基于python的键值对分层分组
我有这样一份清单:基于python的键值对分层分组,python,python-2.7,grouping,Python,Python 2.7,Grouping,我有这样一份清单: 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': 'ipho
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'},
]
我想做到这一点:
[
{
'2017-01-02':[{'iphone5':['feature1']}, {'iphone7':['feature2']}]
},
{
'2017-01-03': [{'iphone6':['feature2']}, {'iphone7':['feature3']}]
},
{
'2017-01-10':[{'iphone7':['feature2', 'feature1']}]
}
]
我需要一个有效的方法,因为它可能需要很多数据
我试着这样做:
data = sorted(data, key=itemgetter('date'))
date = itertools.groupby(data, key=itemgetter('date'))
但我对“日期”键的值一无所获
稍后,我将迭代此结构以构建HTML
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)
总结果
[{'2017-01-02': [{'iphone7': ['feature2']}, {'iphone5': ['feature1']}]},
{'2017-01-03': [{'iphone6': ['feature2']}, {'iphone7': ['feature3']}]},
{'2017-01-10': [{'iphone7': ['feature2', 'feature1']}]}]
你可以试试这个,这是我的方法,td是一个dict来存储{iphone:index}来检查dict列表中是否存在新项:
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']}]}
]
事实上,我认为数据结构可以简化,也许您不需要这么多嵌套。您可以使用defaultdict高效、干净地完成这项工作。不幸的是,这是一个非常高级的用法,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']}]}]
如果你可以使用它,你应该考虑。带有默认设置的字典可能会帮助…@ Aikanaro,我认为只有在数据按日期排序之前,这才可能起作用。