我正在尝试编写一个由任意数量的 lambda 函数组成的函数。
我有两个简单的 lambda 函数。
f = lambda x: x + 1
g = lambda x: x**2
我对组合函数的尝试是这样的:
def compose(*functions):
composed = lambda x: x # Function that returns input
for function in reversed(functions):
composed = lambda x: function(composed(x))
return composed
我的想法是遍历可变数量的函数,每次都让 composed
函数在其中包含一个新函数。
然后我可以创建一个由 f
Và g
组成的函数
c = compose(f, g)
因此调用 c(5)
应该返回 f(g(5))
,即 26。但是,我得到了
RecursionError: maximum recursion depth exceeded
我认为引入一个中间变量可能会解决问题。
def compose(*functions):
composed = lambda x: x # Function that returns input
for function in reversed(functions):
intermediate = lambda x: function(composed(x))
composed = intermediate
return composed
但是报同样的错误。
有办法解决这个问题吗?
首先,我认为您的方法将受到后期闭包绑定(bind)的影响,因为 lambda 中的 function
只会在迭代结束时采用 function 的最后一个值。其次,由于第一个原因,composed
最终只会递归调用自己; composed
- lambda - 调用 composed
的最后一个值 - 本身!
一个可能的解决方法是在每次迭代时将 composed
Và function
绑定(bind)到 lambda
:
def compose(*functions):
composed = lambda x: x
for function in reversed(functions):
composed = lambda x, function=function, composed=composed: function(composed(x))
return composed
print(compose(f, g)(5))
# 26
但是您的整体问题看起来像是 functools.reduce
的一个很好的用例:
from functools import reduce
def compose(*functions):
def inner(v):
return reduce(lambda x, y: y(x), reversed(functions), v)
return inner
print(compose(f, g)(5))
# 26
Tôi là một lập trình viên xuất sắc, rất giỏi!