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

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 2

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 3

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 4

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 5

Your code looks fine to me.

Perhaps you think the code is "clunky" because of using exceptions? Note that Python programmers tend to use exceptions liberally when it improves code readability, thanks to its low performance penalty.

Answer 6

Try this.

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