On Thu, Jun 19, 2014 at 12:48 AM, Nicholas Cannon <nicholascann...@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 int >> 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 allow it >> to check and use floating points. If you need the source code i am happy 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 > possible 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 test if > it is normal int with the isnumeric function. Unfortunately this function > 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 oppose 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. -- https://mail.python.org/mailman/listinfo/python-list
