在 Haskell 中过滤斐波那契数列
我正在尝试过滤包含斐波那契数字的列表。
我需要的只是奇数，并且小于或等于N。
这就是我到目前为止所拥有的：
fib n | n == 0 = 0
| n == 1 = 1
| otherwise = fib (n-1) + fib (n-2)
fibs n = [a | a <- [fib x | x <- [1..]], odd a, a < n]
这会给我想要的东西，但同时该解决方案不起作用，因为我不知道如何停止从 fib 函数检索元素。当然，这是因为x <- [1..]。
我考虑过两种选择：
- 在
x <- [1..]中设置限制（取决于n） - 定义
fibs递归这样我就可以知道何时停止（在写问题时思考）
我该怎么做？
我不是在寻找一种有效的方法
编辑：
这是我最后得到的两个解决方案：
fib n | n == 0 = 0
| n == 1 = 1
| otherwise = fib (n-1) + fib (n-2)
fibsAux n k xs | a < n = fibsAux n (k+1) (xs ++ [a])
| otherwise = xs
where
a = fib k
fibs n = filter odd $ fibsAux n 0 []
一个是使用@hammar的建议：
fibs x = takeWhile (< x) [a | a <- [fib x | x <- [1..]], odd n]
I'm trying to filter a list that contains the Fibonacci numbers.
What I need is only odd numbers, and less or equal than N.
This is what I have so far:
fib n | n == 0 = 0
| n == 1 = 1
| otherwise = fib (n-1) + fib (n-2)
fibs n = [a | a <- [fib x | x <- [1..]], odd a, a < n]
That will give me what I want, but at the same time that solution won't work because I don't know how to stop retrieving elements from fib function. Of course, that is because of x <- [1..].
I have thought about two options:
- Placing a limit (that depends on
n) inx <- [1..] - Defining
fibsrecursive so I can know when to stop (thought about it while writing the question)
How could I do this?
I'm not looking for an efficient way
Edit:
Here are the two solutions I had at the end:
fib n | n == 0 = 0
| n == 1 = 1
| otherwise = fib (n-1) + fib (n-2)
fibsAux n k xs | a < n = fibsAux n (k+1) (xs ++ [a])
| otherwise = xs
where
a = fib k
fibs n = filter odd $ fibsAux n 0 []
and the one using the suggestion of @hammar:
fibs x = takeWhile (< x) [a | a <- [fib x | x <- [1..]], odd n]
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看看 Data.List 中的
takeWhile函数（并由 Prelude 重新导出）。例如，请注意，即使列表是无限的，一旦找到不满足谓词的元素，该列表就会终止。
Have a look at the
takeWhilefunction from Data.List (and re-exported by the Prelude). For example,Note that even though the list is infinite, this terminates once it finds an element that does not satisfy the predicate.
还有一种更有效的方法来计算斐波那契数：
可以对其进行过滤以仅给出奇数：
可以将其截断为小于或等于 N：
There's also a more efficient way to compute Fibonacci numbers:
Which can be filtered to give only the odd numbers:
Which can be truncated to less than or equal to N: