Haskell - Create list by adding element from one list only if its equal to element at same index in another list

Question

Here is the problem I want to solve:

The two lists [1,2,3,4] [2,2,3,4] is to become [2,3,4] , because elements at index position zero are not equal. So elements are compared for equality with respect to their index. You can assume equal length lists as input.

So I created a function that solved this with recursion:

oneHelper :: (Eq t) => [t] -> [t] -> [t]
oneHelper [] [] = [] 
oneHelper (x:xs) (y:ys) = if x == y then [x] ++ oneHelper xs ys
                          else oneHelper xs ys

Then I tried to solve it with list comprehension like this:

test a b = [x | x <- a, y <- b, x == y]

which just gives me [2,2,3,4] with example input above used. I feel like there is a neat way of solving this with a list comprehension, or just a more neat way than the solution I came up with in general, but I am struggling to reach that solution. Does anyone see something better than what I did for the problem?

Thanks

willeM_ Van Onsem · Accepted Answer · 2019-11-22 11:20:53Z

1

If you use the two generators, you will iterate over all possible combinations of the elements in the first list (a) and second list (b).

You probably want to use zip :: [a] -> [b] -> [(a, b)] here where we iterate over the two lists concurrently:

test a b = [x | (x, y) <- zip a b, x == y]

answered Nov 22, 2019 at 11:20

willeM_ Van Onsem

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1 Comment

clickedok Over a year ago

"if you use the two generators, you will iterate over all possible combinations..", Thanks for the enlightenment.

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