Comments on: Replace in Haskell

By: Anonymous

Anonymous — Tue, 05 Aug 2008 06:37:37 +0000

legend!

By: Joseph

Joseph — Sat, 24 Nov 2007 16:13:50 +0000

The first function here is missing a check that can cause infinite recursion if an empty `find` list is given

 replace "abc" "" "A"

it needs something like...

replace _ [] list = list

The second function given only replaces the first instance of the string, and if the `find` list is empty then it is equivalent to

> to:xs

the fourth line could be changed to...

then to ++ subst from to (drop (length from) xs)

but then it has the same problem as the first function, and needs another check

substr [] _ list = list -- or something like that

I chose to write my replace as follows

replace::(Eq a) => [a] -> [a] -> [a] -> [a]
replace [] newSub list = join newSub list
replace oldSub newSub list = _replace list where
	_replace list@(h:ts) = if isPrefixOf oldSub list
		then newSub ++ _replace (drop len list)
		else h : _replace ts
	_replace [] = []
	len = length oldSub

I decided that an empty match list would mean inserting the new element between all of the elements in the list, but maybe the new element should be inserted on the beginning of the list as well.

join::[a] -> [a] -> [a]
join glue [h] = [h]
join glue (h:ts) = h : glue ++ join glue ts
join _ [] = []

another function that works nicely with replace

strip::(Eq a) => [a] -> [a] -> [a]
strip old = replace old []

By: Cale Gibbard

Cale Gibbard — Sat, 20 Jan 2007 22:52:24 +0000

Yeah, the general rule is to always order parameters to functions in order of increasing likelihood of variation. In this case, for instance, it seems more likely that the same substitutions will be used on many different lists, rather than many different substitutions on the same initial list.

By: Thomas David Baker

Thomas David Baker — Wed, 17 Jan 2007 20:19:59 +0000

Thanks jethr0. I see you’ve moved the list to act on to the last of the arguments. Someone else on #haskell suggested that too, to improve composability.

By: jethr0

jethr0 — Tue, 16 Jan 2007 14:09:51 +0000

sorry, too smart (sic!) for my own good. my “isPrefixOf” is broken, instead use “Data.List.isPrefixOf”

By: jethr0

jethr0 — Tue, 16 Jan 2007 13:51:27 +0000

“head” and “tail” should be used sparingly, because “head” is a partially defined function (see “head []“).

subst _    _  [       ] = []
subst from to xs@(a:as) =
    if isPrefixOf from xs
        then to ++ drop (length from) xs
        else a : subst from to as
    where isPrefixOf as bs = and $ zipWith (==) as bs