Cartesian product of 2 lists in Haskell

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最佳答案

This is very easy with list comprehensions. To get the cartesian product of the lists xs and ys, we just need to take the tuple (x,y) for each element x in xs and each element y in ys.

This gives us the following list comprehension:

cartProd xs ys = [(x,y) | x <- xs, y <- ys]

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Well, one very easy way to do this would be with list comprehensions:

cartProd :: [a] -> [b] -> [(a, b)]
cartProd xs ys = [(x, y) | x <- xs, y <- ys]

Which I suppose is how I would do this, although I'm not a Haskell expert (by any means).

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something like:

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

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The right way is using list comprehensions, as other people have already pointed out, but if you wanted to do it without using list comprehensions for any reason, then you could do this:

cartProd :: [a] -> [b] -> [(a,b)]
cartProd xs [] = []
cartProd [] ys = []
cartProd (x:xs) ys = map (\y -> (x,y)) ys ++ cartProd xs ys

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As other answers have noted, using a list comprehension is the most natural way to do this in Haskell.

If you're learning Haskell and want to work on developing intuitions about type classes like Monad, however, it's a fun exercise to figure out why this slightly shorter definition is equivalent:

import Control.Monad (liftM2)


cartProd :: [a] -> [b] -> [(a, b)]
cartProd = liftM2 (,)

You probably wouldn't ever want to write this in real code, but the basic idea is something you'll see in Haskell all the time: we're using liftM2 to lift the non-monadic function (,) into a monad—in this case specifically the list monad.

If this doesn't make any sense or isn't useful, forget it—it's just another way to look at the problem.

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If your input lists are of the same type, you can get the cartesian product of an arbitrary number of lists using sequence (using the List monad). This will get you a list of lists instead of a list of tuples:

> sequence [[1,2,3],[4,5,6]]
[[1,4],[1,5],[1,6],[2,4],[2,5],[2,6],[3,4],[3,5],[3,6]]

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Yet another way, using the do notation:

cartProd :: [a] -> [b] -> [(a,b)]
cartProd xs ys = do x <- xs
y <- ys
return (x,y)

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Yet another way to accomplish this is using applicatives:

import Control.Applicative


cartProd :: [a] -> [b] -> [(a,b)]
cartProd xs ys = (,) <$> xs <*> ys

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There is a very elegant way to do this with Applicative Functors:

import Control.Applicative


(,) <$> [1,2,3] <*> [4,5,6]
-- [(1,4),(1,5),(1,6),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6)]

The basic idea is to apply a function on "wrapped" arguments, e.g.

(+) <$> (Just 4) <*> (Just 10)
-- Just 14

In case of lists, the function will be applied to all combinations, so all you have to do is to "tuple" them with (,).

See http://learnyouahaskell.com/functors-applicative-functors-and-monoids#applicative-functors or (more theoretical) http://www.soi.city.ac.uk/~ross/papers/Applicative.pdf for details.

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Here is my implementation of n-ary cartesian product:

crossProduct :: [[a]] -> [[a]]
crossProduct (axis:[]) = [ [v] | v <- axis ]
crossProduct (axis:rest) = [ v:r | v <- axis, r <- crossProduct rest ]

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It's a job for sequenceing. A monadic implementation of it could be:

cartesian :: [[a]] -> [[a]]
cartesian [] = return []
cartesian (x:xs) = x >>= \x' -> cartesian xs >>= \xs' -> return (x':xs')


*Main> cartesian [[1,2,3],[4,5,6]]
[[1,4],[1,5],[1,6],[2,4],[2,5],[2,6],[3,4],[3,5],[3,6]]

As you may notice, the above resembles the implementation of map by pure functions but in monadic type. Accordingly you can simplify it down to

cartesian :: [[a]] -> [[a]]
cartesian = mapM id


*Main> cartesian [[1,2,3],[4,5,6]]
[[1,4],[1,5],[1,6],[2,4],[2,5],[2,6],[3,4],[3,5],[3,6]]

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Other answers assume that the two input lists are finite. Frequently, idiomatic Haskell code includes infinite lists, and so it is worthwhile commenting briefly on how to produce an infinite Cartesian product in case that is needed.

The standard approach is to use diagonalization; writing the one input along the top and the other input along the left, we could write a two-dimensional table that contained the full Cartesian product like this:

   1  2  3  4 ...
a a1 a2 a3 a4 ...
b b1 b2 b3 b4 ...
c c1 c2 c3 c4 ...
d d1 d2 d3 d4 ...


.  .  .  .  . .
.  .  .  .  .  .
.  .  .  .  .   .

Of course, working across any single row will give us infinitely elements before it reaches the next row; similarly going column-wise would be disastrous. But we can go along diagonals that go down and to the left, starting again in a bit farther to the right each time we reach the edge of the grid.

a1


a2
b1


a3
b2
c1


a4
b3
c2
d1

...and so on. In order, this would give us:

a1 a2 b1 a3 b2 c1 a4 b3 c2 d1 ...

To code this in Haskell, we can first write the version that produces the two-dimensional table:

cartesian2d :: [a] -> [b] -> [[(a, b)]]
cartesian2d as bs = [[(a, b) | a <- as] | b <- bs]

An inefficient method of diagonalizing is to then iterate first along diagonals and then along depth of each diagonal, pulling out the appropriate element each time. For simplicity of explanation, I'll assume that both the input lists are infinite, so we don't have to mess around with bounds checking.

diagonalBad :: [[a]] -> [a]
diagonalBad xs =
[ xs !! row !! col
| diagonal <- [0..]
, depth <- [0..diagonal]
, let row = depth
col = diagonal - depth
]

This implementation is a bit unfortunate: the repeated list indexing operation !! gets more and more expensive as we go, giving quite a bad asymptotic performance. A more efficient implementation will take the above idea but implement it using zippers. So, we'll divide our infinite grid into three shapes like this:

a1 a2 / a3 a4 ...
/
/
b1 / b2 b3 b4 ...
/
/
/
c1 c2 c3 c4 ...
---------------------------------
d1 d2 d3 d4 ...


.  .  .  . .
.  .  .  .  .
.  .  .  .   .

The top left triangle will be the bits we've already emitted; the top right quadrilateral will be rows that have been partially emitted but will still contribute to the result; and the bottom rectangle will be rows that we have not yet started emitting. To begin with, the upper triangle and upper quadrilateral will be empty, and the bottom rectangle will be the entire grid. On each step, we can emit the first element of each row in the upper quadrilateral (essentially moving the slanted line over by one), then add one new row from the bottom rectangle to the upper quadrilateral (essentially moving the horizontal line down by one).

diagonal :: [[a]] -> [a]
diagonal = go [] where
go upper lower = [h | h:_ <- upper] ++ case lower of
[]         -> concat (transpose upper')
row:lower' -> go (row:upper') lower'
where upper' = [t | _:t <- upper]

Although this looks a bit more complicated, it is significantly more efficient. It also handles the bounds checking that we punted on in the simpler version.

But you shouldn't write all this code yourself, of course! Instead, you should use the universe package. In Data.Universe.Helpers, there is (+*+), which packages together the above cartesian2d and diagonal functions to give just the Cartesian product operation:

Data.Universe.Helpers> "abcd" +*+ [1..4]
[('a',1),('a',2),('b',1),('a',3),('b',2),('c',1),('a',4),('b',3),('c',2),('d',1),('b',4),('c',3),('d',2),('c',4),('d',3),('d',4)]

You can also see the diagonals themselves if that structure becomes useful:

Data.Universe.Helpers> mapM_ print . diagonals $ cartesian2d "abcd" [1..4]
[('a',1)]
[('a',2),('b',1)]
[('a',3),('b',2),('c',1)]
[('a',4),('b',3),('c',2),('d',1)]
[('b',4),('c',3),('d',2)]
[('c',4),('d',3)]
[('d',4)]

If you have many lists to product together, iterating (+*+) can unfairly bias certain lists; you can use choices :: [[a]] -> [[a]] for your n-dimensional Cartesian product needs.

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Just adding one more way for the enthusiast, using only recursive pattern matching.

cartProd :: [a]->[b]->[(a,b)]
cartProd _ []=[]
cartProd [] _ = []
cartProd (x:xs) (y:ys) = [(x,y)] ++ cartProd [x] ys  ++ cartProd xs ys ++  cartProd xs [y]

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Recursive pattern matching with out List comprehension

crossProduct [] b=[]
crossProduct (x : xs) b= [(x,b)] ++ crossProduct xs b


cartProd _ []=[]
cartProd x (u:uv) = crossProduct x u ++ cartProd x uv

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If all you want is the Cartesian product, any of the above answers will do. Usually, though, the Cartesian product is a means to an end. Usually, this means binding the elements of the tuple to some variables, x and y, then calling some function f x y on them. If this is the plan anyway, you might be better off just going full monad:

do
x <- [1, 2]
y <- [6, 8, 10]
pure f x y

This will produce the list [f 1 6, f 1 8, f 1 10, f 2 6, f 2 8, f 2 10].