non-numbers, try: except: is actually slower than regular expressions. For strings of valid numbers, regex is slower. So, the appropriate method depends on your input.

If you find that you are in a performance bind, you can use a new third-party module called fastnumbers that provides a function called isfloat. Full disclosure, I am the author. I have included its results in the timings below.

---

from __future__ import print_function
import timeit

prep_base = '''\
x = 'invalid'
y = '5402'
z = '4.754e3'
'''

prep_try_method = '''\
def is_number_try(val):
    try:
        float(val)
        return True
    except ValueError:
        return False

'''

prep_re_method = '''\
import re
float_match = re.compile(r'[-+]?\d*\.?\d+(?:[eE][-+]?\d+)?$').match
def is_number_re(val):
    return bool(float_match(val))

'''

fn_method = '''\
from fastnumbers import isfloat

'''

print('Try with non-number strings', timeit.timeit('is_number_try(x)',
    prep_base + prep_try_method), 'seconds')
print('Try with integer strings', timeit.timeit('is_number_try(y)',
    prep_base + prep_try_method), 'seconds')
print('Try with float strings', timeit.timeit('is_number_try(z)',
    prep_base + prep_try_method), 'seconds')
print()
print('Regex with non-number strings', timeit.timeit('is_number_re(x)',
    prep_base + prep_re_method), 'seconds')
print('Regex with integer strings', timeit.timeit('is_number_re(y)',
    prep_base + prep_re_method), 'seconds')
print('Regex with float strings', timeit.timeit('is_number_re(z)',
    prep_base + prep_re_method), 'seconds')
print()
print('fastnumbers with non-number strings', timeit.timeit('isfloat(x)',
    prep_base + 'from fastnumbers import isfloat'), 'seconds')
print('fastnumbers with integer strings', timeit.timeit('isfloat(y)',
    prep_base + 'from fastnumbers import isfloat'), 'seconds')
print('fastnumbers with float strings', timeit.timeit('isfloat(z)',
    prep_base + 'from fastnumbers import isfloat'), 'seconds')
print()

---

Try with non-number strings 2.39108395576 seconds
Try with integer strings 0.375686168671 seconds
Try with float strings 0.369210958481 seconds

Regex with non-number strings 0.748660802841 seconds
Regex with integer strings 1.02021503448 seconds
Regex with float strings 1.08564686775 seconds

fastnumbers with non-number strings 0.174362897873 seconds
fastnumbers with integer strings 0.179651021957 seconds
fastnumbers with float strings 0.20222902298 seconds

As you can see

• try: except: was fast for numeric input but very slow for an invalid input
• regex is very efficient when the input is invalid
• fastnumbers wins in both cases

Answer 10

This answer provides step by step guide having function with examples to find the string is:

• Positive integer
• Positive/negative - integer/float
• How to discard "NaN" (not a number) strings while checking for number?

Check if string is positive integer

You may use str.isdigit() to check whether given string is positive integer.

Sample Results:

# For digit
>>> '1'.isdigit()
True
>>> '1'.isalpha()
False

Check for string as positive/negative - integer/float

str.isdigit() returns False if the string is a negative number or a float number. For example:

# returns `False` for float
>>> '123.3'.isdigit()
False
# returns `False` for negative number
>>> '-123'.isdigit()
False

If you want to also check for the negative integers and float, then you may write a custom function to check for it as:

def is_number(n):
    try:
        float(n)   # Type-casting the string to `float`.
                   # If string is not a valid `float`, 
                   # it'll raise `ValueError` exception
    except ValueError:
        return False
    return True

Sample Run:

>>> is_number('123')    # positive integer number
True

>>> is_number('123.4')  # positive float number
True
 
>>> is_number('-123')   # negative integer number
True

>>> is_number('-123.4') # negative `float` number
True

>>> is_number('abc')    # `False` for "some random" string
False

Discard "NaN" (not a number) strings while checking for number

The above functions will return True for the "NAN" (Not a number) string because for Python it is valid float representing it is not a number. For example:

>>> is_number('NaN')
True

In order to check whether the number is "NaN", you may use math.isnan() as:

>>> import math
>>> nan_num = float('nan')

>>> math.isnan(nan_num)
True

Or if you don't want to import additional library to check this, then you may simply check it via comparing it with itself using ==. Python returns False when nan float is compared with itself. For example:

# `nan_num` variable is taken from above example
>>> nan_num == nan_num
False

Hence, above function is_number can be updated to return False for "NaN" as:

def is_number(n):
    is_number = True
    try:
        num = float(n)
        # check for "nan" floats
        is_number = num == num   # or use `math.isnan(num)`
    except ValueError:
        is_number = False
    return is_number

Sample Run:

>>> is_number('Nan')   # not a number "Nan" string
False

>>> is_number('nan')   # not a number string "nan" with all lower cased
False

>>> is_number('123')   # positive integer
True

>>> is_number('-123')  # negative integer
True

>>> is_number('-1.12') # negative `float`
True

>>> is_number('abc')   # "some random" string
False

PS: Each operation for each check depending on the type of number comes with additional overhead. Choose the version of is_number function which fits your requirement.

Answer 11

Just Mimic C#

In C# there are two different functions that handle parsing of scalar values:

• Float.Parse()
• Float.TryParse()

float.parse():

def parse(string):
    try:
        return float(string)
    except Exception:
        throw TypeError

Note: If you're wondering why I changed the exception to a TypeError, here's the documentation.

float.try_parse():

def try_parse(string, fail=None):
    try:
        return float(string)
    except Exception:
        return fail;

Note: You don't want to return the boolean 'False' because that's still a value type. None is better because it indicates failure. Of course, if you want something different you can change the fail parameter to whatever you want.

To extend float to include the 'parse()' and 'try_parse()' you'll need to monkeypatch the 'float' class to add these methods.

If you want respect pre-existing functions the code should be something like:

def monkey_patch():
    if(!hasattr(float, 'parse')):
        float.parse = parse
    if(!hasattr(float, 'try_parse')):
        float.try_parse = try_parse

SideNote: I personally prefer to call it Monkey Punching because it feels like I'm abusing the language when I do this but YMMV.

Usage:

float.parse('giggity') // throws TypeException
float.parse('54.3') // returns the scalar value 54.3
float.tryParse('twank') // returns None
float.tryParse('32.2') // returns the scalar value 32.2

And the great Sage Pythonas said to the Holy See Sharpisus, "Anything you can do I can do better; I can do anything better than you."

Answer 12

I know this is particularly old but I would add an answer I believe covers the information missing from the highest voted answer that could be very valuable to any who find this:

For each of the following methods connect them with a count if you need any input to be accepted. (Assuming we are using vocal definitions of integers rather than 0-255, etc.)

x.isdigit() works well for checking if x is an integer.

x.replace('-','').isdigit() works well for checking if x is a negative.(Check - in first position)

x.replace('.','').isdigit() works well for checking if x is a decimal.

x.replace(':','').isdigit() works well for checking if x is a ratio.

x.replace('/','',1).isdigit() works well for checking if x is a fraction.

Answer 13

Casting to float and catching ValueError is probably the fastest way, since float() is specifically meant for just that. Anything else that requires string parsing (regex, etc) will likely be slower due to the fact that it's not tuned for this operation. My $0.02.

Answer 14

You can use Unicode strings, they have a method to do just what you want:

>>> s = u"345"
>>> s.isnumeric()
True

Or:

>>> s = "345"
>>> u = unicode(s)
>>> u.isnumeric()
True

http://www.tutorialspoint.com/python/string_isnumeric.htm

http://docs.python.org/2/howto/unicode.html

Answer 15

I wanted to see which method is fastest. Overall the best and most consistent results were given by the check_replace function. The fastest results were given by the check_exception function, but only if there was no exception fired - meaning its code is the most efficient, but the overhead of throwing an exception is quite large.

Please note that checking for a successful cast is the only method which is accurate, for example, this works with check_exception but the other two test functions will return False for a valid float:

huge_number = float('1e+100')

Here is the benchmark code:

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))

Here are the results with Python 2.7.10 on a 2017 MacBook Pro 13:

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

Here are the results with Python 3.6.5 on a 2017 MacBook Pro 13:

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

Here are the results with PyPy 2.7.13 on a 2017 MacBook Pro 13:

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

Answer 16

So to put it all together, checking for Nan, infinity and complex numbers (it would seem they are specified with j, not i, i.e. 1+2j) it results in:

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

Answer 17

The input may be as follows:

a="50" b=50 c=50.1 d="50.1"

---

1-General input:

The input of this function can be everything!

Finds whether the given variable is numeric. Numeric strings consist of optional sign, any number of digits, optional decimal part and optional exponential part. Thus +0123.45e6 is a valid numeric value. Hexadecimal (e.g. 0xf4c3b00c) and binary (e.g. 0b10100111001) notation is not allowed.

is_numeric function

import ast
import numbers              
def is_numeric(obj):
    if isinstance(obj, numbers.Number):
        return True
    elif isinstance(obj, str):
        nodes = list(ast.walk(ast.parse(obj)))[1:]
        if not isinstance(nodes[0], ast.Expr):
            return False
        if not isinstance(nodes[-1], ast.Num):
            return False
        nodes = nodes[1:-1]
        for i in range(len(nodes)):
            #if used + or - in digit :
            if i % 2 == 0:
                if not isinstance(nodes[i], ast.UnaryOp):
                    return False
            else:
                if not isinstance(nodes[i], (ast.USub, ast.UAdd)):
                    return False
        return True
    else:
        return False

test:

>>> is_numeric("54")
True
>>> is_numeric("54.545")
True
>>> is_numeric("0x45")
True

is_float function

Finds whether the given variable is float. float strings consist of optional sign, any number of digits, ...

import ast

def is_float(obj):
    if isinstance(obj, float):
        return True
    if isinstance(obj, int):
        return False
    elif isinstance(obj, str):
        nodes = list(ast.walk(ast.parse(obj)))[1:]
        if not isinstance(nodes[0], ast.Expr):
            return False
        if not isinstance(nodes[-1], ast.Num):
            return False
        if not isinstance(nodes[-1].n, float):
            return False
        nodes = nodes[1:-1]
        for i in range(len(nodes)):
            if i % 2 == 0:
                if not isinstance(nodes[i], ast.UnaryOp):
                    return False
            else:
                if not isinstance(nodes[i], (ast.USub, ast.UAdd)):
                    return False
        return True
    else:
        return False

test:

>>> is_float("5.4")
True
>>> is_float("5")
False
>>> is_float(5)
False
>>> is_float("5")
False
>>> is_float("+5.4")
True

what is ast?

---

2- If you are confident that the variable content is String:

use str.isdigit() method

>>> a=454
>>> a.isdigit()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'int' object has no attribute 'isdigit'
>>> a="454"
>>> a.isdigit()
True

---

3-Numerical input:

detect int value:

>>> isinstance("54", int)
False
>>> isinstance(54, int)
True
>>> 

detect float:

>>> isinstance("45.1", float)
False
>>> isinstance(45.1, float)
True

Answer 18

str.isnumeric()

Return True if all characters in the string are numeric characters, and there is at least one character, False otherwise. Numeric characters include digit characters, and all characters that have the Unicode numeric value property, e.g. U+2155, VULGAR FRACTION ONE FIFTH. Formally, numeric characters are those with the property value Numeric_Type=Digit, Numeric_Type=Decimal or Numeric_Type=Numeric.

str.isdecimal()

Return True if all characters in the string are decimal characters and there is at least one character, False otherwise. Decimal characters are those that can be used to form numbers in base 10, e.g. U+0660, ARABIC-INDIC DIGIT ZERO. Formally a decimal character is a character in the Unicode General Category “Nd”.

Both available for string types from Python 3.0.

Answer 19

I needed to determine if a string cast into basic types (float,int,str,bool). After not finding anything on the internet I created this:

def str_to_type (s):
    """ Get possible cast type for a string

    Parameters
    ----------
    s : string

    Returns
    -------
    float,int,str,bool : type
        Depending on what it can be cast to

    """    
    try:                
        f = float(s)        
        if "." not in s:
            return int
        return float
    except ValueError:
        value = s.upper()
        if value == "TRUE" or value == "FALSE":
            return bool
        return type(s)

Example

str_to_type("true") # bool
str_to_type("6.0") # float
str_to_type("6") # int
str_to_type("6abc") # str
str_to_type(u"6abc") # unicode       

You can capture the type and use it

s = "6.0"
type_ = str_to_type(s) # float
f = type_(s) 

Answer 20

I did some speed test. Lets say that if the string is likely to be a number the try/except strategy is the fastest possible.If the string is not likely to be a number and you are interested in Integer check, it worths to do some test (isdigit plus heading '-'). If you are interested to check float number, you have to use the try/except code whitout escape.

Answer 21

I think your solution is fine, but there is a correct regexp implementation.

There does seem to be a lot of regexp hate towards these answers which I think is unjustified, regexps can be reasonably clean and correct and fast. It really depends on what you're trying to do. The original question was how can you "check if a string can be represented as a number (float)" (as per your title). Presumably you would want to use the numeric/float value once you've checked that it's valid, in which case your try/except makes a lot of sense. But if, for some reason, you just want to validate that a string is a number then a regex also works fine, but it's hard to get correct. I think most of the regex answers so far, for example, do not properly parse strings without an integer part (such as ".7") which is a float as far as python is concerned. And that's slightly tricky to check for in a single regex where the fractional portion is not required. I've included two regex to show this.

It does raise the interesting question as to what a "number" is. Do you include "inf" which is valid as a float in python? Or do you include numbers that are "numbers" but maybe can't be represented in python (such as numbers that are larger than the float max).

There's also ambiguities in how you parse numbers. For example, what about "--20"? Is this a "number"? Is this a legal way to represent "20"? Python will let you do "var = --20" and set it to 20 (though really this is because it treats it as an expression), but float("--20") does not work.

Anyways, without more info, here's a regex that I believe covers all the ints and floats as python parses them.

# Doesn't properly handle floats missing the integer part, such as ".7"
SIMPLE_FLOAT_REGEXP = re.compile(r'^[-+]?[0-9]+\.?[0-9]+([eE][-+]?[0-9]+)?$')
# Example "-12.34E+56"      # sign (-)
                            #     integer (12)
                            #           mantissa (34)
                            #                    exponent (E+56)

# Should handle all floats
FLOAT_REGEXP = re.compile(r'^[-+]?([0-9]+|[0-9]*\.[0-9]+)([eE][-+]?[0-9]+)?$')
# Example "-12.34E+56"      # sign (-)
                            #     integer (12)
                            #           OR
                            #             int/mantissa (12.34)
                            #                            exponent (E+56)

def is_float(str):
  return True if FLOAT_REGEXP.match(str) else False

Some example test values:

True  <- +42
True  <- +42.42
False <- +42.42.22
True  <- +42.42e22
True  <- +42.42E-22
False <- +42.42e-22.8
True  <- .42
False <- 42nope

Running the benchmarking code in @ron-reiter's answer shows that this regex is actually faster than the normal regex and is much faster at handling bad values than the exception, which makes some sense. Results:

check_regexp with good floats: 18.001921
check_regexp with bad floats: 17.861423
check_regexp with strings: 17.558862
check_correct_regexp with good floats: 11.04428
check_correct_regexp with bad floats: 8.71211
check_correct_regexp with strings: 8.144161
check_replace with good floats: 6.020597
check_replace with bad floats: 5.343049
check_replace with strings: 5.091642
check_exception with good floats: 5.201605
check_exception with bad floats: 23.921864
check_exception with strings: 23.755481

Answer 22

RyanN suggests

If you want to return False for a NaN and Inf, change line to x = float(s); return (x == x) and (x - 1 != x). This should return True for all floats except Inf and NaN

But this doesn't quite work, because for sufficiently large floats, x-1 == x returns true. For example, 2.0**54 - 1 == 2.0**54

Answer 23

This code handles the exponents, floats, and integers, wihtout using regex.

return True if str1.lstrip('-').replace('.','',1).isdigit() or float(str1) else False

Answer 24

import re
def is_number(num):
    pattern = re.compile(r'^[-+]?[-0-9]\d*\.\d*|[-+]?\.?[0-9]\d*$')
    result = pattern.match(num)
    if result:
        return True
    else:
        return False


​>>>: is_number('1')
True

>>>: is_number('111')
True

>>>: is_number('11.1')
True

>>>: is_number('-11.1')
True

>>>: is_number('inf')
False

>>>: is_number('-inf')
False

Answer 25

In a most general case for a float, one would like to take care of integers and decimals. Let's take the string "1.1" as an example.

I would try one of the following:

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

Speed:

► All the aforementioned methods have similar speeds.

%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)

Answer 26

Here's my simple way of doing it. Let's say that I'm looping through some strings and I want to add them to an array if they turn out to be numbers.

try:
    myvar.append( float(string_to_check) )
except:
    continue

Replace the myvar.apppend with whatever operation you want to do with the string if it turns out to be a number. The idea is to try to use a float() operation and use the returned error to determine whether or not the string is a number.

Answer 27

I was working on a problem that led me to this thread, namely how to convert a collection of data to strings and numbers in the most intuitive way. I realized after reading the original code that what I needed was different in two ways:

1 - I wanted an integer result if the string represented an integer

2 - I wanted a number or a string result to stick into a data structure

so I adapted the original code to produce this derivative:

def string_or_number(s):
    try:
        z = int(s)
        return z
    except ValueError:
        try:
            z = float(s)
            return z
        except ValueError:
            return s

Answer 28

I also used the function you 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 your function, that will return false on those type of input and will not fail "1e3" variants:

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

Answer 29

User helper function:

def if_ok(fn, string):
  try:
    return fn(string)
  except Exception as e:
    return None

then

if_ok(int, my_str) or if_ok(float, my_str) or if_ok(complex, my_str)
is_number = lambda s: any([if_ok(fn, s) for fn in (int, float, complex)])

Answer 30

You can generalize the exception technique in a useful way by returning more useful values than True and False. For example this function puts quotes round strings but leaves numbers alone. Which is just what I needed for a quick and dirty filter to make some variable definitions for R.

import sys

def fix_quotes(s):
    try:
        float(s)
        return s
    except ValueError:
        return '"{0}"'.format(s)

for line in sys.stdin:
    input = line.split()
    print input[0], '<- c(', ','.join(fix_quotes(c) for c in input[1:]), ')'

Answer 31

Try this.

 def is_number(var):
    try:
       if var == int(var):
            return True
    except Exception:
        return False

Answer 32

Sorry for the Zombie thread post - just wanted to round out the code for completeness...

# is_number() function - Uses re = regex library
# Should handle all normal and complex numbers
# Does not accept trailing spaces. 
# Note: accepts both engineering "j" and math "i" but only the imaginary part "+bi" of a complex number a+bi
# Also accepts inf or NaN
# Thanks to the earlier responders for most the regex fu

import re

ISNUM_REGEXP = re.compile(r'^[-+]?([0-9]+|[0-9]*\.[0-9]+)([eE][-+]?[0-9]+)?[ij]?$')

def is_number(str):
#change order if you have a lot of NaN or inf to parse
    if ISNUM_REGEXP.match(str) or str == "NaN" or str == "inf": 
        return True 
    else:
        return False
# A couple test numbers
# +42.42e-42j
# -42.42E+42i

print('Is it a number?', is_number(input('Gimme any number: ')))

Gimme any number: +42.42e-42j

Is it a number? True

Answer 33

def is_float(s):
    if s is None:
        return False

    if len(s) == 0:
        return False

    digits_count = 0
    dots_count = 0
    signs_count = 0

    for c in s:
        if '0' <= c <= '9':
            digits_count += 1
        elif c == '.':
            dots_count += 1
        elif c == '-' or c == '+':
            signs_count += 1
        else:
            return False

    if digits_count == 0:
        return False

    if dots_count > 1:
        return False

    if signs_count > 1:
        return False

    return True

Answer 34

I have a similar problem. Instead of defining a isNumber function, I want to convert a list of strings to floats, something that in high-level terms would be:

[ float(s) for s in list if isFloat(s)]

It is a given we can not really separate the float(s) from the isFloat(s) functions: these two results should be returned by the same function. Also, if float(s) fails, the whole process fails, instead of just ignoring the faulty element. Plus, "0" is a valid number and should be included in the list. When filtering out bad elements, be certain not to exclude 0.

Therefore, the above comprehension must be modified somehow to:

• if any element in the list cannot be converted, ignore it and don't throw an exception
• avoid calling float(s) more than once for each element (one for the conversion, the other for the test)
• if the converted value is 0, it should still be present in the final list

I propose a solution inspired in the Nullable numerical types of C#. These types are internally represented by a struct that has the numerical value and adds a boolean indicating if the value is valid:

def tryParseFloat(s):
    try:
        return(float(s), True)
    except:
        return(None, False)

tupleList = [tryParseFloat(x) for x in list]
floats = [v for v,b in tupleList if b]

Answer 35

use following it handles all cases:-

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')