Question

我正在写一个求和的迭代解决方案，它似乎给出了正确的答案。但是我的老师告诉我，它给non-commutative combine operations给出了错误的结果。我去了谷歌，但我仍然不确定这到底意味着什么...

这是我写的递归代码：

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

这是我写的迭代版本，给出了non-commutative combine operations的错误结果- 编辑：当lambda x,y: x- y用于合并器时，accumulate_iter给出错误的结果

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

希望有人可以为此迭代版本的accumulate

提供解决方案

Answer 1

当组合器是可交换的时，您accumulate_iter可以很好地工作，但是当组合器是不可交换的时，您会得到不同的结果。这是因为递归accumulate从后到前组合了元素，但是迭代版本从前到后组合了元素。

所以我们需要做的是使accumulate_iter从后面合并，然后是重写的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]

Python求和高阶函数

1 个答案: