[python] Detect whether a Python string is a number or a letter

Check if string is positive digit (integer) and alphabet

You may use str.isdigit() and str.isalpha() to check whether given string is positive integer and alphabet respectively.

Sample Results:

# For alphabet
>>> 'A'.isdigit()
False
>>> 'A'.isalpha()
True

# For digit
>>> '1'.isdigit()
True
>>> '1'.isalpha()
False

Check for strings as positive/negative - integer/float

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

Discard "NaN" (not a number) strings while checking for number

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

Allow Complex Number like "1+2j" to be treated as valid number

The above function will still return you False for the complex numbers. If you want your is_number function to treat complex numbers as valid number, then you need to type cast your passed string to complex() instead of float(). Then your is_number function will look like:

def is_number(n):
    is_number = True
    try:
        #      v type-casting the number here as `complex`, instead of `float`
        num = complex(n)
        is_number = num == num
    except ValueError:
        is_number = False
    return is_number

Sample Run:

>>> is_number('1+2j')    # Valid 
True                     #      : complex number 

>>> is_number('1+ 2j')   # Invalid 
False                    #      : string with space in complex number represetantion
                         #        is treated as invalid complex number

>>> is_number('123')     # Valid
True                     #      : positive integer

>>> is_number('-123')    # Valid 
True                     #      : negative integer

>>> is_number('abc')     # Invalid 
False                    #      : some random string, not a valid number

>>> is_number('nan')     # Invalid
False                    #      : not a number "nan" string

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.