Vim search and replace, adding a constant

Question

I know this is a long shot, but I have a huge text file and I need to add a given number to other numbers matching some criteria.

Eg.

identifying text 1.1200
identifying text 1.1400

and I'd like to transform this (by adding say 1.15) to

identifying text 2.2700
identifying text 2.2900

Normally I'd do this in Python, but it's on a Windows machine where I can't install too many things. I've got Vim though :)

Luc Hermitte · Accepted Answer · 2010-11-16T13:36:15.187

18

Here is a simplification and a fix on hobbs' solution:

:%s/identifying text \zs\d\+\(.\d\+\)\=/\=(1.15+str2float(submatch(0)))/

Thanks to \zs, there is no need to recall the leading text. Thanks to str2float() a single addition is done on the whole number (in other words, 1.15 + 2.87 will give the expected result, 4.02, and not 3.102).

Of course this solution requires a recent version of Vim (7.3?)

edited Nov 16 '10 at 13:36

answered Nov 16 '10 at 12:20

Luc Hermitte

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+1. *I* can even get this to work (and that's saying a lot) but is there a way to retain the trailing zero's. – Lieven Keersmaekers Nov 16 '10 at 13:07
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With `printf()` may be ? (:h printf()) -> `...\=printf('%.4f', 1.15+str2float(......))` – Luc Hermitte Nov 16 '10 at 13:34
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Some thing I had to look up: `\=` allows the replacement string to be an expression. ([link](http://vim.wikia.com/wiki/Using_an_expression_in_substitute_command)) – Bryce Guinta Oct 22 '16 at 21:46
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To add two integer numbers without a trailing decimal, use `str2nr`. For more functions, see `:h functions` – Bryce Guinta Oct 22 '16 at 21:50

score 12 · Answer 2 · answered Nov 16 '10 at 11:43

12

You can do a capturing regex and then use a vimscript expression as a replacement, something like

:%s/\(identifying text \)\(\d\+\)\.\(\d\+\)/
  \=submatch(1) . (submatch(2) + 1) . "." . (submatch(3) + 1500)

(only without the linebreak).

answered Nov 16 '10 at 11:43

hobbs

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score 3 · Answer 3 · answered Nov 16 '10 at 11:33

Your number format seems to be a fixed one, so it's easy to convert to int and come back (remove the dot) add 11500 and put the dot back.

:%s/\.//
:%normal11500^A " type C-V then C-a
:%s/....$/.&/

If you don't want to do that on all the lines but only the one which match 'identifying text' replace all the % by 'g/indentifying text/'

score 1 · Answer 4 · answered Nov 16 '10 at 11:20

1

For integers you can just use n^A to add n to a number (and n^X to subtract it). I doubt whether that works for fractional numbers though.

answered Nov 16 '10 at 11:20

Colin Fine

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score 1 · Answer 5 · answered Nov 16 '10 at 12:01

Well this might not be a solution for vim but I think awk can help:

cat testscript | LC_ALL=C awk '{printf "%s %s %s %s %.3f\n", $1,$2,$3,$4,$5+1.567   }'

and the test

this is a number 1.56
this is a number 2.56
this is a number 3.56

I needed the LC_ALL=C for the correct conversion of the floating point separator, and maybe there is a more elegant solution for printing the beginning/ rest of the string. And the result looks like:

this is a number 3.127
this is a number 4.127
this is a number 5.127

Dam! Just read that it is a windows machine, so forget about awk :D — joecks, Nov 16 '10 at 15:39

score 1 · Answer 6 · answered Nov 16 '10 at 22:53

Using macro

qa .......................... start record macro 'a'
/iden<Enter> ................ search 'ident*' press Enter
2w .......................... jump 2 words until number one  (before dot)
Ctrl-a ...................... increases the number
2w .......................... jump to number after dot
1500 Ctrl-a ................. perform increases 1500 times
q ........................... stop record to macro 'a'

if you have 300 lines with this pattern just now making

300@a

The OP wants to increment one number ("**a** given number" <- singular) with 1.15, and not two separate numbers. — Luc Hermitte, Nov 17 '10 at 07:03

Vim search and replace, adding a constant

6 Answers6