Steven D'Aprano wrote: > Shwinn Ricci wrote: >> When comparing a given value with a database of values, but allowing for >> freedom due to variation at say, the thousandth digit, how does one >> generalize the precision to this degree? I don't want to truncate, so is >> there a round() function that takes into account what decimal place the >> value (database value) needs to be rounded to? > > Thousandth digit??? Python doesn't support floating point numbers with a > thousand digits! I think about seventeen is about the limit.
Not only that. The idea of n-digit precision is flawed because assuming an actual value of 1.00000 0.999 is closer than 1.01 but the latter has two correct digits and the former has none. > Yes, there is a round function: > > > >>> x = 123.45678 > >>> y = 123.45731 > >>> x == y > False > >>> round(x, 3) == round(y, 3) > True I prefer abs(x-y) < eps for similar reasons as stated above. 1.8 is as close to 1.4 as 1.0, but round(1.8) != round(1.4) whereas round(1.4) == round(1.0). _______________________________________________ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
