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,它给带来了错误的结果non-commutative combine operations
。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
这是我编写的迭代版本,给出了错误的结果non-commutative combine operations
-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
希望有人可以为该版本的迭代版本提供解决方案 accumulate
accumulate_iter
当组合器是可交换的时,您可以很好地工作,但是当组合器是不可交换的时,它会提供不同的结果。这是因为递归从头到尾accumulate
组合了元素,而迭代版本则从头到尾组合了元素。
因此,我们需要做的是accumulate_iter
从后面进行合并,然后重写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]
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