如何在python中编写函数n次

Question

我知道如何通过将两个函数作为输入并输出其合成函数来合成两个函数，但如何返回合成函数f（f（…f（x））？谢谢

def compose2(f, g):
    return lambda x: f(g(x))

def f1(x):
    return x * 2
def f2(x):
    return x + 1

f1_and_f2 = compose2(f1, f2)
f1_and_f2(1)

Answer 1

如果你想让它成为一行作文，

那就试试这个，

def f1(x):
    return x * 2
def f2(x):
    return x + 1

>>> out = lambda x, f=f1, f2=f2: f1(f2(x)) # a directly passing your input(1) with 2 function as an input (f1, f2) with default so you dont need to pass it as an arguments
>>> out(1)
4
>>> 

>>> def compose(f1, n):
...   def func(x):
...     while n:
...       x = f1(x)
...       n = n-1
...     return x
...   return func

>>> d = compose(f1, 2)
>>> d(2)
8
>>> d(1)
4
>>> 

Answer 2

请注意，这主要是从https://stackoverflow.com/a/16739439/2750819复制的，但我想清楚地说明如何将其应用于任何一个函数n次。

def compose (*functions): 
    def inner(arg): 
        for f in reversed(functions): 
            arg = f(arg) 
        return arg 
    return inner

n = 10
def square (x): 
    return x ** 2

square_n = [square] * n
composed = compose(*square_n)
composed(2)

输出

179769313486231590772930519078902473361797697894230657273430081157732675805500963132708477322407536021120113879871393357658789768814416622492847430639474124377767893424865485276302219601246094119453082952085005768838150682342462881473913110540827237163350510684586298239947245938479716304835356329624224137216

Answer 3

在嵌套函数中使用循环：

def compose(f, n):
    def fn(x):
        for _ in range(n):
            x = f(x)
        return x
    return fn

fn将具有闭包，该闭包保留对调用compose的f和n的引用。