isFloat: Without Exception-Handling

Rajarshi Guha rajarshi at presidency.com
Tue Jun 10 17:18:46 CEST 2003


On Wed, 18 Sep 2002 12:47:40 -0400, Steve Holden wrote:

> "Rajarshi Guha" <rajarshi at presidency.com> wrote in message
> news:pan.2002.09.18.11.40.50.766802.29524 at presidency.com...
>> On Wed, 18 Sep 2002 11:41:47 -0400, Thomas Guettler wrote:
>>
>> > Hi!
>> >
>> > Is there a way to write the following method without using
>> > exceptions?
>> >
>> > def isFloat(string):
>> >      is_float=1
>> >      try:
>> >          float(string)
>> >      except:
>> >          is_float=0
>> >      return is_float
>> >
>> > print isFloat("asdf") # --> 0
>> > print isFloat("0.1")  # --> 1
>>
>>
>> def isFloat(string):
>>   if type(string) == type(1.0):
>>     return 1
>>   else
>>     return 0
>>
>> This should do what you want without exceptions
> 
> I doubt that VERY much. Please test your code before submitting, or
> ensure you note it wasn't tested ...
> 
>>>> def isFloat(s):
> ...   if type(s) == type(1.0):
> ...     return 1
> ...   else:
> ...     return 0
> ...
>>>> isFloat("banana")
> 0
>>>> isFloat("1.2")
> 0

If its called as isFloat(1.2) it works. 
(My idea was that it would called with a variable and not a literl)





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