在Haskell中组合地图和长度

Question

我有两个问题,使用map和length第一个应该给我回字数,但它只计算列表中的元素.

countWords :: [String]-> Int
countWords xs = length (map (words) xs)

countWords ["asd  qwe", "-- Foo", "", "\thello world "] => 4--instead it have six words

第二个是棘手的,因为它应该为整个列表返回一个int.我只计算各个元素的字符,而不是整数.

countChars :: [String]-> [Int] --it should be Int
countChars xs = map (\w -> length (w)) xs 

countChars ["asd  qwe", "-- Foo", "", "\thello world "] => [8,6,0,13]--it should give back the sum of this list which is 27

Answer 1

对于第二个,您只需要调用sum结果.

countChars xs = sum (map (\w -> length (w)) xs)

也可以改写为

countChars xs = sum $ map length xs 

对于第一个,我们必须计算每个元素中的单词数,最后对结果求和.

words会给你一个单词列表,所以在做完之后map (words) xs(不需要用btw的单词括起来),你会得到以下内容:

map words ["asd  qwe", "-- Foo", "", "\thello world "] 
=> 
[["asd","qwe"],["--","Foo"],[],["hello","world"]]

你要做的第一件事是获得每个子列表的长度,你可以适应你的 map

map (\x -> length (words x)) xs

现在,结果是:

[2,2,0,2]

通过对结果运行求和,得到6.所以最终的结果是

countWords :: [String]-> Int
countWords xs = sum $ map (\x -> length (words x)) xs

使用一些语法糖,您可以执行以下操作,但我发现大多数初学者都对此感到困惑:

countWords xs = sum $ map (length . words) xs

甚至更好

countWords = sum . map (length . words)