Question

我有一个看起来像这样的元组：

(
    ('Category 1', 40),
    ('Category 1 | Sub-Category 1', 20),
    ('Category 1 | Sub-Category 2', 20),
    ('Category 1 | Sub-Category 2 | Sub-Sub-Category 1', 5),
    ('Category 1 | Sub-Category 2 | Sub-Sub-Category 2', 15),
    ('Category 2', 20),
    ('Category 2 | Sub-Category 1', 15),
    ('Category 2 | Sub-Category 2', 5)
)

我想把它变成一个看起来像这样的字典：

{
    'Category 1': {
        'count': 40,
        'children': {
            'Sub-Category 1': {'count': 20, 'children': []},
            'Sub-Category 2': {
                'count': 20,
                'children': {
                    'Sub-Sub-Category 1': {'count': 5, 'children': []},
                    'Sub-Sub-Category 2': {'count': 15, 'children': []}
                }
            }
        }
    },
    'Category 2': {
        'count': 20,
        'children': {
            'Sub-Category 1': {'count': 15, 'children': []},
            'Sub-Category 2': {'count': 5, 'children': []},
        }
    }
}

有任意数量的子类别。我很难想到Pythonic这样做的方式。有什么建议吗？

编辑：如果有其他人遇到这类问题而想要一个解决方案，这就是我（最终）想出来的。我会发布答案，但不能，因为问题被关闭（叹气）。

from itertools import groupby

def categoriesdict(value, depth=0):
    categories = {}
    for name, children in groupby(value, lambda c: c[0].split(' | ')[depth]):
        # assumes that the first child is the group info
        categories[name] = {
            'count': children.next()[1],
            'children': categoriesdict(children, depth + 1)
        }
    return categories

Answer 1

对于每个2元组，拆分第一个元组[el.strip() for el in path.split('|')]，然后按照该路径创建字典和子词典。

~~我会在几分钟内编辑一些代码。~~

d = {'count': 0, 'children': {}}
for (path, count) in els:
    path = [el.strip() for el in path.split('|')]
    here = d
    for el in path:
        print(el)
        if el not in here['children']:
            here['children'][el] = {'count': 0, 'children': {}}
        here = here['children'][el]
    here['count'] = count

将平面Python元组转换为嵌套字典？

1 个答案: