fold

问题 This question already has answers here : Outputting the contents of a list of a custom data type (2 answers) Closed 5 years ago . I have avgRatingsForDirector :: String -> [Film] -> [Int] avgRatingsForDirector _ [] = [] avgRatingsForDirector requestedDirector ((Film _ director _ ((_, rating):ratings)):restOfFilms) | requestedDirector == director = [rating] ++ avgRatingsForDirector requestedDirector restOfFilms | otherwise = avgRatingsForDirector requestedDirector restOfFilms This outputs a

问题 the type is defined as follows: gap :: (Eq a) => a -> a -> [a] -> Maybe Int I have been stuck on this problem for more than an hour and have no idea how to approach the problem. I am aware that it requires the use of fold and am familiar with that topic. Please take into consideration that either foldl or foldr must be used. The output when called ought to look like this gap 3 8 [1..10] =Just 5 gap 8 3 [1..10] =Nothing gap 'h' 'l' "hello" =Just 2 gap 'h' 'z' "hello" =Nothing 回答1: You might

问题 I'm working through "Haskell Programming From First Principles". In the chapter on Folding Lists, exercise 5f, when I evaluate foldr const 'a' [1..5] I get No instance for (Num Char) arising from the literal ‘1’ However, with foldl const 'a' [1..5] I get 'a' . I get that the folds are lazy, foldr doesn't traverse the spine and foldl does. But even looking at the definitions of foldr and foldl, foldr f z [] = z foldr f z (x:xs) = f x (foldr f z xs) foldl f z [] = z foldl f z (x:xs) = foldl f

问题 I'm trying to learn haskell by solving some online problems and training exercises. Right now I'm trying to make a function that'd remove adjacent duplicates from a list. Sample Input "acvvca" "1456776541" "abbac" "aabaabckllm" Expected Output "" "" "c" "ckm" My first though was to make a function that'd simply remove first instance of adjacent duplicates and restore the list. module Test where removeAdjDups :: (Eq a) => [a] -> [a] removeAdjDups [] = [] removeAdjDups [x] = [x] removeAdjDups

问题 I'm trying to declare a function, string list -> string, that with the input for instance ["Chicago","city","USA"] should return "Chicago city USA" . What I did so far was this: fun gather ts = foldr op ^ "" ts; This seems to be somewhat along the lines, however the problem is, I would like to include the spaces between the words, as this function would return "ChigagocityUSA" . 回答1: Yes, the problem is that ^ is a function that for two strings "foo" and "bar" returns "foobar", although you

问题 Trying to learn to think like a functional programmer a little more---I'd like to transform a data set with what I think is either a fold or a reduce operation. In R, I would think of this as a reshape operation, but I'm not sure how to translate that thinking. My data is a json string that looks like this: s = '[ {"query":"Q1", "detail" : "cool", "rank":1,"url":"awesome1"}, {"query":"Q1", "detail" : "cool", "rank":2,"url":"awesome2"}, {"query":"Q1", "detail" : "cool", "rank":3,"url":

问题 I am trying to dive deep in the folds, considering it seems a very powerful asset to me. However, can you help me with this: foldr (/) 2 [1,2,3] -- (1/(2/(3/2))), result 0,75 {where 2 is base) foldr1 (/) [2,2,3] -- (1/(2/(3/2))), result 3.00 {where 2 is base) I think I am seriously overseeing an essential difference between the folds. Thx 回答1: foldr :: (a -> b -> b) -> b -> [a] -> b has as implementation: foldr :: (a -> b -> b) -> b -> [a] -> b foldr _ z [] = z foldr f z (x:xs) = f x (foldr f

问题 Is there standard implementation of fold (reduce, aggregate etc) for one dimensional vector in Octave? If no, is there any way to express fold without using a loop statement? 回答1: The miscellaneous package provides the function reduce. For example, octave:6> reduce(@(x,y)(x*y), [1:5]) ans = 120 If you look at the source code for reduce , you'll see that it is a fairly simple Octave function that is implemented with a for loop, so it won't be more efficient than implementing the reduction with

问题 I have to make an implementation of a Binary Tree instantiate a typeclass: class Set s where add :: (Eq a) => a -> s a -> s a remove :: (Eq a) => a -> s a -> s a exists :: (Eq a) => a -> s a -> Bool fold :: (a -> b -> b) -> s a -> b -> b data BTree k v = Empty | Node k v (BTree k v) (BTree k v) deriving (Show) All went well until I had to implement a fold for a binary tree. The issue I'm having is that I don't really know how to keep the type declaration of my function with a signature like

问题 Problem: using fold , take from the list elements which are on the even positions: GHCi> evenOnly [1..10] [2,4,6,8,10] GHCi> evenOnly ['a'..'z'] "bdfhjlnprtvxz" evenOnly :: [a] -> [a] evenOnly = undefined I decided at first to get a list of alternating 0 -es and 1 -s: [0,1,0,1..] Prelude> let g = iterate (\x -> (x + 1) `mod` 2) 0 Prelude> take 10 $ g [0,1,0,1,0,1,0,1,0,1] Then zip it with the original list, getting a list of pairs: [(x1, 0), (x2,1), (x3,0) .. (xn, ?)] : Prelude> zip g [1,2,3