[Tutor] Rounding to n significant digits?

[Dick Moores]

At 20:18 7/2/2004, Tim Peters wrote:
>[Dick Moores]
> > ...
> > No, the 3 examples I gave are exactly what I want:
> > float = 123.456789, n = 4, returns 123.5
> > float = .000000123456789, n = 2, returns .00000012
> > float = 123456789, n = 5, returns 123460000
>
>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)
>
>print round_to_n(123.456789, 4)
>print round_to_n(.000000123456789, 2)
>print round_to_n(123456789, 5)
>"""
>
>That displays
>
>123.5
>1.2e-007
>123460000.0
>
>Be sure to the read the appendix on floating-point issues in the
>Python Tutorial too!  You clearly have decimal digits in mind, but
>that's not how your computer's floating-point hardware works.  The
>appendix talks about the consequences of that.

*That* is beautiful. Thank you! And the appendix you mentioned is 
enlightening.

But I realize that I don't need to return a float. Just the string will 
do. Therefore, this revision of your function will do exactly what I was 
thinking of, even if I didn't say so:

"""
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 as_string
"""
print round_to_n(123.456789, 4)
print round_to_n(.000000123456789, 2)
print round_to_n(123456789, 5)

That displays

1.235e+002
1.2e-007
1.2346e+008  (this is much better than getting the ".0" of 123460000.0, 
which implies accuracy to 10 significant digits instead of 5.)

Dick Moores