Python求和高阶函数

Question

我正在写一个求和的迭代解，它似乎给出了正确的答案。但是我的导师告诉我，它给non-commutative combine operations带来了错误的结果。我去了谷歌，但我仍然不确定它到底是什么意思……

下面是我写的递归代码：

def sum(term, a, next, b):
    # First recursive version
    if a > b:
        return 0
    else:
        return term(a) + sum(term, next(a), next, b)

def accumulate(combiner, base, term, a, next, b):
    # Improved version
    if a > b:
        return base
    else:
        return combiner(term(a), accumulate(combiner, base, term, next(a), next, b))

print(sum(lambda x: x, 1, lambda x: x, 5))
print(accumulate(lambda x,y: x+y, 0, lambda x: x, 1, lambda x: x, 5))
# Both solution equate to - 1 + 2 + 3 + 4 + 5 

这是我编写的迭代版本，它给出了错误的non-commutative combine operations编辑结果:当lambda x,y: x- y用于组合器时，accumulate_iter给出了错误的结果

def accumulate_iter(combiner, null_value, term, a, next, b):
    while a <= b:
        null_value = combiner(term(a), null_value)
        a = next(a)
    return null_value

希望有人能为这个迭代版本的accumulate提供一个解决方案

Answer 1

当合并器是可交换的时，你的accumulate_iter工作得很好，但当合并器是非交换的时，它会给出不同的结果。这是因为递归accumulate从后面到前面组合元素，而迭代版本从前面到后面组合元素。

所以我们需要做的是让accumulate_iter从后面组合起来，下面是一个重写的accumulate_iter

def accumulate_iter(a, b, base, combiner, next, term):
    # we want to combine from behind, 
    # but it's hard to do that since we are iterate from ahead
    # here we first go through the process, 
    # and store the elements encounted into a list
    l = []
    while a <= b:
        l.append(term(a))
        a = next(a)
    l.append(base)
    print(l)

    # now we can combine from behind!
    while len(l)>1:
        l[-2] = combiner(l[-2], l[-1])
        l.pop()
    return l[0]