Yeah, the question is so open that it's hard to answer. Was thinking recommending to inline the desired list all together, but thought it too impolite :).

But the `concatMap` is the simplest one (to replace all "z"s that is), and if the OP is interested, then for lists >>= is defined as concatMap, so:

a = ["x", "y", "z"]
b = ["f", "g"]

f :: String -> [String]
f str = case str of
  "z" -> b
  _ -> [str]

answer = a >>= f

(Didn't run this through the compiler)


On Mon, Nov 6, 2017 at 12:20 AM, Richard A. O'Keefe <ok@cs.otago.ac.nz> wrote:


On 4/11/17 11:48 PM, mirone wrote:
Hello everyone!
If I have a list A : ["x", "y", "z"]
and a list B: ["f", "g"].
Here is my question:
What's the simplest way to replace "z" in the list A by list B, and get
["x", "y", "f", "g"] ?

f _ _ = ["x","y","f","g"]

That's a joke, but it really isn't clear what the question means.
One possibility is

f a b = init a ++ b

f [1,2,3,4,5] [9,8,7] => [1,2,3,4,9,8,7]

If you mean that you want to replace "z" by ... wherever it occurs,
something like
-- (f needle straws haystack) replaces each occurrence of needle as
-- an element of haystack with the elements of straws.
f :: Eq t => t -> [t] -> ([t] -> [t])
f needle straws =
  concatMap (\straw -> if straw == needle then straws else [straw])

f "z" ["f","g"] ["x","z","y","z"] => ["x","f","g","y","f","g"]

Your question admits of too many generalisations for us to be really
helpful.

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Markus Läll