"Christopher Spears" <cspears2...@yahoo.com> wrote
I have a float variable that is very long.
float_a = 1.16667
Thats not really very long!
However, I want to pass the value of float_a to float_b,
but I want the float to be accurate to two decimal points.
float_a = 1.16667
print "%.2f" % float_a
1.17
Rounding issues will give you all sorts of issues here.
However if you really want to round it up or down to 2 digits
then the round() function is what you want.
But I'd be interested in why you need to lose precision
like that, it's not that common a requirement. Usually controlling
the presentation is sufficient (and preferred).
float_b = "%.2f" % float_a
float_b
'1.17'
type(float_b)
<type 'str'>
This doesn't work because it yields a string.
It works in that it returns a string representation of what you want.
It doesn't produce a float because *string formatting* can only
produce strings.
HTH,
--
Alan Gauld
Author of the Learn to Program web site
http://www.alan-g.me.uk/
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