Python - Tuple分组字典

Question

所以我有这个从我的数据库返回的元组，这表明用户每天总共登录多少次。

mytuple = 
(
    ('jim', datetime.date(2017, 2, 10), 1L), 
    ('jeff', datetime.date(2017, 2, 11), 1L), 
    ('jeff', datetime.date(2017, 2, 12), 1L), 
    ('jeff', datetime.date(2017, 2, 13), 1L), 
    ('jeff', datetime.date(2017, 2, 14), 26L) 
) 

如何以这种方式将它格式化为Dict/Json？

mydict = { 
    { 
     name: 'jeff', 
     data: [ 
      [(2017, 2, 11), 1], 
      [(2017, 2, 12), 1], 
      [(2017, 2, 13), 1], 
      [(2017, 2, 14), 26] 
     ] 
    }, 
    { 
     name: 'jim', 
     data: [ 
      [(2017, 2, 10), 1] 
     ] 
    } 
    } 

Answer 1

import datetime 
import json 

mytype = (
    ('jim', datetime.date(2017, 2, 10), 1L), 
    ('jeff', datetime.date(2017, 2, 11), 1L), 
    ('jeff', datetime.date(2017, 2, 12), 1L), 
    ('jeff', datetime.date(2017, 2, 13), 1L), 
    ('jeff', datetime.date(2017, 2, 14), 26L) 
) 

mydict = {} 

for name, date, count in mytype: 
    if name not in mydict: 
     mydict[name] = [] 
    mydict[name].append(((date.year, date.month, date.day), count)) 

print mydict 
print json.dumps(mydict) 

Answer 2

您可以使用下列内容：

import datetime 
import json 

mytype = (
    ('jim', datetime.date(2017, 2, 10), 1L), 
    ('jeff', datetime.date(2017, 2, 11), 1L), 
    ('jeff', datetime.date(2017, 2, 12), 1L), 
    ('jeff', datetime.date(2017, 2, 13), 1L), 
    ('jeff', datetime.date(2017, 2, 14), 26L) 
) 

output = [] 

def date_to_ymd_tuple(input_date): 
    return (input_date.year, input_date.month, input_date.day) 

for user in set(item[0] for item in mytype): 
    output.append({'name': user, 
        'data': [[date_to_ymd_tuple(x[1]), x[2]] for x in mytype if x[0] == user] 
        }) 

print output 

with open('output.txt', 'w') as f: 
    json.dump(output, f) 

将写出

[{"data": [[[2017, 2, 10], 1]], "name": "jim"}, {"data": [[[2017, 2, 11], 1], [[2017, 2, 12], 1], [[2017, 2, 13], 1], [[2017, 2, 14], 26]], "name": "jeff"}] 

Answer 3

您还可以使用itertools库，它有groupby功能做到这一点。希望这会更快，一旦你有更大的数据集。但是，输出将是列表而不是字典，与您所要求的不完全相同。

import datetime 
from itertools import groupby 

mytuple = [('jim', datetime.date(2017, 2, 10), 1), 
    ('jeff', datetime.date(2017, 2, 11), 1), 
    ('jeff', datetime.date(2017, 2, 12), 1), 
    ('jeff', datetime.date(2017, 2, 13), 1), 
    ('jeff', datetime.date(2017, 2, 14), 26)] 

group_dict = [] 
for key, group in groupby(mytuple, lambda x: x[0]): 
    group_dict.append({'name': key, 'data': [[g[1], g[2]] for g in group]}) 

这里你做groupby，并使用元组的第一个键为它提供函数，例如，这里lambda x: x[0]。在这种情况下，数据是各组指标1和2

您也可以使用元组的格式变换日期时间的因素太多，像以下：

group_dict = [] 
for key, group in groupby(mytuple, lambda x: x[0]): 
    data = [[(g[1].year, g[1].month, g[1].day), g[2]] for g in group] 
    group_dict.append({'name': key, 'data': data})