[Tutor] Rounding to n significant digits? (original) (raw)

Tim Peters tim.peters at gmail.com
Sat Jul 3 15:49:09 EDT 2004

[Dick Moores] ...

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 roundton(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" asstring = format % x return asstring """ print roundton(123.456789, 4) print roundton(.000000123456789, 2) print roundton(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.)

Then it's time to learn about one of the more obscure features of string formats: if you put an asterisk in a string format where a precision specifier is expected, the actual precision to use will be taken from the argument tuple. I realize that's confusing; that's why I called it obscure . It should be clearer from this rewrite of your rewrite of my original round_to_n function:

def round_to_n(x, n): if n < 1: raise ValueError("number of significant digits must be >= 1") return "%.*e" % (n-1, x)

That's probably the end of the line for this problem .

A related but harder problem is to get a string rounded to n significant digits, but where the exponent is constrained to be a multiple of 3. This is often useful in engineering work. For example, in computer work, 1e6 and 1e3 are natural units (mega and kilo), but 1e4 isn't -- if I have 15,000 of something, I want to see that as 15e3, not as 1.5e4. I don't know an easy way to get that in Python (or in most other programming languages).

More information about the Tutor mailing list