如何重复一次功能n次

Jam*_*ith 5 python higher-order-functions

我正在尝试在python中编写一个函数,如:

def repeated(f, n):
    ...
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where f是一个带有一个参数的函数,n是一个正整数.

例如,如果我将square定义为:

def square(x):
    return x * x
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我打来电话

repeated(square, 2)(3)
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这将是3次,2次.

Joh*_*iss 22

应该这样做:

 def repeated(f, n):
     def rfun(p):
         return reduce(lambda x, _: f(x), xrange(n), p)
     return rfun

 def square(x):
     print "square(%d)" % x
     return x * x

 print repeated(square, 5)(3)
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输出:

 square(3)
 square(9)
 square(81)
 square(6561)
 square(43046721)
 1853020188851841
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还是没有lambda

def repeated(f, n):
    def rfun(p):
        acc = p
        for _ in xrange(n):
            acc = f(acc)
        return acc
    return rfun
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car*_*cio 9

使用reduce和lamba.从您的参数开始构建一个元组,然后是您要调用的所有函数:

>>> path = "/a/b/c/d/e/f"
>>> reduce(lambda val,func: func(val), (path,) + (os.path.dirname,) * 3)
"/a/b/c"
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phi*_*mue 3

像这样的东西吗?

def repeat(f, n):
     if n==0:
             return (lambda x: x)
     return (lambda x: f (repeat(f, n-1)(x)))
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