Reducing the precision of numbers - regex vim

Question

My regex is quite rusty. How could vim be used to change the precision of a decimal number.

For example:

Changing 30.2223221 8188.2121213212 to 30.22 8188.21

score 18 · Answer 1 · answered Mar 05 '11 at 06:08

18

Using just VimL:

:%s/\d\+\.\d\+/\=printf('%.2f',str2float(submatch(0)))/g

answered Mar 05 '11 at 06:08

Raimondi

score 13 · Accepted Answer · answered Mar 04 '11 at 22:26

13

It's likely possible using vim internal search/replace, but I would use "perldo":

:perldo s/(\d+\.\d+)/sprintf "%.2f", $1/eg

answered Mar 04 '11 at 22:26

Pontus

I modified this regex to also work with negative numbers and e-notation: ([-]?\d+\.\d+(e[+-]\d)?) – Reimund Apr 25 '16 at 23:13

score 5 · Answer 3 · answered Mar 05 '11 at 06:16

5

If you just want to truncate the last digit instead of rounding,

:%s/(\d+\.\d\d)\d+/\1/g

answered Mar 05 '11 at 06:16

geekosaur

1

The above did not work for me, but adding a few slashes fixes it... :%s/\(\d\+\.\d\d\)\d\+/\1/g – Charles Mar 12 '13 at 19:38

score 1 · Answer 4 · answered Dec 02 '16 at 09:54

1

Based on the previous answers, using VimL, for negative numbers and exponential notation:

:%s/\c\v[-]=\d+\.\d+(e[+-]\d+)=/\=printf('%.2f',str2float(submatch(0)))/g

answered Dec 02 '16 at 09:54

jabellcu

4 Answers4