Possible Duplicate:
Library function to compose a function with itself n times可能的重复:库函数，用自己n次组成一个函数
I need a function to call another function n number of times.
我需要一个函数调用另一个函数n次。
so it would look something like this f n = g(g(g(g(l)))) where n equals to the number of function g nested.
它看起来像这样f n = g(g(g(g(l))) n等于函数g嵌套的个数。
how should I go about this? thanks!
我该怎么做呢?谢谢!
4 个解决方案
#1
56
iterate is a common solution:
迭代是一种常见的解决方案:
> :t iterate
iterate :: (a -> a) -> a -> [a]
So, given a function with a domain the same as its range, a -> a, and an initial input a, produce an infinite list of results in the form:
因此，给定一个与域相同的函数，a -> a和初始输入a，会产生一个无限的结果列表:
iterate f a --> [a, f(a), f(f(a)), ...]
And you can access the nth element of the list using !!:
您可以使用!! !访问列表的第n个元素!
iterate f a !! n
NB iterate f a !! 0 == a.
NB迭代fa !!0 = =。
#2
7
This is a function that I use often at the ghci prompt. There are a few ways to write it, none of which I am particularly fond of, but they are all reasonably clean:
这是我在ghci提示符中经常使用的函数。有几种写法，没有一种是我特别喜欢的，但它们都相当干净:
fpow n f x = iterate f x !! n
fpow n f = foldr (.) id $ replicate n f
fpow n = foldr (.) id . replicate n -- just eta the above
fpow 0 f = id
fpow n f = f . fpow (n-1) f
The middle two appeal to me because my brain has chunked foldr (.) id to mean "compose a list of functions".
中间的两个对我很有吸引力，因为我的大脑把foldr (.) id分成了“组成一个功能列表”的意思。
I kinda just wish it were in the prelude :-).
我只是希望它是在序曲:)。
#3
6
f 0 = l
f n = g (f (n-1))
But more functional would be:
但功能更强的是:
f 0 l = l
f n l = g (f (n-1) l)
This could also be done with folds or morfisms, but this is easier to understand.
这也可以通过折叠或morfisms来完成，但这更容易理解。
For example here's using a hylomorphism, but it doesn't make it clearer really:
例如，这里使用的是低同构，但这并不能使它更清晰:
f g l = hylo l (.) (\n -> (g, n-1)) (==0)
It says some thing like compose (.) g(l) until n==0
它说了一些东西，比如写(。)g(l)直到n= 0。
#4
3
Can be done using fold:
可以使用折叠:
applyNTimes :: Int -> (a -> a) -> a -> a
applyNTimes n f val = foldl (\s e -> e s) val [f | x <- [1..n]]
#1
56
iterate is a common solution:
迭代是一种常见的解决方案:
> :t iterate
iterate :: (a -> a) -> a -> [a]
So, given a function with a domain the same as its range, a -> a, and an initial input a, produce an infinite list of results in the form:
因此，给定一个与域相同的函数，a -> a和初始输入a，会产生一个无限的结果列表:
iterate f a --> [a, f(a), f(f(a)), ...]
And you can access the nth element of the list using !!:
您可以使用!! !访问列表的第n个元素!
iterate f a !! n
NB iterate f a !! 0 == a.
NB迭代fa !!0 = =。
#2
7
This is a function that I use often at the ghci prompt. There are a few ways to write it, none of which I am particularly fond of, but they are all reasonably clean:
这是我在ghci提示符中经常使用的函数。有几种写法，没有一种是我特别喜欢的，但它们都相当干净:
fpow n f x = iterate f x !! n
fpow n f = foldr (.) id $ replicate n f
fpow n = foldr (.) id . replicate n -- just eta the above
fpow 0 f = id
fpow n f = f . fpow (n-1) f
The middle two appeal to me because my brain has chunked foldr (.) id to mean "compose a list of functions".
中间的两个对我很有吸引力，因为我的大脑把foldr (.) id分成了“组成一个功能列表”的意思。
I kinda just wish it were in the prelude :-).
我只是希望它是在序曲:)。
#3
6
f 0 = l
f n = g (f (n-1))
But more functional would be:
但功能更强的是:
f 0 l = l
f n l = g (f (n-1) l)
This could also be done with folds or morfisms, but this is easier to understand.
这也可以通过折叠或morfisms来完成，但这更容易理解。
For example here's using a hylomorphism, but it doesn't make it clearer really:
例如，这里使用的是低同构，但这并不能使它更清晰:
f g l = hylo l (.) (\n -> (g, n-1)) (==0)
It says some thing like compose (.) g(l) until n==0
它说了一些东西，比如写(。)g(l)直到n= 0。
#4
3
Can be done using fold:
可以使用折叠:
applyNTimes :: Int -> (a -> a) -> a -> a
applyNTimes n f val = foldl (\s e -> e s) val [f | x <- [1..n]]