How do I check if a string represents a numeric value in Python?
def is_number(s):
try:
float(s)
return True
except ValueError:
return False
The above works, but it seems clunky.
If what you are testing comes from user input, it is still a string even if it represents an int
or a float
. See How can I read inputs as numbers? for converting the input, and Asking the user for input until they give a valid response for ensuring that the input represents an int
or float
(or other requirements) before proceeding.
x = float('0.00'); if x: use_float(x);
you've now got a bug in your code. Truthy values are the reason these functions raise an exception rather than returning None
in the first place. A better solution is just to avoid the utility function and surround the call to float in a try catch
when you want to use it. - anyone For non-negative (unsigned) integers only, use isdigit()
:
>>> a = "03523"
>>> a.isdigit()
True
>>> b = "963spam"
>>> b.isdigit()
False
Documentation for isdigit()
: Python2, Python3
For Python 2 Unicode strings:
isnumeric()
.
Answered 2023-09-21 08:07:29
isdigit()
and int()
have different opinions about what is an integer e.g., for the Unicode character u'\u00b9'
: u'¹'.isdigit()
is True
but int(u'¹')
raises ValueError. - anyone 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
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.
Answered 2023-09-21 08:07:29
try
clause, so I would put the return True
in an else
clause of the try
. One of the reasons is that with the code in the question, if I had to review it, I would have to check that the second statement in the try
clause cannot raise a ValueError: granted, this does not require too much time or brain power, but why use any when none is needed? - anyone IsNumeric()
I either end up with a try/catch or another wrapping a try/catch. Ugh - anyone if is_number(s): x = float(x) else: // fail
is the same number of lines of code as the try: x = float(x) catch TypeError: # fail
. This utility function is an entirely unecessary abstraction. - anyone 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 if string is a number. """
try:
float(s)
return True
except ValueError:
return False
import re
def is_number_regex(s):
""" Returns True if string is a number. """
if re.match("^\d+?\.\d+?$", s) is None:
return s.isdigit()
return True
def is_number_repl_isdigit(s):
""" Returns True if string is a number. """
return s.replace('.','',1).isdigit()
If the string is not a number, the except-block is quite slow. But more importantly, the try-except method is the only approach that handles scientific notations correctly.
funcs = [
is_number_tryexcept,
is_number_regex,
is_number_repl_isdigit
]
a_float = '.1234'
print('Float notation ".1234" is not supported by:')
for f in funcs:
if not f(a_float):
print('\t -', f.__name__)
Float notation ".1234" is not supported by:
is_number_regex
scientific1 = '1.000000e+50' scientific2 = '1e50'
print('Scientific notation "1.000000e+50" is not supported by:') for f in funcs: if not f(scientific1): print('\t -', f.name)
print('Scientific notation "1e50" is not supported by:') for f in funcs: if not f(scientific2): print('\t -', f.name)
Scientific notation "1.000000e+50" is not supported by:
import timeit
test_cases = ['1.12345', '1.12.345', 'abc12345', '12345']
times_n = {f.__name__:[] for f in funcs}
for t in test_cases:
for f in funcs:
f = f.__name__
times_n[f].append(min(timeit.Timer('%s(t)' %f,
'from __main__ import %s, t' %f)
.repeat(repeat=3, number=1000000)))
where the following functions were tested
from re import match as re_match
from re import compile as re_compile
def is_number_tryexcept(s):
""" Returns True if string is a number. """
try:
float(s)
return True
except ValueError:
return False
def is_number_regex(s):
""" Returns True if string is a number. """
if re_match("^\d+?\.\d+?$", s) is None:
return s.isdigit()
return True
comp = re_compile("^\d+?\.\d+?$")
def compiled_regex(s):
""" Returns True if string is a number. """
if comp.match(s) is None:
return s.isdigit()
return True
def is_number_repl_isdigit(s):
""" Returns True if string is a number. """
return s.replace('.','',1).isdigit()
Answered 2023-09-21 08:07:29
'1.5e-9'
or on negatives. - anyone s.replace()
calls. For example, s.replace('.','',1).replace('e-','',1).replace('e','',1).isdigit()
handles exponentiation. To then handle negatives, just left-strip the first character if a dash. For example, s.lstrip('-').replace('.','',1).replace('e-','',1).replace('e','',1).isdigit()
. Yes, I have exhaustively tested that one-liner and can confirm it behaves as expected. - anyone 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.
Answered 2023-09-21 08:07:29
NaN
might be a good value to return (rather than False
) if the text passed is not in fact a representation of a number. Checking for it is kind of a pain (Python's float
type really needs a method for it) but you can use it in calculations without producing an error, and only need to check the result. - anyone 'inf'
. Either inf
or NaN
can also be prefixed with a +
or -
and still be accepted. - anyone x-1 == x
is true for large floats smaller than inf
. From Python 3.2 you can use math.isfinite
to test for numbers that are neither NaN nor infinite, or check both math.isnan
and math.isinf
prior to that. - anyone 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
Answered 2023-09-21 08:07:29
1.234e56
(which might also be written as +1.234E+56
and several more variants). - anyone re.match(r'^[+-]*(0[xbo])?[0-9A-Fa-f]*\.?[0-9A-Fa-f]*(E[+-]*[0-9A-Fa-f]+)$', 'str')
should do a better job of determining a number (but not all, I'm not claiming that). I don't recommend using this, far better to use the Questioner's original code. - anyone 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
Answered 2023-09-21 08:07:29
float()
stuff completely and just check for the complex()
call to succeed. Everything parsed by float()
can be parsed by complex()
. - anyone complex('(01989)')
will return (1989+0j)
. But float('(01989)')
will fail. So I think using complex
is not good idea. - anyone complex()
accepts (
- and )
-delimited syntax – presumably to account for the composite vector addition in the imaginary plane underlying complex numbers, but still. As @plhn suggests, using complex()
here invites false positives. Don't do this in production code. Honestly, s.lstrip('-').replace('.','',1).replace('e-','',1).replace('e','',1).isdigit()
remains the optimal solution for most use cases. - anyone 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.
Answered 2023-09-21 08:07:29
hasattr()
which is just a getattr()
call wrapped in a try/except
. Still, exception handling is slower than normal flow control, so using it for something that is going to be true most of the time can result in a performance penalty. - anyone 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
Answered 2023-09-21 08:07:29
1.234e56
and similar. Also, I'd be interested how you'd find out that 99999999999999999999e99999999999999999999
is not a number. Trying to parse it finds out quickly. - anyone >>> '--1234'.lstrip('-').replace('.','',1).isdigit()
returns true; perhaps not what's expected - anyone This answer provides step by step guide having function with examples to find the string is:
You may use str.isdigit()
to check whether given string is positive integer.
Sample Results:
# For digit
>>> '1'.isdigit()
True
>>> '1'.isalpha()
False
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
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.
Answered 2023-09-21 08:07:29
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 inputfastnumbers
wins in both casesAnswered 2023-09-21 08:07:29
prep_code_basis
and prep_code_re_method
would have prevented my mistake. - anyone isfloat
function? - anyone str(s).strip('-').replace('.','',1).isdigit()
which is approximately 10x slower! - anyone timeit.timeit
runs the statement 1 million times. I was confused why these numbers seemed so slow. - anyone 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.
Answered 2023-09-21 08:07:29
x.replace('/','',1).isdigit()
or otherwise dates such as 4/7/2017 would be misinterpreted as numbers. - anyone x.replace('.','',1).isdigit()
rather than x.replace('.','').isdigit()
, and similarly, a more delicate check for things that aren't negative numbers such as ranges like 5-9 ("5 to 9") in contrast with -59 (the actual negative 59). - anyone In C# there are two different functions that handle parsing of scalar values:
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."
Answered 2023-09-21 08:07:29
!
instead of not
might be a minor error, but you definitely can't assign attributes to the built-in float
in CPython. - anyone 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 $0.02.
Answered 2023-09-21 08:07:29
You can use Unicode strings, they have a method to do just what you want:
>>> s = u"345"
>>> s.isnumeric()
True
Or:
>>> s = "345"
>>> u = unicode(s)
>>> u.isnumeric()
True
Answered 2023-09-21 08:07:29
s.isdecimal()
checks if s
string is a non-negative integer. s.isnumeric()
includes characters that int()
rejects. - anyone I wanted to see which method is fastest. Overall the best and most consistent results were given by the check_replace
function. The fastest results were given by the check_exception
function, but only if there was no exception fired - meaning its code is the most efficient, but the overhead of throwing an exception is quite large.
Please note that checking for a successful cast is the only method which is accurate, for example, this works with check_exception
but the other two test functions will return False for a valid float:
huge_number = float('1e+100')
Here is the benchmark code:
import time, re, random, string
ITERATIONS = 10000000
class Timer:
def __enter__(self):
self.start = time.clock()
return self
def __exit__(self, *args):
self.end = time.clock()
self.interval = self.end - self.start
def check_regexp(x):
return re.compile("^\d*\.?\d*$").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
Answered 2023-09-21 08:07:29
The input may be as follows:
a="50"
b=50
c=50.1
d="50.1"
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?
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
detect int value:
>>> isinstance("54", int)
False
>>> isinstance(54, int)
True
>>>
detect float:
>>> isinstance("45.1", float)
False
>>> isinstance(45.1, float)
True
Answered 2023-09-21 08:07:29
ast
"? - anyone is_numeric("String 1")
Wrapped the method in try/except and works. - anyone 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
Answered 2023-09-21 08:07:29
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)
Answered 2023-09-21 08:07:29
1.1.1
, which is not a valid number. - anyone 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
Answered 2023-09-21 08:07:29
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.
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.
Answered 2023-09-21 08:07:29
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)
Answered 2023-09-21 08:07:29
str_to_type("123e-4")
returns int
while float("123e-4")
returns 0.0123
. Also, str_to_type("NaN")
returns int
. - anyone if "." not in s:
to if "." not in s and "e" not in s and "N" not in s:
fixes the cases I show here, but it makes the code clunkier and I'm not sure it will catch all the cases. - anyone i = int(s); return int
into a first try/except, then put f = float(s); return float
inside a second try/except, all the the except block of the first try/except. Like the structure of user1508746's answer, with different return values. - anyone 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.
Answered 2023-09-21 08:07:29
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
Answered 2023-09-21 08:07:29
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
Answered 2023-09-21 08:07:29
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
Answered 2023-09-21 08:07:29
1e6
to represent a number? - anyone This code handles the exponents, floats, and integers, wihtout using regex.
return True if str1.lstrip('-').replace('.','',1).isdigit() or float(str1) else False
Answered 2023-09-21 08:07:29
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.
Answered 2023-09-21 08:07:29
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
Answered 2023-09-21 08:07:29
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)])
Answered 2023-09-21 08:07:29
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
Answered 2023-09-21 08:07:29