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

The best answers to the question “How do I check if a string is a number (float)?” in the category Dev.

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, but also seems clunky. However, I haven’t found a better method because calling `float()` in the main function is even worse.

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.

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

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.

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()
`````` 