[Tutor] What does "TypeError: 'int' object is not iterable"mean?

[Richard D. Moores]

On Fri, Oct 22, 2010 at 11:27, Alan Gauld <alan.gauld at btinternet.com> wrote:
>
> "Richard D. Moores" <rdmoores at gmail.com> wrote
>
>>   return ("%%.%sf" % n) % floatt
>>
>> which works fine, but a question remains: n is an integer. Why the 's'
>> in '%sf'?
>
> Its arbitrary. You could use %d just as easily.
> %s will put the numner into the string, so will %d.
> %d is probably a bit more picky about what it inserts - so might
> actually be a better choice in this case since
>
> float2n(1.234567, 3.5)
>
> Doesn't work as you might hope... but changing it to use %d might -
> depending
> on your definition of better! And what you might hope! :-)
>
> OTOH %s handles this better:
>
> float2n(1.234567, '3')

If I use n = 2.5 and replace %s with %d,

my function is

def float2n_decimals(floatt, n):
    """
    Given a float (floatt), return floatt to n decimal places.

    E.g., with n = 2, 81.34567 -> 81.35
    """
    return ("%%.%df" % n) % floatt

This works (the d converts the 2.5 to an int, 2), but I get a
deprecation warning: "DeprecationWarning: integer argument expected,
got float
  rate = rate1 = round(float(rate2),n)"

Of course, n = 2.5 makes no sense in my script.

Dick