RegEx expression to keep first appearance of word grouping - regex

Question

I have the following RegEx (\[[^]]+])(?=.*\1)
which identifies the first set of appearances of a duplicate word group inside a string (each word group is enclosed between [ ] brackets). However, I am trying to come up with a RegEx that identifies the last set of appearances of a duplicate word group. Reason being, I need to remove duplicate word groups while retaining the order in which each group appears in the overall string.
Using the following string as an example whereby only [John Smith] and [Jane Doe] are duplicate word groups:
[John Smith][John Smith][Mr. Smith][Jane Doe][Mrs. Doe][John Smith][Jane Doe][Doe][John][Smith John][John Smith Sr]
After using my RegEx in a RegEx Replace formula, I get the below:
[Mr. Smith][Mrs. Doe][John Smith][Jane Doe][Doe][John][Smith John][John Smith Sr]"
However, I need my RegEx Replace formula to give me:
[John Smith][Mr. Smith][Jane Doe][Mrs. Doe][Doe][John][Smith John][John Smith Sr]
I have tried many ways to achieve the latter with no luck. Thanks in advance.

Answer 1

Considering infinite-width lookbehinds:
(\[[^\][]+])(?<=\1.*\1)
See regex proof.
EXPLANATION
--------------------------------------------------------------------------------
( group and capture to \1:
--------------------------------------------------------------------------------
\[ '['
--------------------------------------------------------------------------------
[^\][]+ any character except: '\]', '[' (1 or
more times (matching the most amount
possible))
--------------------------------------------------------------------------------
] ']'
--------------------------------------------------------------------------------
) end of \1
--------------------------------------------------------------------------------
(?<= look behind to see if there is:
--------------------------------------------------------------------------------
\1 what was matched by capture \1
--------------------------------------------------------------------------------
.* any character except \n (0 or more times
(matching the most amount possible))
--------------------------------------------------------------------------------
\1 what was matched by capture \1
--------------------------------------------------------------------------------
) end of look-behind

Answer 2

Many regex engines do not support variable-length lookbehinds, but most support variable-length lookaheads. When working with an engine that supports variable-length lookaheads, but not variable-length lookbehinds, one approach that often works is to reverse the string, modify the (reversed) string with a regex and then reverse the resulting string. That approach could be used here.
Suppose, for example, the string were
"[John Smith][John Smith][Mr. Smith][Jane Doe][John Smith][Jane Doe][Doe]"
Reversing the string produces
"]eoD[]eoD enaJ[]htimS nhoJ[]eoD enaJ[]htimS .rM[]htimS nhoJ[]htimS nhoJ["
We now convert matches of the following expression to empty strings:
(\][^[]*\[)(?=.*\1)
which produces
"]eoD[]eoD enaJ[]htimS .rM[]htimS nhoJ["
Demo
Lastly, we reverse that string to obtain
"[John Smith][Mr. Smith][Jane Doe][Doe]"
The regular expression can be written in free-spacing mode to make it self-documenting:
( # begin capture group 1
\] # match ']'
[^[]* # match one or more (as many as possible) chars other than '['
\[ # match '['
) # end capture group 1
(?= # begin a positive lookahead
.* # match one or more (as many as possible) chars
\1 # match the content of capture group 1
) # end the positive lookahead
At first glance this may seem a kludge, but since reversing a string is so easy in any language it does provide a useful work-around in some cases. Mind you, doing what you want to do here in code is pretty easy in most languages. In Ruby, for example, you could write (str being a variable holding the string)
str.scan(/\[[^\]]*\]/).uniq.join