筛选列表列表
我对 Haskell 很陌生，才刚刚开始学习。 我正在使用“为了伟大的利益而学习 Haskell！”开始教程，并看到解决“3n+1”问题的示例：
chain :: (Integral a) => a -> [a]
chain 1 = [1]
chain n
| even n = n:chain (n `div` 2)
| odd n = n:chain (n*3 + 1)
numLongChains :: Int
numLongChains = length (filter isLong (map chain [1..100]))
where isLong xs = length xs > 15
所以，numLongChains 计算所有长度为 15 步的链，对于从 1 到 100 的所有数字。
现在，我想要我自己的：
numLongChains' :: [Int]
numLongChains' = filter isLong (map chain [1..100])
where isLong xs = length xs > 15
所以现在，我不想计数这些链，但返回这些链的过滤列表。 但现在编译时出现错误：
无法将预期类型“Int”与实际类型“[a0]”匹配 预期类型：Int ->布尔 实际类型：[a0] ->布尔 在`filter'的第一个参数中，即`isLong' 表达式中：filter isLong(map chain [1 .. 100])
可能是什么问题？
I'm very new with Haskell, only starting to learn it.
I'm using "Learn You a Haskell for Great Good!" tutorial for start, and saw example of solving "3n+1" problem:
chain :: (Integral a) => a -> [a]
chain 1 = [1]
chain n
| even n = n:chain (n `div` 2)
| odd n = n:chain (n*3 + 1)
numLongChains :: Int
numLongChains = length (filter isLong (map chain [1..100]))
where isLong xs = length xs > 15
so, numLongChains counts all chains that longer 15 steps, for all numbers from 1 to 100.
Now, I wanna my own:
numLongChains' :: [Int]
numLongChains' = filter isLong (map chain [1..100])
where isLong xs = length xs > 15
so now, I wanna not to count these chains, but return filtered list with these chains.
But now I get error when compiling:
Couldn't match expected type `Int' with actual type `[a0]' Expected type: Int -> Bool Actual type: [a0] -> Bool In the first argument of `filter', namely `isLong' In the expression: filter isLong (map chain [1 .. 100])
What can be the problem?
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numLongChains的类型签名可能不正确。根据您想要执行的操作，需要以下其中一项：Intfilter (>15) (map (length . chain) [1..100])。[[Int]]（Int的链（列表）列表）就可以了。The type signature of
numLongChainsis probably not correct. Depending on what you want to do, one of the following is needed:numLongChainsobviously shall return a number, change the first line tolength $ filter isLong (map chain [1..100])and the type toIntfilter (>15) (map (length . chain) [1..100]).[[Int]](A list of chains (lists) ofInts) and you're fine.FUZxxl 是对的。您需要将函数的类型签名更改为
[[Int]]。当您过滤列表列表并仅选择足够长的列表时，您将返回列表列表。关于阅读 Haskell 编译时调试器/错误的一点说明。这个错误可能看起来很奇怪。它说你有
[a0] -> Bool但你期待的是Int ->布尔。这是因为类型检查器假设，根据 numLongChains 函数的签名，您将需要一个过滤器函数来检查 Int 并返回可接受的列表。过滤列表并返回 [Int] 的唯一方法是使用一个接受Int并返回Bool的函数(Int -> Bool ）。相反，它看到一个检查长度的函数。 Length 接受一个列表，因此它猜测您编写了一个检查列表的函数。([a0] -> Bool)。有时，检查器并不像您希望的那样友好，但如果您足够仔细，您会发现十分之九的难以破译的错误是由此类假设造成的。FUZxxl is right. You are going to want to change the type signature of your function to
[[Int]]. As you are filtering a list of lists and only selecting the ones that are sufficiently long, you will have returned a lists of lists.One note about reading Haskell compile-time debugger/errors. This error may seem strange. It says you had
[a0] -> Boolbut you were expectingInt -> Bool. This is because that the type checker assumes that, from the signature of your numLongChains' function, you are going to need a filter function that checks Ints and returns a list of acceptable ones. The only way to filter over a list and get [Int] back is to have a function that takesInts and returnsBools(Int -> Bool). Instead, it sees a function that checks length. Length takes a list, so it guesses that you wrote a function that checks lists.([a0] -> Bool). Sometimes, the checker is not as friendly as you would like it to be but if you look hard enough, you will see that 9 times out of 10, a hard to decipher error is the result of such as assumptions.