Why isdigit() return false on float?

Question

I would like to check if my value is a float with a point or a comma but isdigit() return false with a point. I would like to know why and how go through it.

> value = "0.0"
> print value.isdigit():
>>> False

My code is :

if "." in value and value.isdigit()
    print "ok"

Answer 1

str.isdigit() will only return true if all characters in the string are digits. . is punctuation, not a digit.

From the Python 3 str.isdigit() documentation:

Formally, a digit is a character that has the property value Numeric_Type=Digit or Numeric_Type=Decimal

(For Python 2, for str objects only the ASCII digits (0 through to 9) are considered, but for unicode objects the same definition applies).

See the official Numeric Property definitions specification; there are 708 Unicode codepoints that match that description.

Simplifying this to general unicode categories, numeric types in Unicode have a category starting with N, but . does not:

>>> import unicodedata
>>> unicodedata.category(u'.')
'Po'

P stands for punctuation here, o for other.

Vice-versa, strings that contain only digits are not always convertable to a float or number either:

>>> unicodedata.name(u'\u2080')
'SUBSCRIPT ZERO'
>>> unicodedata.category(u'\u2080')
'No'
>>> unicodedata.digit(u'\u2080')
0
>>> u'\u2080'.isdigit()
True
>>> float(u'\u2080')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
UnicodeEncodeError: 'decimal' codec can't encode character u'\u2080' in position 0: invalid decimal Unicode string

So a subscript-zero is not really 0 as far as float() is concerned, but it is a digit.

If you want to test if a string is a valid floating point number, use float and catch the ValueError:

def is_float(string):
    try:
        float(string)
        return True
    except ValueError:
        return False