List 哈斯克尔:我如何附加到元组列表中?
对不起,我刚开始学哈斯克尔。我不明白如何在每个子列表中附加一个元组 我目前的名单是:List 哈斯克尔:我如何附加到元组列表中?,list,haskell,append,tuples,List,Haskell,Append,Tuples,对不起,我刚开始学哈斯克尔。我不明白如何在每个子列表中附加一个元组 我目前的名单是: [[("String", "String", 2.0)],[("String", "String", 2.0),("String", "String", 2.0)]] 我想在每个子列表中添加一个元组,因此我的新列表是: [[("String", "String", 2.0),("String","String",0.0)],[("String", "String", 2.0),("String", "Stri
[[("String", "String", 2.0)],[("String", "String", 2.0),("String", "String", 2.0)]]
我想在每个子列表中添加一个元组,因此我的新列表是:
[[("String", "String", 2.0),("String","String",0.0)],[("String", "String", 2.0),("String", "String", 2.0),("String","String",0.0)]]
我试图写一个函数,但它一直给我一个错误
append :: a -> [a] -> [a]
append i (x:xs) = x : append i xs
append i [] = i : []
那么我会:
map append ("String","String",0.0)
这给了我很多错误,比如不正确的“类型匹配”。我不知道如何在Haskell中追加。类型的[(“String”,“String”,2.0)],[(“String”,“String”,2.0),(“String”,“String”,“String”,2.0)]
是分数t=>[(String,String,t)]
,或者简化[(String,String,Double)]
因此,为了将元素附加到每个嵌套列表中,我们需要一个函数(String,String,Double)->[[(String,String,Double)]->[[(String,String,Double)]
。换句话说,a->[[a]]->[[a]]]
:
append :: a -> [[a]] -> [[a]]
append x xs = map (\xss -> xss ++ [x]) xs
最后
> append ("String","String",0.0) [[("String", "String", 2.0)],[("String", "String", 2.0),("String", "String", 2.0)]]
[[("String","String",2.0),("String","String",0.0)],[("String","String",2.0),("String","String",2.0),("String","String",0.0)]]
顺便说一下,注意在列表末尾添加元素是O(n),因此可以考虑将其添加到开始:
append' :: a -> [[a]] -> [[a]]
append' x = map ((:) x)
我在打电话,所以无法给出正确的答案,但您需要的是
映射(append(“String”,“String”,0.0))
(x:)
更简单