How to create dict with childrens from dict with parent_id

问题

How to get from this dict:

    cats = [
    {'parent_id': False, 'id': 1, 'title': u'All'},
    {'parent_id': False, 'id': 2, 'title': u'Toys'},
    {'parent_id': 2, 'id': 3, 'title': u'Toypads'},
    {'parent_id': 3, 'id': 4, 'title': u'Green'},
    ]

Something like this?

cats = [
{'parent_id': False, 'id': 1, 'title': u'All'},
{'parent_id': False,
 'children': [{'parent_id': 2,
               'children': [{'parent_id': 3, 'id': 4,
                             'title': u'Green'}],
               'id': 3, 'title': u'Toypads'},
              [{'parent_id': 3, 'id': 4, 'title': u'Green'}]],
 'id': 2, 'title': u'Toys'}
]

I need it to build a menu\sub-menu in Jinja2. I wrote a very bad code. It would be a more elegant solution.

    q = dict(zip([i['id'] for i in cats], cats))

    from collections import defaultdict
    parent_map = defaultdict(list)

    for item in q.itervalues():
        parent_map[item['parent_id']].append(item['id'])

    def tree_level(parent):
        for item in parent_map[parent]:
            yield q[item]
            sub_items = list(tree_level(item))
            if sub_items:
                for ca in cats:
                    if ca['id'] == item:
                        cats[cats.index(ca)]['children'] = sub_items
                        for s_i in sub_items:
                            try:
                                for ca_del_child in cats:
                                    if ca_del_child['id'] == s_i['id']:
                                        del cats[cats.index(ca_del_child)]
                            except:
                                pass
                yield sub_items
    for i in list(tree_level(False)):
        pass

回答1:

Here is a fairly concise solution:

cats = [{'parent_id': False, 'id': 1, 'title': u'All'},
        {'parent_id': False, 'id': 2, 'title': u'Toys'},
        {'parent_id': 2, 'id': 3, 'title': u'Toypads'},
        {'parent_id': 3, 'id': 4, 'title': u'Green'},]

cats_dict = dict((cat['id'], cat) for cat in cats)

for cat in cats:
    if cat['parent_id'] != False:
        parent = cats_dict[cat['parent_id']]
        parent.setdefault('children', []).append(cat)

cats = [cat for cat in cats if cat['parent_id'] == False]

Note that the comparisons to False are generally not necessary, but they should be used here in case you had a cat with 0 as its id or parent_id. In this case I would use None instead of False for a cat with no parent.

回答2:

It can be done as follows:

# Step 1: index all categories by id and add an element 'children'
nodes = {}
for cat in cats:
    nodes[cat['id']] = cat
    cat['children'] = []

# Step 2: For each category, add it to the parent's children
for index, cat in nodes.items():
    if cat['parent_id']:
        nodes[cat['parent_id']]['children'].append(cat)

# Step 3: Keep only those that do not have a parent
cats = [c for c in nodes.values() if not c['parent_id']]

Note that each node will have an attribute called 'children', which may be an empty list or a list with one or more nodes. If you don't want empty children lists, you can simply remove them between step 2 and step from each category in nodes.

Also note that the above assumes a node with a given parent_id actually exists. Lastly, note that if not c['parent_id'] will also be true when a node with id zero exists so you'll need to keep than in mind if that can occur.

来源：https://stackoverflow.com/questions/10047643/how-to-create-dict-with-childrens-from-dict-with-parent-id

标签

python

dictionary

parent-child