重复函数python
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Repeat function python
提问by username_HI
I'm stuck at higher-order functions in python. I need to write a repeat function repeatthat applies the function fntimes on a given argument x.
我被困在 python 中的高阶函数。我需要编写一个重复函数repeat，将函数fn时间应用于给定参数x。
For example, repeat(f, 3, x)is f(f(f(x))).
例如，repeat(f, 3, x)是f(f(f(x)))。
This is what I have:
这就是我所拥有的：
def repeat(f,n,x):
if n==0:
return f(x)
else:
return repeat(f,n-1,x)
When I try to assert the following line:
当我尝试断言以下行时：
plus = lambda x,y: repeat(lambda z:z+1,x,y)
assert plus(2,2) == 4
It gives me an AssertionError. I read about How to repeat a function n timesbut I need to have it done in this way and I can't figure it out...
它给了我一个AssertionError. 我读过如何重复函数 n 次，但我需要以这种方式完成它，但我无法弄清楚......
采纳答案by jonrsharpe
You have two problems:
你有两个问题：
- You are recursing the wrong number of times (if
n == 1, the function should be called once); and - You aren't calling
fon the returned value from the recursive call, so the function is only ever applied once.
- 您递归的次数错误（如果
n == 1，该函数应该被调用一次）；和 - 您没有调用
f递归调用的返回值，因此该函数只应用一次。
Try:
尝试：
def repeat(f, n, x):
if n == 1: # note 1, not 0
return f(x)
else:
return f(repeat(f, n-1, x)) # call f with returned value
or, alternatively:
或者，或者：
def repeat(f, n, x):
if n == 0:
return x # note x, not f(x)
else:
return f(repeat(f, n-1, x)) # call f with returned value
(thanks to @Kevin for the latter, which supports n == 0).
（感谢@Kevin 支持后者n == 0）。
Example:
例子：
>>> repeat(lambda z: z + 1, 2, 2)
4
>>> assert repeat(lambda z: z * 2, 4, 3) == 3 * 2 * 2 * 2 * 2
>>>
回答by A_User
You've got a very simple error there, in the else block you are just passing x along without doing anything to it. Also you are applying x when n == 0, don't do that.
你在那里有一个非常简单的错误，在 else 块中你只是传递 x 而没有对其做任何事情。此外，当 n == 0 时，您正在应用 x，不要这样做。
def repeat(f,n,x):
"""
>>> repeat(lambda x: x+1, 2, 0)
2
"""
return repeat(f, n-1, f(x)) if n > 0 else x
