I'm fairly new to Python and am trying to find a simple way to round
floats to a specific number of significant digits. I found an old
post on this list with exactly the same problem:
<http://mail.python.org/pipermail/tutor/2004-July/030268.html>
> Is there something (a function?) in Python 2.3.4 that will round a > result to n significant digits, or do I need to roll my own? I don't > see one in the math module.>> I mean something like rounding(float, n) that would do this:> float = 123.456789, n = 4, returns 123.5> float = .000000123456789, n = 2, returns .00000012> float = 123456789, n = 5, returns 123460000>> Thanks,>> Dick Moores
And another post gave this solution:
<http://mail.python.org/pipermail/tutor/2004-July/030311.html>
> I expect the easiest way to do this in Python is to convert to > string using an %e format, then convert that back to float again. > Like this:>> def round_to_n(x, n):> if n < 1:> raise ValueError("number of significant digits must be >= 1")> # Use %e format to get the n most significant digits, as a string.> format = "%." + str(n-1) + "e"> as_string = format % x> return float(as_string)
Converting to a string seemed like an awkward hack to me, so I came up
with a mathematical solution:
> import math>> def round_figures(x, n):> """Returns x rounded to n significant figures."""> return round(x, int(n - math.ceil(math.log10(abs(x)))))>> print round_figures(123.456789,4)> print round_figures(.000000123456789,2)> print round_figures(123456789,5)> print round_figures(0.987,3)> print round_figures(0.987,2)> print round_figures(0.987,1)> print round_figures(-0.002468,2)>> 123.5> 1.2e-07> 123460000.0> 0.987> 0.99> 1.0> -0.0025
Since the built-in round(x,n) can do rounding in the 10's and 100's
places just as easy as 0.1's and 0.01's, my function just counts how
many digits are in use and rounds off n digits away.
I thought others might find this solution useful. Or somebody else
might share a nicer way.
Richard Wagner