How do I parse a string to a float or int?

Question

In Python, how can I parse a numeric string like \"545.2222\" to its corresponding float value, 545.2222? Or parse the string \"31\" t

Answer 1

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'>

Answer 2

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

Answer 3

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

Answer 4

This is a corrected version of https://stackoverflow.com/a/33017514/5973334

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

Answer 5

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')