对于以下有序词典,我如何仅打印 1)“价格”及其价值 2)按照降序对其对应的“ room_id”进行排名
[OrderedDict([('room_id', '1133718'), ('survey_id', '1280'), ('host_id', '6219420'), ('room_type', 'Shared room'), ('country', ''), ('city', 'Singapore'), ('borough', ''), ('neighborhood', 'MK03'), ('reviews', '9'), ('overall_satisfaction', '4.5'), ('accommodates', '12'), ('bedrooms', '1.0'), ('bathrooms', ''), ('price', '74.0'), ('minstay', ''), ('last_modified', '2017-05-17 09:10:25.431659'), ('latitude', '1.293354'), ('longitude', '103.769226'), ('location', '0101000020E6100000E84EB0FF3AF159409C69C2F693B1F43F')]), OrderedDict([('room_id', '3179080'), ('survey_id', '1280'), ('host_id', '15295886'), ('room_type', 'Shared room'), ('country', ''), ('city', 'Singapore'), ('borough', ''), ('neighborhood', 'TS17'), ('reviews', '15'), ('overall_satisfaction', '5.0'), ('accommodates', '12'), ('bedrooms', '1.0'), ('bathrooms', ''), ('price', '77.0'), ('minstay', ''), ('last_modified', '2017-05-17 09:10:24.216548'), ('latitude', '1.310862'), ('longitude', '103.858828'), ('location', '0101000020E6100000E738B709F7F659403F1BB96E4AF9F43F')]), OrderedDict([('room_id', '15303457'), ('survey_id', '1280'), ('host_id', '97053568'), ('room_type', 'Shared room'), ('country', ''), ('city', 'Singapore'), ('borough', ''), ('neighborhood', 'MK05'), ('reviews', '0'), ('overall_satisfaction', '0.0'), ('accommodates', '14'), ('bedrooms', '1.0'), ('bathrooms', ''), ('price', '60.0'), ('minstay', ''), ('last_modified', '2017-05-17 09:10:16.969900'), ('latitude', '1.333744'), ('longitude', '103.764612'), ('location', '0101000020E610000044882B67EFF0594093C7D3F20357F53F')])]
答案 0 :(得分:1)
要按price
订购,可以将sorted
与自定义键一起使用:
res = sorted(L, key=lambda x: float(x['price']), reverse=True)
要将结果提取为price和room_id的组合,可以使用列表理解:
res_id_price = [(x['room_id'], x['price']) for x in res]
如果您希望打印并且不需要列表:
print(*((x['room_id'], x['price']) for x in res), sep='\n')
('3179080', '77.0')
('1133718', '74.0')
('15303457', '60.0')
答案 1 :(得分:1)
给出OrderedDicts列表:
from collections import OrderedDict
li=[OrderedDict([('room_id', '1133718'), ('survey_id', '1280'), ('host_id', '6219420'), ('room_type', 'Shared room'), ('country', ''), ('city', 'Singapore'), ('borough', ''), ('neighborhood', 'MK03'), ('reviews', '9'), ('overall_satisfaction', '4.5'), ('accommodates', '12'), ('bedrooms', '1.0'), ('bathrooms', ''), ('price', '74.0'), ('minstay', ''), ('last_modified', '2017-05-17 09:10:25.431659'), ('latitude', '1.293354'), ('longitude', '103.769226'), ('location', '0101000020E6100000E84EB0FF3AF159409C69C2F693B1F43F')]), OrderedDict([('room_id', '3179080'), ('survey_id', '1280'), ('host_id', '15295886'), ('room_type', 'Shared room'), ('country', ''), ('city', 'Singapore'), ('borough', ''), ('neighborhood', 'TS17'), ('reviews', '15'), ('overall_satisfaction', '5.0'), ('accommodates', '12'), ('bedrooms', '1.0'), ('bathrooms', ''), ('price', '77.0'), ('minstay', ''), ('last_modified', '2017-05-17 09:10:24.216548'), ('latitude', '1.310862'), ('longitude', '103.858828'), ('location', '0101000020E6100000E738B709F7F659403F1BB96E4AF9F43F')]), OrderedDict([('room_id', '15303457'), ('survey_id', '1280'), ('host_id', '97053568'), ('room_type', 'Shared room'), ('country', ''), ('city', 'Singapore'), ('borough', ''), ('neighborhood', 'MK05'), ('reviews', '0'), ('overall_satisfaction', '0.0'), ('accommodates', '14'), ('bedrooms', '1.0'), ('bathrooms', ''), ('price', '60.0'), ('minstay', ''), ('last_modified', '2017-05-17 09:10:16.969900'), ('latitude', '1.333744'), ('longitude', '103.764612'), ('location', '0101000020E610000044882B67EFF0594093C7D3F20357F53F')])]
您可以这样做:
>>> li2=[OrderedDict([('room_id',od['room_id']),('price',od['price'])]) for od in li]
>>> sorted(li2, key=lambda od: float(od['price']), reverse=True)
[OrderedDict([('room_id', '3179080'), ('price', '77.0')]), OrderedDict([('room_id', '1133718'), ('price', '74.0')]), OrderedDict([('room_id', '15303457'), ('price', '60.0')])]
如果要格式化以进行打印:
>>> li3=sorted(li2, key=lambda od: float(od['price']), reverse=True)
>>> print("\n".join(["Room ID: {} Price: {}".format(od['room_id'], od['price']) for od in li3]))
Room ID: 3179080 Price: 77.0
Room ID: 1133718 Price: 74.0
Room ID: 15303457 Price: 60.0
答案 2 :(得分:1)
@Arjun ,您也可以尝试以下代码来解决问题。
该解决方案使用列表理解和 sorted()函数及其关键字参数 key 之一的概念。
from collections import OrderedDict
dicts = [
OrderedDict([
('room_id', '1133718'), ('survey_id', '1280'),
('host_id', '6219420'), ('room_type', 'Shared room'),
('country', ''), ('city', 'Singapore'), ('borough', ''),
('neighborhood', 'MK03'), ('reviews', '9'),
('overall_satisfaction', '4.5'), ('accommodates', '12'),
('bedrooms', '1.0'), ('bathrooms', ''), ('price', '74.0'),
('minstay', ''), ('last_modified', '2017-05-17 09:10:25.431659'),
('latitude', '1.293354'), ('longitude', '103.769226'),
('location', '0101000020E6100000E84EB0FF3AF159409C69C2F693B1F43F')]),
OrderedDict([('room_id', '3179080'), ('survey_id', '1280'),
('host_id', '15295886'), ('room_type', 'Shared room'),
('country', ''), ('city', 'Singapore'), ('borough', ''),
('neighborhood', 'TS17'), ('reviews', '15'),
('overall_satisfaction', '5.0'), ('accommodates', '12'),
('bedrooms', '1.0'), ('bathrooms', ''), ('price', '77.0'),
('minstay', ''), ('last_modified', '2017-05-17 09:10:24.216548'),
('latitude', '1.310862'), ('longitude', '103.858828'),
('location', '0101000020E6100000E738B709F7F659403F1BB96E4AF9F43F')]),
OrderedDict([('room_id', '15303457'), ('survey_id', '1280'),
('host_id', '97053568'), ('room_type', 'Shared room'),
('country', ''), ('city', 'Singapore'), ('borough', ''),
('neighborhood', 'MK05'), ('reviews', '0'),
('overall_satisfaction', '0.0'), ('accommodates', '14'),
('bedrooms', '1.0'), ('bathrooms', ''), ('price', '60.0'),
('minstay', ''), ('last_modified', '2017-05-17 09:10:16.969900'),
('latitude', '1.333744'), ('longitude', '103.764612'),
('location', '0101000020E610000044882B67EFF0594093C7D3F20357F53F')])];
output = '\n'.join([ 'room_id: {}, price: {}'.format(
item['room_id'], item['price']
) for item in sorted(dicts, key=lambda d: int(d['room_id']))])
print(output);
输出»
room_id: 1133718, price: 74.0
room_id: 3179080, price: 77.0
room_id: 15303457, price: 60.0