Haskell列表理解0和1

Question

I am trying to write a function

row :: Int -> Int -> [Int]
row n v

that returns a list of n integers, all 0 's, except for the v th element, which needs to be a 1 .

For instance,

row 0 0 = []
row 5 1 = [1,0,0,0,0]
row 5 3 = [0,0,1,0,0]

I am new to Haskell and having a lot of difficulty with this. In particular I can't figure out how to make it repeat 0 's. I understand the concept of building a list from let's say [1..n] , but I just get [1,2,3,4,5]

Any help with this would be greatly appreciated. Thank you.

Answer 1

尝试：

let row n v = map (\x -> if x == v then 1 else 0) [1..n]

Answer 2

Here a "monadic" solution:

row n v = [(v-1, 0), (1, 1), (n-v, 0)] >>= (uncurry replicate)

The replicate function repeats a given value a number of times, eg replicate (v-1) 0 gives a list of v-1 zeros. The uncurry is used to modify the replicate in order to accept a tuple instead of two single arguments. The funny operator >>= is the heart of a monad; for lists it is the same as concatMap with flipped arguments.

Answer 3

With a comprehensive list :

 row n v = [if x == v then 1 else 0 | x <- [1..n]]

Or using fromEnum (thanks dave4420)

 row n v = [fromEnum (x == v) | x <- [1..n]]

Answer 4

这应该也有效：

row n v = replicate (v-1)­ 0 ++ [1] ++ repl­icate (n-v)­ 0

Answer 5

And yet another solution, recursively building up the list:

row :: Int -> Int -> [Int]
row 0 _ = []
row n 1 = 1 : (row (n-1) 0)
row n m = 0 : (row (n-1) (m-1))

And a more readable one, where zeros are "taken":

row :: Int -> Int -> [Int]
row 0 _ = []
row n m = take (m - 1) zeros ++ [1] ++ take (n - m) zeros
    where zeros = (iterate id 0)

Answer 6

A simple recursive loop with two temporary variables c and lst . c is for counting and lst is list which we have to return.

row :: Int -> Int -> [ Int ]
row 0 0 = []
row n v = rowHelp n v 1 [] where
    rowHelp n v c lst
            | c > n = lst
            | v == c = rowHelp n v ( c + 1 ) ( lst ++ [ 1 ] )
            | otherwise = rowHelp n v ( c + 1 ) ( lst ++ [ 0 ] )

~
~

Answer 7

the fun with haskell is that it let's you write your program very much the way you would express the algorithm. So try:

row n v = [if (x `mod` v==0) then 1 else 0  | x <- [1..n] ]

At first you create a list from 1,2 to n. Then you check if the number is divisible by v, if it is, 1 is inserted in the output list, if not 0.

Examples:

> row 0 0
[]
> row 5 1
[1,0,0,0,0]
> row 5 3
[0,0,1,0,0]
> row 15 3
[0,0,1,0,0,1,0,0,1,0,0,1,0,0,1]

HTH Chris

Answer 8

I like to demonstrate a top down approach, based on Chris's solution:

row n v = result           
    where
        result = take n numbers        -- our result will have a length of n
        numbers = map trans [1,2,..]   -- and is some transformation of
                                       -- the list of natural numbers
        trans e
           | e `mod` v == 0  = 1       -- let every v-th element be 1
           | otherwise       = 0       -- 0 otherwise

This style emphasizes the idea in functional programming that one writes down what a certain value like row nv is supposed to be , rather than trying to write down what a function does . In reminiscence of a well known joke about the lesser known pragramming language Sartre one could say that in pure functional programming functions do nothing, they just are .