列表乘以嵌套列表

Question

考虑列表清单

thisList = [[1], [1, 1], [1, 1, 1]] 

我怎么会多thisList，这样它会产生另一个列表

anotherList = [[1], [2, 2], [3, 3, 3]] 

我所做的功能

reps = [1] : map (\ns -> head ns:ns) reps 

哪产生thisList

感谢所有帮助

Answer 1

你可以这样做：

zipWith (\x -> map (const x)) [1..] thisList 

用法示例：

Prelude> let thisList = [[1], [1,1], [1,1,1]] 
Prelude> zipWith (\x -> map (const x)) [1..] thisList 
[[1],[2,2],[3,3,3]] 

或者简单：

zipWith (map . const) [1..] thisList 

---

这是很容易。 zipWith f as bs相当于map (uncurry f) $ zip as bs。因此，我们有：

zip [1..] thisList == [(1, [1]), (2, [1,1,]), (3, [1,1,1])] 

然后我们应用f每对等：

map (const 1) [1] == [const 1 1] == [1] 
map (const 2) [1,1] == [const 2 1, const 2 1] == [2,2] 
map (const 3) [1,1,1] == [const 3 1, const 3 1, const 3 1] == [3,3,3] 

---

如果你的意思是长度n的子表应该由长度n的[n, n, ..., n]更换，如：

thisList = [[1, 1], [1], [1], [1, 1, 1]] 
result = [[2,2], [1], [1], [3,3,3]] 

然后，你必须改变的办法：

map (\xs -> let len = length xs in replicate len len) thisList 

示例：你为什么不使用`[1，2，3]`对列表符号

Prelude> let thisList = [[1, 1], [1], [1], [1, 1, 1]] 
Prelude> map (\xs -> let len = length xs in replicate len len) thisList 
[[2,2],[1],[1],[3,3,3]]