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Today's Topics:
1. Using Functions with Fold (A. Mc.)
2. Re: Using Functions with Fold (Francesco Ariis)
----------------------------------------------------------------------
Message: 1
Date: Sun, 31 Jan 2021 20:49:39 -0800
From: "A. Mc." <47dragonf...@gmail.com>
To: beginners@haskell.org
Subject: [Haskell-beginners] Using Functions with Fold
Message-ID:
<CAOsti3=yXK96dTfm2c1LWhce8oX+9Dk=245uuezo0spdvda...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
Hello,
I feel like I need to go back to basics, so I've made a toy problem that
takes a list of Ints, say [10, 20, 30, 40, 50, 60] and, in theory, iterates
through the list in successive pairs i.e. [10, 20], [20, 30], [30, 40],
[40, 50], [50, 60] and looks for a specific pair set, say 30, 40, and
accumulates the number of times that pair appears in the list (for example
purpose, there are no repeats, but assuming there were....). What I've
done is the following:
---take starting at position 2, in increments of 2 with overlap allowed
takeAtTwo :: [Int] -> Int -> [Int]
takeAtTwo list position = take (2 + position) list
---expects range of 0 to length list - 2
grabTwo :: [Int] -> Int -> [Int]
grabTwo list position = drop position (takeAtTwo list position)
-- and this does NOT work, half-pseudocode
--- pass in a pair of Int, say [30, 40] and sum up the number of appearances
numPair list = foldr (\pair acc -> if grabTwo list iterator == pair then
acc + 1 else acc) 0
However, I am unsure how to get this to work properly with fold.
Suggestions? Thanks in advance and thank you for your time.
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Message: 2
Date: Mon, 1 Feb 2021 06:07:32 +0100
From: Francesco Ariis <fa...@ariis.it>
To: beginners@haskell.org
Subject: Re: [Haskell-beginners] Using Functions with Fold
Message-ID: <20210201050732.GA21439@extensa>
Content-Type: text/plain; charset=utf-8
Il 31 gennaio 2021 alle 20:49 A. Mc. ha scritto:
> I feel like I need to go back to basics, so I've made a toy problem that
> takes a list of Ints, say [10, 20, 30, 40, 50, 60] and, in theory, iterates
> through the list in successive pairs i.e. [10, 20], [20, 30], [30, 40],
> [40, 50], [50, 60] and looks for a specific pair set, say 30, 40, and
> accumulates the number of times that pair appears in the list (for example
> purpose, there are no repeats, but assuming there were....). What I've
> done is the following:
>
> ---take starting at position 2, in increments of 2 with overlap allowed
> takeAtTwo :: [Int] -> Int -> [Int]
> takeAtTwo list position = take (2 + position) list
>
> ---expects range of 0 to length list - 2
> grabTwo :: [Int] -> Int -> [Int]
> grabTwo list position = drop position (takeAtTwo list position)
>
> -- and this does NOT work, half-pseudocode
> --- pass in a pair of Int, say [30, 40] and sum up the number of appearances
> numPair list = foldr (\pair acc -> if grabTwo list iterator == pair then
> acc + 1 else acc) 0
>
> However, I am unsure how to get this to work properly with fold.
> Suggestions? Thanks in advance and thank you for your time.
You might have better luck with first generating the pairs and then
passing them to another function.
λ> prova as = zip as (tail as)
λ> prova [10,20..60]
[(10,20),(20,30),(30,40),(40,50),(50,60)]
Counting now is a simple matter of
λ> length . filter (==(30,40)) $ it
1
-- or use a fold if you prefer
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