使用python在键值对中进行分层分组

Question

我有一个这样的列表：

data = [
{'date':'2017-01-02', 'model': 'iphone5', 'feature':'feature1'},
{'date':'2017-01-02', 'model': 'iphone7', 'feature':'feature2'},
{'date':'2017-01-03', 'model': 'iphone6', 'feature':'feature2'},
{'date':'2017-01-03', 'model': 'iphone6', 'feature':'feature2'},
{'date':'2017-01-03', 'model': 'iphone7', 'feature':'feature3'},
{'date':'2017-01-10', 'model': 'iphone7', 'feature':'feature2'},
{'date':'2017-01-10', 'model': 'iphone7', 'feature':'feature1'},
]

我想实现这个：

[
   {
      '2017-01-02':[{'iphone5':['feature1']}, {'iphone7':['feature2']}]
   },
   {
      '2017-01-03': [{'iphone6':['feature2']}, {'iphone7':['feature3']}]
   },
   {
      '2017-01-10':[{'iphone7':['feature2', 'feature1']}]
   }
]

我需要一种有效的方法，因为它可能包含大量数据。

我正在尝试这个：

data = sorted(data, key=itemgetter('date'))
date = itertools.groupby(data, key=itemgetter('date'))

但对于“日期”键的值，我一无所获。

稍后我将遍历此结构以构建 HTML。

Answer 1

您可以使用defaultdict 高效而干净地完成此操作。不幸的是，这是一个相当高级的用法，而且很难阅读。

from collections import defaultdict
from pprint import pprint

# create a dictionary whose elements are automatically dictionaries of sets
result_dict = defaultdict(lambda: defaultdict(set))

# Construct a dictionary with one key for each date and another dict ('model_dict') 
# as the value.
# The model_dict has one key for each model and a set of features as the value.
for d in data:
    result_dict[d["date"]][d["model"]].add(d["feature"])

# more explicit version:
# for d in data:
#     model_dict = result_dict[d["date"]]   # created automatically if needed
#     feature_set = model_dict[d["model"]]  # created automatically if needed
#     feature_set.add(d["feature"])

# convert the result_dict into the required form
result_list = [
    {   
        date: [
            {phone: list(feature_set)} 
                for phone, feature_set in sorted(model_dict.items())
        ]
    } for date, model_dict in sorted(result_dict.items())
]

pprint(result_list)
# [{'2017-01-02': [{'iphone5': ['feature1']}, {'iphone7': ['feature2']}]},
#  {'2017-01-03': [{'iphone6': ['feature2']}, {'iphone7': ['feature3']}]},
#  {'2017-01-10': [{'iphone7': ['feature2', 'feature1']}]}]

Answer 2

你可以试试这个，这是我的方法，td 是一个存储{ iphone : index } 的字典，用于检查字典列表中是否存在新项目：

from itertools import groupby
from operator import itemgetter

r = []
for i in groupby(sorted(data, key=itemgetter('date')), key=itemgetter('date')):
    td, tl = {}, []
    for j in i[1]:
        if j["model"] not in td:
            tl.append({j["model"]: [j["feature"]]})
            td[j["model"]] = len(tl) - 1
        elif j["feature"] not in tl[td[j["model"]]][j["model"]]:
            tl[td[j["model"]]][j["model"]].append(j["feature"])
    r.append({i[0]: tl})

结果：

[
  {'2017-01-02': [{'iphone5': ['feature1']}, {'iphone7': ['feature2']}]},
  {'2017-01-03': [{'iphone6': ['feature2']}, {'iphone7': ['feature3']}]},
  {'2017-01-10': [{'iphone7': ['feature2', 'feature1']}]}
]

其实我觉得数据结构可以简化，也许你不需要那么多嵌套。

Answer 3

total_result = list()
result = dict()
inner_value = dict()

for d in data:
    if d["date"] not in result:
        if result:
            total_result.append(result)
        result = dict()
        result[d["date"]] = set()
        inner_value = dict()

    if d["model"] not in inner_value:
        inner_value[d["model"]] = set()

    inner_value[d["model"]].add(d["feature"])
    tmp_v = [{key: list(inner_value[key])} for key in inner_value]
    result[d["date"]] = tmp_v

total_result.append(result)

total_result

[{'2017-01-02': [{'iphone7': ['feature2']}, {'iphone5': ['feature1']}]},
 {'2017-01-03': [{'iphone6': ['feature2']}, {'iphone7': ['feature3']}]},
 {'2017-01-10': [{'iphone7': ['feature2', 'feature1']}]}]