Question

我给出了一个字符串列表和一个长度限制N。我必须编写一个函数来累积列表中的连续字符串，直到下一个字符串超过N限制。我必须返回一个列表列表，其中每个列表是不超过N个字符总数的最大连续子字符串。有关示例，请参阅下面的测试用例。此外，如果列表中的任何单个字符串长于N，我必须打印一条有用的消息并返回。

def break_lst(lst, size):
    def len_lst(l):
        return len("".join(l))

    result = []
    sublst = []

    for i, v in enumerate(lst):
        sublst.append(v)
        ls = len_lst(sublst)

        if ls == size:
            result.append(sublst)
            sublst = []
        elif ls > size:
            prev_sublst = sublst[:-1]
            if not prev_sublst or len_lst(prev_sublst) > size:
                raise Exception("Error: use a bigger size than " + str(size))
            else:
                result.append(prev_sublst)
                sublst = []

    return result

if __name__ == "__main__":
    lst = ["1", "22", "333", "4444", "55555", "666666", "7777777", "88888888"]

    for i in range(17):
        try:
            print(i, break_lst(lst, size=i))
        except Exception as e:
            print(e)

以上代码不仅丑陋而且还有错误，它给了我这个输出：

Error: use a bigger size than 0
Error: use a bigger size than 1
Error: use a bigger size than 2
Error: use a bigger size than 3
Error: use a bigger size than 4
Error: use a bigger size than 5
Error: use a bigger size than 6
Error: use a bigger size than 7
8 [['1', '22', '333'], ['55555'], ['7777777']]
9 [['1', '22', '333'], ['55555'], ['7777777']]
10 [['1', '22', '333', '4444'], ['55555'], ['7777777']]
11 [['1', '22', '333', '4444'], ['666666']]
12 [['1', '22', '333', '4444'], ['666666']]
13 [['1', '22', '333', '4444'], ['666666', '7777777']]
14 [['1', '22', '333', '4444'], ['666666', '7777777']]
15 [['1', '22', '333', '4444', '55555'], ['666666', '7777777']]
16 [['1', '22', '333', '4444', '55555']]

当预期输出应为：

Error: use a bigger size than 0
Error: use a bigger size than 1
Error: use a bigger size than 2
Error: use a bigger size than 3
Error: use a bigger size than 4
Error: use a bigger size than 5
Error: use a bigger size than 6
Error: use a bigger size than 7
8 [['1', '22', '333'], ['4444'], ['55555'], ['666666'], ['7777777'], ['88888888']]
9 [['1', '22', '333'], ['4444', '55555'], ['666666'], ['7777777'], ['88888888']]
10 [['1', '22', '333', '4444'], ['55555'], ['666666'], ['7777777'], ['88888888']]
11 [['1', '22', '333', '4444'], ['55555', '666666'], ['7777777'], ['88888888']]
12 [['1', '22', '333', '4444'], ['55555', '666666'], ['7777777'], ['88888888']]
13 [['1', '22', '333', '4444'], ['55555', '666666'], ['7777777'], ['88888888']]
14 [['1', '22', '333', '4444'], ['55555', '666666'], ['7777777'], ['88888888']]
15 [['1', '22', '333', '4444', '55555'], ['666666', '7777777'], ['88888888']]
16 [['1', '22', '333', '4444', '55555'], ['666666', '7777777'], ['88888888']]

有什么建议吗？

Answer 1

问题出在本节：

    elif ls > size:
        prev_sublst = sublst[:-1]
        if not prev_sublst or len_lst(prev_sublst) > size:
            raise Exception("Error: use a bigger size than " + str(size))
        else:
            result.append(prev_sublst)
            sublst = []

您已确定下一个列表超出了给定的长度。您已正确备份了一个元素并记录了最大化的子列表。但是，您没有找到了v元素的主页，它只会超出您的大小限制。

对于初学者，请尝试将其放入新的子列表中：

        else:
            result.append(prev_sublst)
            sublst = [v]     # <=== the change is here.

现在，检查一下你的逻辑 - 在迭代到下一个列表之前，你需要验证这个单项列表。

关于

的一个小评论

for i, v in enumerate(lst):

当你从未使用过enumerate时，为什么会遇到i的麻烦？

如何将字符串列表拆分为len（＆＃34;＆＃34; .join（sublist））＆lt; = N？

1 个答案: