0

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。我很难考虑用 Pythonic 的方式来做这件事。有什么建议么?

编辑:如果其他人遇到这种问题并想要解决方案,这就是我(最终)想出的。我会发布作为答案,但由于问题已关闭(叹息),我不能。

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
4

1 回答 1

2

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

我将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
于 2013-09-04T21:46:42.317 回答