[Tutor] What does "TypeError: 'int' object is not iterable"mean?
Richard D. Moores
rdmoores at gmail.com
Fri Oct 22 22:56:51 CEST 2010
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 -
> 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,
rate = rate1 = round(float(rate2),n)"
Of course, n = 2.5 makes no sense in my script.
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