I have a data type:
data Numbers = Numbers {a::Int, b::Int}
How can I construct [Numbers] in order to get the same effect as
[[a,b] | a <- [1,2], b <- (filter (/=a) [1,2])]
so the result will be similar to [[1,2],[2,1]]
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2 Answers
You have to use Numbers as the constructor (note: [] is also a constructor, only with a specific syntax sugar, so there is no fundamental difference).
data Numbers = Numbers {a::Int, b::Int}
deriving Show
main = print [ Numbers a b | a <- [1, 2], b <- filter (/=a) [1, 2] ]
> main
[Numbers {a = 1, b = 2},Numbers {a = 2, b = 1}]
3 Comments
PinkiePie-Z
if it is a really big list, like a<-[1..100] b<-[1..100], how can i implement it in a hs file ? can i do sth like [Numbers] = [Numbers a b | a<-[1..100], b<- filter (/=a) [1..100]]
huon
@Z.pyyyy, yes, sort of,
x = [Numbers a b | a <- [1..100], b <- filter (/= a) [1..100]].
Satvik
@Z.pyyyy Read some haskell turorial like YAHT or real world haskell.
This seems to be nothing but a selection with removal. You can find efficient code to do that in an older question.
If it's certainly about exactly two elements, then this implementation will be efficient:
do x:ys <- tails [1..3]
y <- ys
[(x, y), (y, x)]
1 Comment
AndrewC
I think the question's quite clear that it's about more than two elements, but yes, that works in this case.