On Fri, Dec 26, 2008 at 1:52 PM, <da...@bag.python.org> wrote:
> I'm a newbee trying 3.0 Please help with math.sqrt()
>
> At the command line this function works correctly
> >>> import math
> n = input("enter a number > ")
raw_input() was renamed input() in Python 3.0, and it returns a
*string*, not a *number*.
Therefore, you need to convert the string to an int or float.
Change the line to:
n = float(input("enter a number > "))
And it should work just fine.
> s = math.sqrt(n)
> An entry of 9 or 9.0 will yield 3.0
>
> Yet the same code in a script gives an error message
> Script1
> import math
> n = input("enter a number > ")
> s = math.sqrt(n)
>
> Traceback (most recent call last) :
> File "<stdin>", line 1, in <module>
> File "script1.py" line 3 in <module>
> s = math.sqrt(n)
> TypeError : a float is required
You're currently giving it a string, not a number, which is
nonsensical, hence the TypeError. I presume ints would be coerced to
floats by the function.
> Entering 9 or 9.0 gives same error message.
>
> According to the math module the results of all
> functions are floats. However it says nothing about
> inputs.
>
> Strangely the above code runs fine in version 2.5 ( ? )
> and will handle large integers.
>
> I've read the documentation for 3.0 including the section
> "Floating Point Arithmetic: Issues & Limitations" and it
> helps nada.
You should read "What's New in Python 3.0", it covers changes such as
the one you've encountered -
http://docs.python.org/3.0/whatsnew/3.0.html
Cheers,
Chris
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