Question

我试图将我的Javascript撰写函数转换为Python，但我遇到了最后一个问题。如何推广未知数量的函数？

阅读this后，我想避免使用reduce()方法。

的javascript：

/* compose functions */
const comp1 = (fn2, fn1) => arr       => fn2(fn1(arr))                            // 2 functions, one array
const comp2 = (fn2, fn1) => (...arrs) => fn2(fn1(...arrs))                        // 2 functions, multiple arrays
const comp3 = (...fns)   => (...arrs) => fns.reduceRight((v,f) => f(v), arrs)     // multiple functions, multiple arrays

蟒：

/* compose functions */
comp1 = lambda fn2,fn1: lambda arr:   fn2(fn1(arr))   # 2 functions, 1 array
comp2 = lambda fn2,fn1: lambda *arrs: fn2(fn1(*arrs)) # 2 functions, multiple arrays
comp3 = lambda *fns:    lambda *arrs: ????

赞赏所有改进......

Answer 1

如果你真的不想使用reduce，我认为这是正确的方法，你可以有一个循环。尽管如此，这会阻止您使用lambda。

def compose(*fns):

    def composed(*args, **kwargs):
        output = fns[-1](*args, **kwargs)
        for fn in reversed(fns[:-1]):
            output = fn(output)

        return output

    return composed

我想再次声明没有理由不使用reduce，因为上面只是reduce的具体实现。组成是减少。

from functools import reduce # Only required in Python3 where reduce was moved

def compose(f1, f2):
    return lambda *args, **kwargs: f1(f2(*args, **kwargs))

def compose_many(*fns):
    return reduce(compose, fns, lambda x: x)

Answer 2

这是一个很好的可变参数，用于促进从右到左的功能组合 - 它适用于0,1或更多功能

def identity (x):
  return x

def compose (f = identity, *gs):
  if not gs:
    return f
  else:
    return lambda x: f (compose (*gs) (x))

def double (x):
  return x * 2

def add1 (x):
  return x + 1


# identity (5)    
compose () (5)                            # 5

# print (5)
compose (print) (5)                       # 5

# print (add1 (5))
compose (print, add1) (5)                 # 6

# print (add1 (double (5)))
compose (print, add1, double) (5)         # 11

# print (add1 (double (double (5))))
compose (print, add1, double, double) (5) # 21

要使其支持可变参数，只需更新lambda - 粗体

中的更改

def identity (x):
  return x

def compose (f = identity, *gs):
  if not gs:
    return f
  else:
    return lambda *xs: f (compose (*gs) (*xs))

def mult (x,y):
  return x * y

def double (x):
  return mult (2, x)

# print (double (double (mult (2, 3))))
compose (print, double, double, mult) (2,3) # 24

将功能从javascript转换为python，功能方式

2 个答案: