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
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