list - 在 Haskell 中生成具有不同数量字符的字符串数组

Question

我需要这样的东西：

[[i]++[j]| i <- ['a'..'d'], j <- ['a'..'d']] where I get the output:
["aa","ab","ac","ad","ba","bb","bc","bd","ca","cb","cc","cd","da","db","dc","dd"]

我需要的是一种更动态的方式来生成这个输出数组。所以如果得到一个整数值 3 我应该看起来像：

[[i]++[j]++[k]| i <- ['a'..'d'], j <- ['a'..'d'], k <- ['a'..'d']

Answer 1

replicateM n ['a'..'d']

n是重复的次数。

Answer 2

为了澄清 Louis Wasserman 的回答：

replicateM i m相当于：

do a1 <- m
   a2 <- m
   a3 <- m
   ...
   ai <- m
   return [a1, a2, a3, ..., ai]

例如，replicateM 3 getLine是一个IO [String]将获得三行并为您提供字符串列表的操作：

do line1 <- getLine
   line2 <- getLine
   line3 <- getLine
   return [line1, line2, line3]

现在，请记住列表推导式只是花哨的语法[] Monad：

[[i, j] | i <- ['a'..'d'], j <- ['a'..'d']]

=

do i <- ['a'..'d']
   j <- ['a'..'d']
   return [i, j]

这正是replicateM 2 ['a'..'d']看起来的样子。

Answer 3

你可以使用以下

seq i xs | i > 0 = [x:ys | x <- xs, ys <- seq (i-1) xs]
         | otherwise = [[]]