问题描述
我想让所有可能的组合与2列表。这里有一个函数可以做到这一点：
I want to make all possible combinations of subgroups with 2 lists. Here is a function that does just this:
getCombinations :: [a] -> [[a]]
getCombinations na = do
a <- na
b <- na
[[a,b]]
如果您将abc传递给此函数，它将返回：
If you pass "abc" to this function, it returns this:
["aa","ab","ac","ba","bb","bc","ca","cb","cc"]
对相同方法的简单修改可以返回3个列表的组合，而不是两个。
A simple modification of the same method could return combinations of 3 lists instead of two.
getCombinations :: [a] -> [[a]]
getCombinations na = do
a <- na
b <- na
c <- na
[[a,b,c]]
传递abc作为参数的结果：
Result of passing "abc" as an argument:
["aaa","aab","aac","aba","abb","abc","aca","acb","acc",
"baa","bab","bac","bba","bbb","bbc","bca","bcb","bcc",
"caa","cab","cac","cba","cbb","cbc","cca","ccb","ccc"]
使其扩展到任意数量的列表最简单的方法是什么？以下是类型声明的样子：
What's the simplest way to make it scale to an arbitrary number of lists? Here is what the type declaration should look like:
getCombinations :: Int -> [a] -> [[a]]
推荐答案
HREF = http://hackage.haskell.org/package/containers-0.5.5.1/docs/Data-Sequence.html#v:replicateM> replicateM ：
What you want is replicateM:
replicateM :: Int -> m a -> m [a]
定义非常简单：
The definition is as simple as:
replicateM n = sequence . replicate n
，所以它是序列 monad在这里做了真正的工作。
so it's sequence on the list monad that's doing the real work here.
这篇关于列表的排列 - 哈斯克尔的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！