将compose函数从javascript转换为python

Question

我试图将我的Javascript撰写函数翻译成Python，但我对最后一个有问题。我如何推广未知数量的函数？

读完本文后，我想避免使用reduce（）方法。

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

蟒蛇：

/* 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: ????

感谢所有的改进。。。

Answer 1

这里有一个很好的变量组合，可以帮助从右到左的函数组合——它适用于0、1或更多函数

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

Answer 2

如果你真的不想使用减少，我相信这是正确的方法，你可以有一个循环。尽管如此，这阻止您使用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，没有理由不使用减少，因为上面只是减少的具体实现。组成是一种减少。

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)