Using isdigit for floats?

Question

a = raw_input('How much is 1 share in that company? ')

while not a.isdigit():
    print("You need to write a number!n")
    a = raw_input('How much is 1 share in that company? ')

This only works if the user enters an integer, but I want it to work even if they enter a float, but not when they enter a string.

So the user should be able to enter both 9 and 9.2, but not abc.

How should I do it?

Asked By: Peter Nolan

||

Source

Answer 1

EAFP

try:
    x = float(a)
except ValueError:
    print("You must enter a number")

Answered By: dan04

Answer 2

Use regular expressions.

import re

p = re.compile('d+(.d+)?')

a = raw_input('How much is 1 share in that company? ')

while p.match(a) == None:
    print "You need to write a number!n"
    a = raw_input('How much is 1 share in that company? ')

Answered By: Peter C

Answer 3

I think @dan04 has the right approach (EAFP), but unfortunately the real world is often a special case and some additional code is really required to manage things—so below is a more elaborate, but also a bit more pragmatic (and realistic):

import sys

while True:
    try:
        a = raw_input('How much is 1 share in that company? ')
        x = float(a)
        # validity check(s)
        if x < 0: raise ValueError('share price must be positive')
    except ValueError, e:
        print("ValueError: '{}'".format(e))
        print("Please try entering it again...")
    except KeyboardInterrupt:
        sys.exit("n<terminated by user>")
    except:
        exc_value = sys.exc_info()[1]
        exc_class = exc_value.__class__.__name__
        print("{} exception: '{}'".format(exc_class, exc_value))
        sys.exit("<fatal error encountered>")
    else:
        break  # no exceptions occurred, terminate loop

print("Share price entered: {}".format(x))

Sample usage:

> python numeric_input.py
How much is 1 share in that company? abc
ValueError: 'could not convert string to float: abc'
Please try entering it again...
How much is 1 share in that company? -1
ValueError: 'share price must be positive'
Please try entering it again...
How much is 1 share in that company? 9
Share price entered: 9.0

> python numeric_input.py
How much is 1 share in that company? 9.2
Share price entered: 9.2

Answered By: martineau

Answer 4

The existing answers are correct in that the more Pythonic way is usually to try...except (i.e. EAFP).

However, if you really want to do the validation, you could remove exactly 1 decimal point before using isdigit().

>>> "124".replace(".", "", 1).isdigit()
True
>>> "12.4".replace(".", "", 1).isdigit()
True
>>> "12..4".replace(".", "", 1).isdigit()
False
>>> "192.168.1.1".replace(".", "", 1).isdigit()
False

Notice that this does not treat floats any different from ints however. You could add that check if you really need it though.

Answered By: Cam Jackson

Answer 5

s = '12.32'
if s.replace('.', '').replace('-', '').isdigit():
    print(float(s))

Note that this will work for negative floats as well.

Answered By: Jigyasu Tailor

Answer 6

Building on dan04’s answer:

def isDigit(x):
    try:
        float(x)
        return True
    except ValueError:
        return False

usage:

isDigit(3)     # True
isDigit(3.1)   # True
isDigit("3")   # True
isDigit("3.1") # True
isDigit("hi")  # False

Answered By: Phlox Midas

Answer 7

import re

string1 = "0.5"
string2 = "0.5a"
string3 = "a0.5"
string4 = "a0.5a"

p = re.compile(r'd+(.d+)?$')

if p.match(string1):
    print(string1 + " float or int")
else:
    print(string1 + " not float or int")

if p.match(string2):
    print(string2 + " float or int")
else:
    print(string2 + " not float or int")

if p.match(string3):
    print(string3 + " float or int")
else:
    print(string3 + " not float or int")

if p.match(string4):
    print(string4 + " float or int")
else:
    print(string4 + " not float or int")

output:
0.5 float or int
0.5a not float or int
a0.5 not float or int
a0.5a not float or int

Answered By: Yakir GIladi Edry

Answer 8

The provided answers fail if the string contains some special characters such as underscore (e.g. ‘1_1’). The following function returns correct answer in all case that I tested.

def IfStringRepresentsFloat(s):
try:
    float(s)
    return str(float(s)) == s
except ValueError:
    return False

Answered By: Hamid Heydarian

Using isdigit for floats?

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