python数据格式处理问题

[nebuung]

show_data = (('2019', '01', '1,601.0'), ('2019', '02', '1,397.0'), ('2019', '03', '2,219.0'), ('2019', '04', '3,229.5'),
             ('2020', '01', '7,060.0'), ('2020', '02', '3,500.5'), ('2020', '03', '5,031.0'), ('2020', '04', '3,623.0'))

求大神指导怎么变成下面的格式

show_data_list = [
    {'year': '2019',
     'sales': [{'month': '01', 'amount': '1,601.0'}, {'month': '02', 'amount': '1,397.0'},
               {'month': '03', 'amount': '2,219.0'}, {'month': '04', 'amount': '3,229.5'}]
     },
    {'year': '2020',
     'sales': [{'month': '01', 'amount': '7,060.0'}, {'month': '02', 'amount': '3,500.5'},
               {'month': '03', 'amount': '5,031.0'}, {'month': '04', 'amount': '3,623.0'}]
     }
]

[陈年椰子]

简单的方法就是遍历

show_data = (('2019', '01', '1,601.0'), ('2019', '02', '1,397.0'), ('2019', '03', '2,219.0'), ('2019', '04', '3,229.5'),
             ('2020', '01', '7,060.0'), ('2020', '02', '3,500.5'), ('2020', '03', '5,031.0'), ('2020', '04', '3,623.0'))


show_data_list = []
show_data_dict = {}
for data in show_data:
    if data[0] in show_data_dict:
        data_temp = show_data_dict[data[0]]
        data_temp[data[1]] = data[2]
    else:
        show_data_dict[data[0]] = {data[1]:data[2]}


for data in show_data_dict:
    data_list_temp = []
    for data1 in show_data_dict[data]:
        data_list_temp.append({'month': data1, 'amount': show_data_dict[data][data1]})
    show_data_list.append({'year': data,'sales': data_list_temp})


for data in show_data_list:
    print(data)

[DuskCrow_]

来个笨办法： show_data = (('2019', '01', '1,601.0'), ('2019', '02', '1,397.0'), ('2019', '03', '2,219.0'), ('2019', '04', '3,229.5'), ('2020', '01', '7,060.0'), ('2020', '02', '3,500.5'), ('2020', '03', '5,031.0'), ('2020', '04', '3,623.0')) show_list = [] for y in set([y[0] for y in show_data]): show_dict = dict.fromkeys(('year','salse')) show_dict['year'] = y data_year = [year_data for year_data in show_data if year_data[0] == y] show_dict['salse'] = [{'month':data[1],'amount':data[2]} for data in data_year] show_list.append(show_dict) print(show_list)

[nebuung]

引用 1 楼 陈年椰子 的回复:

简单的方法就是遍历

show_data = (('2019', '01', '1,601.0'), ('2019', '02', '1,397.0'), ('2019', '03', '2,219.0'), ('2019', '04', '3,229.5'),
             ('2020', '01', '7,060.0'), ('2020', '02', '3,500.5'), ('2020', '03', '5,031.0'), ('2020', '04', '3,623.0'))


show_data_list = []
show_data_dict = {}
for data in show_data:
    if data[0] in show_data_dict:
        data_temp = show_data_dict[data[0]]
        data_temp[data[1]] = data[2]
    else:
        show_data_dict[data[0]] = {data[1]:data[2]}


for data in show_data_dict:
    data_list_temp = []
    for data1 in show_data_dict[data]:
        data_list_temp.append({'month': data1, 'amount': show_data_dict[data][data1]})
    show_data_list.append({'year': data,'sales': data_list_temp})


for data in show_data_list:
    print(data)

谢谢，又学习了一招

[chuifengde]

import more_itertools as mi

show_data = (('2019', '01', '1,601.0'), ('2019', '02', '1,397.0'), ('2019', '03', '2,219.0'), ('2019', '04', '3,229.5'),
             ('2020', '01', '7,060.0'), ('2020', '02', '3,500.5'), ('2020', '03', '5,031.0'), ('2020', '04', '3,623.0'))

keyfun = lambda x: x[0]
valfun = lambda x: (x[1], x[2])

re = mi.groupby_transform(show_data, keyfun, valfun)
result = [{'year': i,'sales': [{'month': k[0],'amount': k[1],} for k in j]} for i, j in re]

print(result)

[nebuung]

加分结帖·