应:

 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)

产出:

 square(3)
 square(9)
 square(81)
 square(6561)
 square(43046721)
 1853020188851841

或lambda -less?

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

Using reduce and lamba. Build a tuple starting with your parameter, followed by all functions you want to call:

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

Something like this?

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

采用一种称为<代码>的惯性ool/代码>来进行这一操作。

<><0>

def square(x):
    """Return the square of a value."""
    return x * x

<><><>>>>

https://docs.python.org/3/library/itertools.html#itertools-recipes”rel=“nofollow noreferer”>itertools recipes:

def repeatfunc(func, times=None, *args):
    """Repeat calls to func with specified arguments.

    Example:  repeatfunc(random.random)
    """
    if times is None:
        return starmap(func, repeat(args))
    return starmap(func, repeat(args, times))

Demo

Optional: You can use a third-party library, more_itertools, that conveniently implements these recipes:

import more_itertools as mit


list(mit.repeatfunc(square, 2, 3))
# [9, 9]

Install via > pip install more_itertools

(如马克宁建议的那样):

from itertools import repeat
from functools import reduce # necessary for python3

def repeated(func, n):
    def apply(x, f):
        return f(x)
    def ret(x):
        return reduce(apply, repeat(func, n), x)
    return ret

您可使用以下方法:

>>> repeated(os.path.dirname, 3)( /a/b/c/d/e/f )
 /a/b/c 

>>> repeated(square, 5)(3)
1853020188851841

(在分别进口<条码>、/code>或定义<条码>后)

我认为,你想要的是职能构成:

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

def square(x):
  return pow(x, 2)

y = compose(square, 3, 2)
print y

此处采用<代码>reduce<>/code>的对应法:

def power(f, p, myapply = lambda init, g:g(init)):
    ff = (f,)*p # tuple of length p containing only f in each slot
    return lambda x:reduce(myapply, ff, x)

def square(x):
    return x * x

power(square, 2)(3)
#=> 81

我称之为“power/code>,因为这实际上是权力功能,其组成取代了重叠。

<代码>myapply在参数清单中作了界定,因此,该表只设定一次。