将具有相同键和值的字典合并到一个字典中

Question

我有一本字典，里面有一个列表，如下所示：

{"items":[{"number":"98", "items": {"code": "X", "color": "Red"}},{"number":"98", "items": {"code": "Y", "color": "Blue"}},{"number":"62", "items": {"code": "B", "color": "Green"}}{"number":"62", "items": {"code": "A", "color": "Yellow"}}]}

有没有一种方法可以将按值匹配的每个 "number" 的项目合并到这样的列表中？

{"items":[{"number":"98","items":[{"code":"X","color":"Red"}, {"code": "Y","color":"Blue"}]},  {"number":"62", "items": [{"code": "B", "color": "Green"},{"code": "A", "color":"Yellow"}]}]}

Answer 1

我能想到的最简单的方法是将所有 'items' 分类到一个中间字典中，并按 'number' 分组。从那里开始，将字典转换为所需的输出是微不足道的。

inp = {"items":[{"number":"98", "items": {"code": "X", "color": "Red"}},{"number":"98", "items": {"code": "Y", "color": "Blue"}},{"number":"62", "items": {"code": "B", "color": "Green"}},{"number":"62", "items": {"code": "A", "color": "Yellow"}}]}
d = dict()
for i in inp['items']:
    d[i['number']] = d.get(i['number'], list()) + [i['items']]
out = {'items': [{'number': n, 'items': d[n]} for n in d.keys()]}

输出：

{'items': [{'number': '98', 'items': [{'code': 'X', 'color': 'Red'}, {'code': 'Y', 'color': 'Blue'}]}, {'number': '62', 'items': [{'code': 'B', 'color': 'Green'}, {'code': 'A', 'color': 'Yellow'}]}]}

将具有相同键和值的字典合并到一个字典中

Merging dictionaries with the same key and values into a single dictionary

list-comprehension

python-2.7

dictionary-comprehension