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From: Ian Kelly <ian.g.kelly@gmail.com>
Date: Thu, 19 Jun 2014 01:23:23 -0600
Subject: Re: how to check if a value is a floating point or not
To: Python <python-list@python.org>
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On Thu, Jun 19, 2014 at 12:48 AM, Nicholas Cannon
<nicholascannon1@gmail.com> wrote:
> On Thursday, June 19, 2014 1:53:31 PM UTC+8, Nicholas Cannon wrote:
>> I am making a calculator and i need it to support floating point values =
but i am using the function isnumeric to check if the user has entered an i=
nt value. I need the same for floating point types so i could implement an =
or in the if statement that checks the values the user has entered and allo=
w it to check and use floating points. If you need the source code i am hap=
py to give it to you. Thank you for your help
>
> I am using python 2.7.7 and i have come up with away but there is still p=
ossible errors for this. What i did was i this
>
> #checks if the user input is an integer value
> def checkint(a):
>         if a.isnumeric():
>                 return True
>         else:
>                 if a.isalpha():
>                         return False
>                 else:
>                         return True
>
> The parameter a is the users input by the raw_input function. I first tes=
t if it is normal int with the isnumeric function. Unfortunately this funct=
ion picks up the decimal as false. This means if the user inputs a float it=
 has to be false. I then test if this input has any alphabetical characters=
 if it does not the user could have only entered  something like 12.5 oppos=
e to abc.d.

unicode.isalpha does not test if the input has *any* alphabetic
characters.  It tests if the input is *only* alphabetic characters.
u'12.5'.isalpha() does return False.  u'abc.d'.isalpha() *also*
returns False, because the decimal point is not alphabetic.

I second Gary Herron's suggestion to just try converting the value and
catch the exception if it fails.  Python already knows how to do this
for you; there's no need to reinvent the wheel.