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Cast to Integer Only If "Lossless"?

I wish to cast a string or a number to an integer only if the casting is "lossless" or, another way to put it, only if the string or number is indeed an integer.
For instance,
3.0 (a float that is indeed an integer) -> 3.
'3.000' (a string that is an integer) -> 3.
3.1 -> exception raised.
'4.2' -> exception raised.
Directly doing int(x) will convert 3.1 to 3.
This is the best I have:
def safe_cast_to_int(x):
int_x = int(x)
if np.issubdtype(type(x), np.floating):
assert int_x == x, \
f"Can't safely cast a non-integer value ({x}) to integer"
return int_x
but I wonder if there is a better or more Pythonic way?

If I understand you correctly, you only want to cast something if it's a whole number.
If that's the case, you could first cast it to a float and then check with float.is_integer() function if it's an integer.
Here are the examples with values of the question.
>>> float('3.0').is_integer()
True
>>> float('3.000').is_integer()
True
>>> float('3.1').is_integer()
False
>>> float('4.2').is_integer()
False

You could convert to float and modulus the data with 1 to check if you want to keep it a float
val = float(src)
val = int(val) if not val%1 else val
Edit: is_integer() is just doing the below for you, but with a bunch of conditions and flags attached before it gets to this line.
o = (floor(x) == x) ? Py_True : Py_False;

If you want things that look like integers, but aren't really integer values, as in (1.03-0.42)*100, then you need to test to see how "near" an integer a value is, and accept anything close. How close you accept as "integer" will depend on your exact use case:
import sys
tests = [
42,
'1.00',
3.2,
(1.03-0.42)*100,
]
for x in tests:
is_close_to_int = abs(int(float(x))-float(x))<0.00000000001
print(f"{x}: {float(x).is_integer()} {is_close_to_int}")
This outputs:
42: True True
1.00: True True
3.2: False False
61.00000000000001: False True
Showing that for many cases float's is_integer helper will do the right thing, but for some edge cases, a person might expect different results.

Using input/variables with exponents [duplicate]

How can I convert a str to float?
"545.2222" → 545.2222
How can I convert a str to int?
"31" → 31
For the reverse, see Convert integer to string in Python and Converting a float to a string without rounding it.
Please instead use How can I read inputs as numbers? to close duplicate questions where OP received a string from user input and immediately wants to convert it, or was hoping for input (in 3.x) to convert the type automatically.

>>> a = "545.2222"
>>> float(a)
545.22220000000004
>>> int(float(a))
545

Python2 method to check if a string is a float:
def is_float(value):
if value is None:
return False
try:
float(value)
return True
except:
return False
For the Python3 version of is_float see: Checking if a string can be converted to float in Python
A longer and more accurate name for this function could be: is_convertible_to_float(value)
What is, and is not a float in Python may surprise you:
The below unit tests were done using python2. Check it that Python3 has different behavior for what strings are convertable to float. One confounding difference is that any number of interior underscores are now allowed: (float("1_3.4") == float(13.4)) is True
val is_float(val) Note
-------------------- ---------- --------------------------------
"" False Blank string
"127" True Passed string
True True Pure sweet Truth
"True" False Vile contemptible lie
False True So false it becomes true
"123.456" True Decimal
" -127 " True Spaces trimmed
"\t\n12\r\n" True whitespace ignored
"NaN" True Not a number
"NaNanananaBATMAN" False I am Batman
"-iNF" True Negative infinity
"123.E4" True Exponential notation
".1" True mantissa only
"1_2_3.4" False Underscores not allowed
"12 34" False Spaces not allowed on interior
"1,234" False Commas gtfo
u'\x30' True Unicode is fine.
"NULL" False Null is not special
0x3fade True Hexadecimal
"6e7777777777777" True Shrunk to infinity
"1.797693e+308" True This is max value
"infinity" True Same as inf
"infinityandBEYOND" False Extra characters wreck it
"12.34.56" False Only one dot allowed
u'四' False Japanese '4' is not a float.
"#56" False Pound sign
"56%" False Percent of what?
"0E0" True Exponential, move dot 0 places
0**0 True 0___0 Exponentiation
"-5e-5" True Raise to a negative number
"+1e1" True Plus is OK with exponent
"+1e1^5" False Fancy exponent not interpreted
"+1e1.3" False No decimals in exponent
"-+1" False Make up your mind
"(1)" False Parenthesis is bad
You think you know what numbers are? You are not so good as you think! Not big surprise.
Don't use this code on life-critical software!
Catching broad exceptions this way, killing canaries and gobbling the exception creates a tiny chance that a valid float as string will return false. The float(...) line of code can failed for any of a thousand reasons that have nothing to do with the contents of the string. But if you're writing life-critical software in a duck-typing prototype language like Python, then you've got much larger problems.

def num(s):
try:
return int(s)
except ValueError:
return float(s)

This is another method which deserves to be mentioned here, ast.literal_eval:
This can be used for safely evaluating strings containing Python expressions from untrusted sources without the need to parse the values oneself.
That is, a safe 'eval'
>>> import ast
>>> ast.literal_eval("545.2222")
545.2222
>>> ast.literal_eval("31")
31

Localization and commas
You should consider the possibility of commas in the string representation of a number, for cases like float("545,545.2222") which throws an exception. Instead, use methods in locale to convert the strings to numbers and interpret commas correctly. The locale.atof method converts to a float in one step once the locale has been set for the desired number convention.
Example 1 -- United States number conventions
In the United States and the UK, commas can be used as a thousands separator. In this example with American locale, the comma is handled properly as a separator:
>>> import locale
>>> a = u'545,545.2222'
>>> locale.setlocale(locale.LC_ALL, 'en_US.UTF-8')
'en_US.UTF-8'
>>> locale.atof(a)
545545.2222
>>> int(locale.atof(a))
545545
>>>
Example 2 -- European number conventions
In the majority of countries of the world, commas are used for decimal marks instead of periods. In this example with French locale, the comma is correctly handled as a decimal mark:
>>> import locale
>>> b = u'545,2222'
>>> locale.setlocale(locale.LC_ALL, 'fr_FR')
'fr_FR'
>>> locale.atof(b)
545.2222
The method locale.atoi is also available, but the argument should be an integer.

float(x) if '.' in x else int(x)

If you aren't averse to third-party modules, you could check out the fastnumbers module. It provides a function called fast_real that does exactly what this question is asking for and does it faster than a pure-Python implementation:
>>> from fastnumbers import fast_real
>>> fast_real("545.2222")
545.2222
>>> type(fast_real("545.2222"))
float
>>> fast_real("31")
31
>>> type(fast_real("31"))
int

Users codelogic and harley are correct, but keep in mind if you know the string is an integer (for example, 545) you can call int("545") without first casting to float.
If your strings are in a list, you could use the map function as well.
>>> x = ["545.0", "545.6", "999.2"]
>>> map(float, x)
[545.0, 545.60000000000002, 999.20000000000005]
>>>
It is only good if they're all the same type.

In Python, how can I parse a numeric string like "545.2222" to its corresponding float value, 542.2222? Or parse the string "31" to an integer, 31?
I just want to know how to parse a float string to a float, and (separately) an int string to an int.
It's good that you ask to do these separately. If you're mixing them, you may be setting yourself up for problems later. The simple answer is:
"545.2222" to float:
>>> float("545.2222")
545.2222
"31" to an integer:
>>> int("31")
31
Other conversions, ints to and from strings and literals:
Conversions from various bases, and you should know the base in advance (10 is the default). Note you can prefix them with what Python expects for its literals (see below) or remove the prefix:
>>> int("0b11111", 2)
31
>>> int("11111", 2)
31
>>> int('0o37', 8)
31
>>> int('37', 8)
31
>>> int('0x1f', 16)
31
>>> int('1f', 16)
31
If you don't know the base in advance, but you do know they will have the correct prefix, Python can infer this for you if you pass 0 as the base:
>>> int("0b11111", 0)
31
>>> int('0o37', 0)
31
>>> int('0x1f', 0)
31
Non-Decimal (i.e. Integer) Literals from other Bases
If your motivation is to have your own code clearly represent hard-coded specific values, however, you may not need to convert from the bases - you can let Python do it for you automatically with the correct syntax.
You can use the apropos prefixes to get automatic conversion to integers with the following literals. These are valid for Python 2 and 3:
Binary, prefix 0b
>>> 0b11111
31
Octal, prefix 0o
>>> 0o37
31
Hexadecimal, prefix 0x
>>> 0x1f
31
This can be useful when describing binary flags, file permissions in code, or hex values for colors - for example, note no quotes:
>>> 0b10101 # binary flags
21
>>> 0o755 # read, write, execute perms for owner, read & ex for group & others
493
>>> 0xffffff # the color, white, max values for red, green, and blue
16777215
Making ambiguous Python 2 octals compatible with Python 3
If you see an integer that starts with a 0, in Python 2, this is (deprecated) octal syntax.
>>> 037
31
It is bad because it looks like the value should be 37. So in Python 3, it now raises a SyntaxError:
>>> 037
File "<stdin>", line 1
037
^
SyntaxError: invalid token
Convert your Python 2 octals to octals that work in both 2 and 3 with the 0o prefix:
>>> 0o37
31

The question seems a little bit old. But let me suggest a function, parseStr, which makes something similar, that is, returns integer or float and if a given ASCII string cannot be converted to none of them it returns it untouched. The code of course might be adjusted to do only what you want:
>>> import string
>>> parseStr = lambda x: x.isalpha() and x or x.isdigit() and \
... int(x) or x.isalnum() and x or \
... len(set(string.punctuation).intersection(x)) == 1 and \
... x.count('.') == 1 and float(x) or x
>>> parseStr('123')
123
>>> parseStr('123.3')
123.3
>>> parseStr('3HC1')
'3HC1'
>>> parseStr('12.e5')
1200000.0
>>> parseStr('12$5')
'12$5'
>>> parseStr('12.2.2')
'12.2.2'

float("545.2222") and int(float("545.2222"))

The YAML parser can help you figure out what datatype your string is. Use yaml.load(), and then you can use type(result) to test for type:
>>> import yaml
>>> a = "545.2222"
>>> result = yaml.load(a)
>>> result
545.22220000000004
>>> type(result)
<type 'float'>
>>> b = "31"
>>> result = yaml.load(b)
>>> result
31
>>> type(result)
<type 'int'>
>>> c = "HI"
>>> result = yaml.load(c)
>>> result
'HI'
>>> type(result)
<type 'str'>

I use this function for that
import ast
def parse_str(s):
try:
return ast.literal_eval(str(s))
except:
return
It will convert the string to its type
value = parse_str('1') # Returns Integer
value = parse_str('1.5') # Returns Float

def get_int_or_float(v):
number_as_float = float(v)
number_as_int = int(number_as_float)
return number_as_int if number_as_float == number_as_int else number_as_float

def num(s):
"""num(s)
num(3),num(3.7)-->3
num('3')-->3, num('3.7')-->3.7
num('3,700')-->ValueError
num('3a'),num('a3'),-->ValueError
num('3e4') --> 30000.0
"""
try:
return int(s)
except ValueError:
try:
return float(s)
except ValueError:
raise ValueError('argument is not a string of number')

You could use json.loads:
>>> import json
>>> json.loads('123.456')
123.456
>>> type(_)
<class 'float'>
>>>
As you can see it becomes a type of float.

You need to take into account rounding to do this properly.
i.e. - int(5.1) => 5
int(5.6) => 5 -- wrong, should be 6 so we do int(5.6 + 0.5) => 6
def convert(n):
try:
return int(n)
except ValueError:
return float(n + 0.5)

To typecast in Python use the constructor functions of the type, passing the string (or whatever value you are trying to cast) as a parameter.
For example:
>>>float("23.333")
23.333
Behind the scenes, Python is calling the objects __float__ method, which should return a float representation of the parameter. This is especially powerful, as you can define your own types (using classes) with a __float__ method so that it can be casted into a float using float(myobject).

Handles hex, octal, binary, decimal, and float
This solution will handle all of the string conventions for numbers (all that I know about).
def to_number(n):
''' Convert any number representation to a number
This covers: float, decimal, hex, and octal numbers.
'''
try:
return int(str(n), 0)
except:
try:
# Python 3 doesn't accept "010" as a valid octal. You must use the
# '0o' prefix
return int('0o' + n, 0)
except:
return float(n)
This test case output illustrates what I'm talking about.
======================== CAPTURED OUTPUT =========================
to_number(3735928559) = 3735928559 == 3735928559
to_number("0xFEEDFACE") = 4277009102 == 4277009102
to_number("0x0") = 0 == 0
to_number(100) = 100 == 100
to_number("42") = 42 == 42
to_number(8) = 8 == 8
to_number("0o20") = 16 == 16
to_number("020") = 16 == 16
to_number(3.14) = 3.14 == 3.14
to_number("2.72") = 2.72 == 2.72
to_number("1e3") = 1000.0 == 1000
to_number(0.001) = 0.001 == 0.001
to_number("0xA") = 10 == 10
to_number("012") = 10 == 10
to_number("0o12") = 10 == 10
to_number("0b01010") = 10 == 10
to_number("10") = 10 == 10
to_number("10.0") = 10.0 == 10
to_number("1e1") = 10.0 == 10
Here is the test:
class test_to_number(unittest.TestCase):
def test_hex(self):
# All of the following should be converted to an integer
#
values = [
# HEX
# ----------------------
# Input | Expected
# ----------------------
(0xDEADBEEF , 3735928559), # Hex
("0xFEEDFACE", 4277009102), # Hex
("0x0" , 0), # Hex
# Decimals
# ----------------------
# Input | Expected
# ----------------------
(100 , 100), # Decimal
("42" , 42), # Decimal
]
values += [
# Octals
# ----------------------
# Input | Expected
# ----------------------
(0o10 , 8), # Octal
("0o20" , 16), # Octal
("020" , 16), # Octal
]
values += [
# Floats
# ----------------------
# Input | Expected
# ----------------------
(3.14 , 3.14), # Float
("2.72" , 2.72), # Float
("1e3" , 1000), # Float
(1e-3 , 0.001), # Float
]
values += [
# All ints
# ----------------------
# Input | Expected
# ----------------------
("0xA" , 10),
("012" , 10),
("0o12" , 10),
("0b01010" , 10),
("10" , 10),
("10.0" , 10),
("1e1" , 10),
]
for _input, expected in values:
value = to_number(_input)
if isinstance(_input, str):
cmd = 'to_number("{}")'.format(_input)
else:
cmd = 'to_number({})'.format(_input)
print("{:23} = {:10} == {:10}".format(cmd, value, expected))
self.assertEqual(value, expected)

Pass your string to this function:
def string_to_number(str):
if("." in str):
try:
res = float(str)
except:
res = str
elif(str.isdigit()):
res = int(str)
else:
res = str
return(res)
It will return int, float or string depending on what was passed.
String that is an int
print(type(string_to_number("124")))
<class 'int'>
String that is a float
print(type(string_to_number("12.4")))
<class 'float'>
String that is a string
print(type(string_to_number("hello")))
<class 'str'>
String that looks like a float
print(type(string_to_number("hel.lo")))
<class 'str'>

There is also regex, because sometimes string must be prepared and normalized before casting to a number:
import re
def parseNumber(value, as_int=False):
try:
number = float(re.sub('[^.\-\d]', '', value))
if as_int:
return int(number + 0.5)
else:
return number
except ValueError:
return float('nan') # or None if you wish
Usage:
parseNumber('13,345')
> 13345.0
parseNumber('- 123 000')
> -123000.0
parseNumber('99999\n')
> 99999.0
And by the way, something to verify you have a number:
import numbers
def is_number(value):
return isinstance(value, numbers.Number)
# Will work with int, float, long, Decimal

a = int(float(a)) if int(float(a)) == float(a) else float(a)

This is a corrected version of Totoro's answer.
This will try to parse a string and return either int or float depending on what the string represents. It might rise parsing exceptions or have some unexpected behaviour.
def get_int_or_float(v):
number_as_float = float(v)
number_as_int = int(number_as_float)
return number_as_int if number_as_float == number_as_int else
number_as_float

If you are dealing with mixed integers and floats and want a consistent way to deal with your mixed data, here is my solution with the proper docstring:
def parse_num(candidate):
"""Parse string to number if possible
It work equally well with negative and positive numbers, integers and floats.
Args:
candidate (str): string to convert
Returns:
float | int | None: float or int if possible otherwise None
"""
try:
float_value = float(candidate)
except ValueError:
return None
# Optional part if you prefer int to float when decimal part is 0
if float_value.is_integer():
return int(float_value)
# end of the optional part
return float_value
# Test
candidates = ['34.77', '-13', 'jh', '8990', '76_3234_54']
res_list = list(map(parse_num, candidates))
print('Before:')
print(candidates)
print('After:')
print(res_list)
Output:
Before:
['34.77', '-13', 'jh', '8990', '76_3234_54']
After:
[34.77, -13, None, 8990, 76323454]

Use:
def num(s):
try:
for each in s:
yield int(each)
except ValueError:
yield float(each)
a = num(["123.55","345","44"])
print a.next()
print a.next()
This is the most Pythonic way I could come up with.

If you don't want to use third party modules the following might be the most robust solution:
def string_to_int_or_float(s):
try:
f = float(s) # replace s with str(s) if you are not sure that s is a string
except ValueError:
print("Provided string '" + s + "' is not interpretable as a literal number.")
raise
try:
i = int(str(f).rstrip('0').rstrip('.'))
except:
return f
return i
It might not be the fastest, but it handles correctly literal numbers where many other solutions fail, such as:
>>> string_to_int_or_float('789.')
789
>>> string_to_int_or_float('789.0')
789
>>> string_to_int_or_float('12.3e2')
1230
>>> string_to_int_or_float('12.3e-2')
0.123
>>> string_to_int_or_float('4560e-1')
456
>>> string_to_int_or_float('4560e-2')
45.6

You can simply do this by
s = '542.22'
f = float(s) # This converts string data to float data with a decimal point
print(f)
i = int(f) # This converts string data to integer data by just taking the whole number part of it
print(i)
For more information on parsing of data types check on python documentation!

This is a function which will convert any object (not just str) to int or float, based on if the actual string supplied looks like int or float. Further if it's an object which has both __float and __int__ methods, it defaults to using __float__
def conv_to_num(x, num_type='asis'):
'''Converts an object to a number if possible.
num_type: int, float, 'asis'
Defaults to floating point in case of ambiguity.
'''
import numbers
is_num, is_str, is_other = [False]*3
if isinstance(x, numbers.Number):
is_num = True
elif isinstance(x, str):
is_str = True
is_other = not any([is_num, is_str])
if is_num:
res = x
elif is_str:
is_float, is_int, is_char = [False]*3
try:
res = float(x)
if '.' in x:
is_float = True
else:
is_int = True
except ValueError:
res = x
is_char = True
else:
if num_type == 'asis':
funcs = [int, float]
else:
funcs = [num_type]
for func in funcs:
try:
res = func(x)
break
except TypeError:
continue
else:
res = x

By using int and float methods we can convert a string to integer and floats.
s="45.8"
print(float(s))
y='67'
print(int(y))

For numbers and characters together:
string_for_int = "498 results should get"
string_for_float = "498.45645765 results should get"
First import re:
import re
# For getting the integer part:
print(int(re.search(r'\d+', string_for_int).group())) #498
# For getting the float part:
print(float(re.search(r'\d+\.\d+', string_for_float).group())) #498.45645765
For easy model:
value1 = "10"
value2 = "10.2"
print(int(value1)) # 10
print(float(value2)) # 10.2

Is there a simple and preferred way of German number string formatting in Python?

I am searching for the proper way of German number formatting (e.g. 1.000,1234) in Python under Windows OS.
I tried locale.setlocale but did not succeed.
Instead, I have written a function to come up with the desired output.
Is there a better way?
def ger_num(number, precision=3):
"""
returns german formatted number as string or an empty string
"""
if number is not None:
try:
my_number = "{:,f}".format(number)
except ValueError:
return ""
decimals, fraction = my_number.split(".")[0], my_number.split(".")[1]
decimals = decimals.replace(",", ".")
if precision:
return decimals + "," + fraction[:precision]
else:
return decimals
else:
return ""

You can use locale.setlocale to set the locale to de and then use locale.format to format your number:
import locale
locale.setlocale(locale.LC_ALL, 'de')
print(locale.format('%.4f', 1000.1234, 1))
This outputs:
1.000,1234

If for some reason locale does not work for you (or is not desired), then the easiest other option would probably be to use string replacement, like suggested in this already mentioned answer (which draws its answer from the PEP-378).
You can always encapsulate that in a function, maybe like this:
def format_number(number, precision=3):
# build format string
format_str = '{{:,.{}f}}'.format(precision)
# make number string
number_str = format_str.format(number)
# replace chars
return number_str.replace(',', 'X').replace('.', ',').replace('X', '.')
This works well for int, float and Decimal:
>>> format_number(1)
'1,000'
>>> format_number(1, 2)
'1,00'
>>> format_number(1, 7)
'1,0000000'
>>> format_number(1234567, 7)
'1.234.567,0000000'
>>> format_number(1234567.9988, 7)
'1.234.567,9988000'
>>> format_number(1234567.9988, 2)
'1.234.568,00'
>>> from decimal import Decimal
>>> format_number(Decimal('1234567.9988'), 2)
'1.234.568,00'
>>> format_number(Decimal('1234567.9988'), 5)
'1.234.567,99880'
>>> format_number(Decimal('1234567.9988'), 0)
'1.234.568'
>>> format_number(Decimal('123456'), 5)
'123.456,00000'

Thanks for the help. If anyone finds this useful I do provide the code for my final solution here.
ger_num.py:
def ger_num(number, digits=2):
'''
retruns <number> as german formattet string (thousands-.),
rounds to n <digits>.
<number> == None OR (!= INT AND != FLOAT) returns '' (empty STR)
'''
import locale
locale.setlocale(locale.LC_ALL, 'de')
if number is None:
return ''
if not isinstance(number, int) and not isinstance(number, float):
return ''
else:
format = '%.'+str(digits)+'f'
return locale.format_string(format, number, 1)
if __name__ == "__main__":
pass

Python - isinstance returns false

I'm having an issue with isinstance().
I'm using Python 2.7.8, and running scripts from the shell.
The array element I'm testing for contains a number, but this function returns false; using number.Numbers:
import numbers
...
print array[x][i]
>> 1
...
print isinstance(array[x][i], numbers.Number)
>>> False
Also tried this, from this post
import types
...
print isinstance(array[x][i], (types.IntType, types.LongType, types.FloatType, types.ComplexType))
>>> False
From the same post, I tried
isinstance(array[x][i], (int, float, long, complex))
I also tried this solution did not work.
All return false.

You don't have a number; you most probably have a string instead, containing the digit '1':
>>> value = '1'
>>> print value
1
>>> print 1
1
This is not a number; it is a string instead. Note that printing that string is indistinguishable from printing an integer.
Use repr() to print out Python representations instead, and / or use the type() function to produce the type object for a given value:
>>> print repr(value)
'1'
>>> print type(value)
<type 'str'>
Now it is clear that the value is a string, not an integer, even though it looks the same when printed.
For actual numeric values, isinstance() together with numbers.Number works as expected:
>>> from numbers import Number
>>> isinstance(value, Number)
False
>>> isinstance(1, Number)
True

How to get number of decimal places

How would I do the following:
>>> num_decimal_places('3.2220')
3 # exclude zero-padding
>>> num_decimal_places('3.1')
1
>>> num_decimal_places('4')
0
I was thinking of doing:
len((str(number) if '.' in str(number) else str(number) + '.').rstrip('0').split('.')[-1])
Is there another, simpler way to do this?

You can use a regex to parse value, capture the decimal digits and count the length of the match, if any:
import re
def num_decimal_places(value):
m = re.match(r"^[0-9]*\.([1-9]([0-9]*[1-9])?)0*$", value)
return len(m.group(1)) if m is not None else 0
this is a bit less "raw" than splitting the string with multiple if else, not sure if simpler or more readable, though.

The best and the most Pythonic way to achieve this is:
import decimal
x = '56.001000'
x = x.rstrip('0') # returns '56.001'
x = decimal.Decimal(x) # returns Decimal('0.001')
x = x.as_tuple().exponent # returns -3
x = abs(x) #returns 3
Above code can be written in simpler way as:
>>> x = '56.001000'
>>> abs(decimal.Decimal(x.rstrip('0')).as_tuple().exponent)
3
Below is the list of used functions for more reference:
str.rstrip(): Return a copy of the string with trailing characters removed.
decimal.Decimal(): Construct a new Decimal object based from value.
x.as_tuple(): Returns a namedtuple of the format: DecimalTuple(sign=0, digits=(1,), exponent=-3)
abs(): Return the absolute value of a number.

You dont need regex, you can convert to float and convert back to string! this automatically will remove the zeroes :
>>> def convertor(s):
... try :
... int(s.rstrip('0').rstrip('.'))
... return 0
... except:
... return len(str(float(s)).split('.')[-1])
...
>>> convertor('4.0230')
3
>>> convertor('4.0')
0
>>> convertor('4')
0

you could also just try something like:
try:
return len(str(num).split('.')[1].rstrip('0'))
except
return 0

By string process:
Check . is present in number string or not.
If Not present then return 0.
If present the split number string by . and get second item from the split result.
Remove 0 from the right side.
Return len of item.
code:
>>> def dCount(no_str):
... if "." in no_str:
... return len(no_str.split(".")[1].rstrip("0"))
... else:
... return 0
...
>>> dCount("2")
0
>>> dCount("2.002")
3
>>> dCount("2.1230")
3
>>> dCount("2.01230")
4
>>>

import re
def f(s):
ls = s.split('.', 1)
if len(ls) == 2 and re.match(r'\d*$', ls[1]):
return len(ls[1].rstrip('0'))
return 0
assert f('12') == 0
assert f('12.') == 0
assert f('12.1') == 1
assert f('12.100006') == 6
assert f('12.1.3') == 0
assert f('12.1abc') == 0
assert f('12.100000') == 1

The Decimal type is perfect for this, you can implement num_decimal_places() as follows:
from decimal import Decimal
def num_decimal_places(value: str):
return -Decimal(value).normalize().as_tuple().exponent
It works as follows: Decimal(value) parses the string, including exponent notation, then .normalize() strips any trailing zeros from the internal representation. .as_tuple().exponent contains the number of decimal places the internally stored integer is shifted to the left, so negative numbers specify the number of places to the right of the decimal.

You could try using the Decimal function in python:
abs(Decimal(string_value).as_tuple().exponent)
as explained in
Easy way of finding decimal places