Use Python 3 str.format to truncate a float to have no more than x number of digits

Question

I am looking for an expression that will truncate a float to at most a certain number of digits. I want to preserve a certain number of decimals, without having unnecessary trailing 0s.

So, this almost works as desired:

"{0:3.f"}.format(number)

For input 3.123000001:

"{0:.3f}".format(3.1230000001)
'3.123'

Great. But for input 3:

"{0:.3f}".format(3)
'3.000'

I now agree that the question is duplicate, but I wasn't able to find the duplicate question when I spent ~5 minutes searching Google/stackoverflow looking for the answer. Is that a sufficient reason to leave this question up, or should I delete it? — J Jones, Feb 07 '19 at 22:38

J Jones · Accepted Answer · 2019-02-11T18:39:46.110

0

I figured out the answer while I was writing the question. Just add .rstrip('0') to the expression. So:

"{0:3.f}".format(number).rstrip('0')

edited Feb 11 '19 at 18:39

answered Feb 07 '19 at 19:44

J Jones

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your right quotation mark are in the wrong position .. also I guess it should be `rstrip` since you don't want to strip the leading 0 if it's between 0 and 1? – Kevin He Feb 07 '19 at 19:47
If you're happy with your own answer, you should accept it. – Marvin Feb 07 '19 at 19:49
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@Marvin, you can't accept your own answer for 2 days. – J Jones Feb 07 '19 at 22:34
@KevinHe, correct on both counts, I was sloppy. I've fixed both. – J Jones Feb 07 '19 at 22:35

Use Python 3 str.format to truncate a float to have no more than x number of digits

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