Truncating floats in Python

Question

Basically to remove digits from a float to have a fixed number of digits after the dot, like:

1.923328437452 -> 1.923

EDIT: I need to output as a string to another function, not print.

Also I want to ignore the lost digits, not round them.

Answer 1

round(1.923328437452, 3)

Python standard types, you'll need to scroll down a bit to get to the round function. Essentially the second number says how many decimal places to round it to.

Answer 2

If you mean when printing, then the following should work:

print '%.3f' % number

Answer 3

The result of round is a float, so watch out:

>>> round(1.923328437452, 3)
1.923
>>> round(1.23456, 3)
1.2350000000000001

You will be better off when using a formatted string:

>>> "%.3f" % 1.923328437452
'1.923'
>>> "%.3f" % 1.23456
'1.235'

Answer 4

I would convert to a string at full precision and then just chop off everything but the first so many characters.

def trunc(f, n):
    '''Truncates/pads a float f to n decimal places without rounding'''
    slen = len('%.*f' % (n, f))
    return str(f)[:slen]

(edited to fix bugs)

In response to your comment: the version above doesn't include the zeros, but here's my old version that did:

def trunc(f, n):
    '''Truncates/pads a float f to n decimal places without rounding'''
    return ('%.*f' % (n + 1, f))[:-1]

The latter does fail on some corner cases, though, like trunc(11.999999, 3)

Answer 5

def trunc(num, digits):
   sp = str(num).split('.')
   return '.'.join([sp[0], sp[:digits]])

This should work. It should give you the truncation you are looking for.

Answer 6

def trunc(f,n): return ('%.16f' % f)[:(n-16)]

Answer 7

Just wanted to mention that the old "make round() with floor()" trick of

round(f) = floor(f+0.5)

can be turned around to make floor() from round()

floor(f) = round(f-0.5)

Although both these rules break around negative numbers, so using it is less than ideal:

def trunc(f, n):
    if f > 0:
        return "%.*f" % (n, (f - 0.5*10**-n))
    elif f == 0:
        return "%.*f" % (n, f)
    elif f < 0:
        return "%.*f" % (n, (f + 0.5*10**-n))