Vim complex regex - regex

Question

I have these strings in a file:
a b
a-b
a / b / c
I want to replace these with:
"a b" => a_b
"a-b" => a_b
"a / b / c" => a_b_c
How do I write the regex ? Please also explain the regex and name the concepts involved.

Answer 1

Yet another way:
:g/^/co.|-s/.*/"&" =>/|+s/\W\+/_/g|-j
Overview:
For every line, :g/^/, copy a line (:copy) and then substitute to add the "..." => on the first line and do a substitution on the non-alpha characters on the next line with _. Then join the two line, -j.
Glory of Details:
:g/{pat}/{cmd} - run {cmd} on each line matching {pat}. Use ^ to match every line
copy . - copy the current line below the current line (.). Short: co.
-1s/.*/.../ - :s the line above (-1). Replace entire line, .*
"&" => - & is the entire match (or \0 in PRCE)
+s/\W\+/_/g - do a global :s on the next line (+1) for all non-alphanumeric characters with _
-j - do a :join starting from the line above with the next line
For more help:
:h :g
:h :copy
:h :s
:h :j
:h :range

Answer 2

This is beyond simple capturing and reordering in the replacement. The modification of the non-alphabetic characters to _ requires a contained substitution of the match. This can be done via :help sub-replace-expr:
:%substitute/.*/\='"' . submatch(0) . '" => ' . substitute(submatch(0), '\A\+', '_', 'g')/
Basically, this matches entire lines, then replaces with the match in double quotes, followed by =>, followed by the match with non-alphabetic character sequences (\A\+) replaced with a single _.
alternative
You can also do this in two separate steps: First duplicating and quoting the line:
:%substitute/.*/"&" => &/
Then, the second copy needs to be modified. To apply the substitution to only match after the => separator, a positive lookbehind (must match after => + any characters) must be given:
:%substitute/\%(=> .*\)\#<=\A\+/_/g

Answer 3

This achieves what you're asking for, although the question is somewhat ambiguous:
%s/\(\a\)\A\+/\1_/g
%s/[find_pattern]/[replace_pattern/g does find and replace for every line (%) in a file, and does any number of matches (g), as opposed to the default behaviour of just the first one.
(\a) captures a group (brackets have to be escaped), containing an alphabetic character.
\A+ means one or more non-alphabetic character
/1 is a backreference to the first captured group in the pattern. In this case the alphabetic character in brackets.
_ is just the literal.
So together it replaces every letter followed by 1 or more non-letters with that letter followed by _. So this only works when the line ends with the last letter.

Answer 4

One way of doing this:
:%s/[\ -]\/*\ */_/g
[\ -] looks for either a space \ (note the space between \ and -) or a dash -.
The asterisk * means 0 or N occurrences. So \/* 0 or N occurrences of slash /; \ * 0 or N occurrences of space. Finally g replace all occurrences in the line.
[Edit]
I had misunderstood the question. Your problem can be solved using multiple sub-expressions in 2 steps.
step 1) Put an underscore before the c
:%s/c/_c/g
step 2) find and replace
:%s/a\([\ -]\/*\ *\)b\(\1\)*\(_\)*\(c\)*/"a\1b\2\4" => a_b\3\4/g
This will give you
"a b" => a_b
"a-b" => a_b
"a / b / c" => a_b_c
Explanation:
\(\) denotes a sub-expression, order of appearance matters so \1 matches to sub-expression one and so forth.
The trick is to add a _ somewhere so we can use it and at the same get information about the length. Because it only appears before c, the subexpression \3 will only match _ for that line.
Now, by replacing by "a\1b\2\4" we skip \3 avoiding to add an underscore.

Answer 5

:%s:[\ /-]\+:_:g
Explanation:
s: : : - Substitute command (with delimiter `:`)
[\ /-] - Match a ` ` (space), `/`, or `-` character
\+ - Match one or more of the previous group consecutively
_ - Replace with one `_` character
g - Replace all matches in line
% - Execute command on every line in file (optional)
I interpreted your question to be very generic. If you need to match more specific patterns, please indicate exactly what needs to be matched.
[Edit]
If you need to match ' / ' exactly, use:
:%s:\ /\ \|[\ -]:_:g
s: : : - Substitute command (with delimiter `:`)
\| - Match left pattern OR right pattern
\ /\ - Match ` / ` exactly
[\ -] - Match a ` ` (space) or `-` character
_ - Replace with one `_` character
g - Replace all matches in line
% - Execute command on every line in file (optional)
[Edit 2]
I misunderstood what you wanted to substitute.
You're making your life very difficult if you're trying to do this with a
single regex. It will get so complicated, at that point you're better off
writing a small function, like some of the other answers. But you should be
able to get away with two substitution commands without it getting too crazy.
One for the first two strings (a b and a-b), and one for the third
(a / b / c).
%s:\v(\a+)[\ -](\a+):"\0"\ =>\ \1_\2
%s:\v(\a+)\s*/\s*(\a+)\s*/\s*(\a+):"\0"\ =>\ \1_\2_\3
Explanation:
%s:\v(\a+)[\ -](\a+):"\0"\ =>\ \1_\2
s: : - Substitute command (with delimiter `:`)
\v - Very Magic mode *
( ) ( ) - Capture contained matches into numbered sub-expressions
\a+ \a+ - Match at least one alphanumeric character
[\ -] - Match either ` ` (space) or `-`
" "\ =>\ _ - Literal text
\0 - Replace with entire matched text
\1 \2 - Replace with first and second `()` sub-expression, respectively
% - Execute command on every line in file (optional)
%s:\v(\a+)\s*/\s*(\a+)\s*/\s*(\a+):"\0"\ =>\ \1_\2_\3
s: : - Substitute command (with delimiter `:`)
\v - Very Magic mode *
( ) ( ) ( ) - Capture contained matches into numbered sub-expressions
\a+ \a+ \a+ - Match at least one alphanumeric character
\s*/\s* \s*/\s* - Match a `/` and any surrounding spaces
" "\ =>\ _ _ - Literal text
\0 - Replace with entire matched text
\1 \2 \3 - Replace with first, second, and third `()` sub-expression, respectively
% - Execute command on every line in file (optional)
* This eliminates the need for a lot of ugly backslashes.
See `:h /magic` and `:h /\v`