35

I have a file with times (minutes and seconds), which looks approximately as follows:

02:53 rest of line 1...
03:10 rest of line 2...
05:34 rest of line 3...
05:35 rest of line 4...
10:02 rest of line 5...
...

I would like to replace the time by its equivalent in seconds. Ideally, I would like to run some magical command like this:

:%s/^\(\d\d\):\(\d\d\) \(.*\)/(=\1*60 + \2) \3/g

...where the (=\1*60 + \2) is the magical part. I know I can insert results of evaluation with the special register =, but is there a way to do this in the subst part of a regex?

Philippe
  • 9,582
  • 4
  • 39
  • 59

3 Answers3

51

Something like this?

:%s/^\(\d\d\):\(\d\d\)/\=submatch(1)*60+submatch(2)/

When the replacement starts with a \= the replacment is interpreted as an expression.

:h sub-replace-expression is copied below

Substitute with an expression           *sub-replace-expression*
                        *sub-replace-\=*
When the substitute string starts with "\=" the remainder is interpreted as an
expression.  This does not work recursively: a substitute() function inside
the expression cannot use "\=" for the substitute string.

The special meaning for characters as mentioned at |sub-replace-special| does
not apply except for "<CR>", "\<CR>" and "\\".  Thus in the result of the
expression you need to use two backslashes to get one, put a backslash before a
<CR> you want to insert, and use a <CR> without a backslash where you want to
break the line.

For convenience a <NL> character is also used as a line break.  Prepend a
backslash to get a real <NL> character (which will be a NUL in the file).

When the result is a |List| then the items are joined with separating line
breaks.  Thus each item becomes a line, except that they can contain line
breaks themselves.

The whole matched text can be accessed with "submatch(0)".  The text matched
with the first pair of () with "submatch(1)".  Likewise for further
sub-matches in ().
Alex Harvey
  • 14,494
  • 5
  • 61
  • 97
FDinoff
  • 30,689
  • 5
  • 75
  • 96
  • 2
    Note that `/` also typically has a special meaning within a `sub-replace-expression` (it ends it), However, if you use `s###` instead of `s///`, you can use `/` (divide) freely within the sub expression. https://stackoverflow.com/a/30072611/712526 – jpaugh Dec 07 '18 at 11:24
6

Use submatch() to refer to a grouped part in the substitution place:

:%s/\v^(\d{2}):(\d{2})>/\=submatch(1) * 60 + submatch(2)/

With your example yields:

173 rest of line 1...
190 rest of line 2...
334 rest of line 3...
335 rest of line 4...
602 rest of line 5...
Birei
  • 35,723
  • 2
  • 77
  • 82
0

Hopefully it would be helpful to someone else, but i had similar problem where i wanted to replace "id" with a different number, in-fact any other number

{"id":1,"first_name":"Ruperto","last_name":"Bonifayipio","gender":"Male","ssn":"318-69-4987"},

used expression

%s/\v(\d+),/\=submatch(1)*1111/g

which results into following new value

{"id":1111,"first_name":"Ruperto","last_name":"Bonifayipio","gender":"Male","ssn":"318-69-4987"},
Ram Dhakne
  • 1