filter out item from ordered dict


filter out item from ordered dict



For the following ordered dictionary, how can I print just the
1)'Price' and its value
2) rank it in descending order with its corresponding 'room_id'


[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')])]




3 Answers
3



@Arjun, you can also try the below code to solve your problem.



The solution uses the concept of list comprehension and sorted() function with one of its keyword argument key.


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: ' + item['room_id'] + ', price: ' + item['price'] for item in sorted(dicts, key=lambda d: int(d['room_id']))])

print(output);



Output »


room_id: 1133718, price: 74.0
room_id: 3179080, price: 77.0
room_id: 15303457, price: 60.0





Thanks! it was helpful!! :)
– Arjun Rajendran
22 hours ago





No problem @Arjun. Thanks for your valuable response.
– Rishikesh Agrawani
21 hours ago



To order by price, you can use sorted with a custom key:


price


sorted


res = sorted(L, key=lambda x: float(x['price']), reverse=True)



To extract the result as combinations of price and room_id, you can use a list comprehension:


res_id_price = [(x['room_id'], x['price']) for x in res]



If you wish to print and don't need a list:


print(*((x['room_id'], x['price']) for x in res), sep='n')

('3179080', '77.0')
('1133718', '74.0')
('15303457', '60.0')



Given a list of OrderedDicts:


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')])]



You can do:


>>> 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')])]



If you want to format that for printing:


>>> 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





Thank You so much, you saved my project :)
– Arjun Rajendran
Jun 29 at 16:57





@ArjunRajendran If this answer sovled your problem, please consider accepting it.
– Mr. T
2 days ago






By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Comments

Post a Comment

Popular posts from this blog

paramiko-expect timeout is happening after executing the command

paramiko-expect timeout is happening after executing the command Can someone help me why timeout is happening after executing the command .I am trying to SSH to a machine and execute a command.I want to store the result of the executed command . I have referred paramiko-timeout but i didn't get my expected result. import traceback try: import paramiko except ImportError: raise Exception("Please install paramiko , pip install paramiko") import pexpect from paramiko_expect import SSHClientInteraction def main(): HOSTNAME = "minion5.net" USERNAME = "root" PASSWORD = "help@432" PROMPT = "root@*:~$s+" try: client = paramiko.SSHClient() client.load_system_host_keys() client.set_missing_host_key_policy(paramiko.AutoAddPolicy()) client.connect(hostname=HOSTNAME, username=USERNAME, password=PASSWORD) with SSHClientInteraction(client, timeout=20, display=True) as interact: ...
Read more

Export result set on Dbeaver to CSV

Export result set on Dbeaver to CSV Normally I use Dbeaver for windows and always export my result set like this: run my query --> select the result --> export the result set --> select export to clipboard --> done This step by step puts my result set in my clipboard and I can paste it wherever I want to work with it. The problem is that now I am using dbeaver for mac and this guide is not working. I can go on until the moment that I select my result set like in the image below: exporting data set But once I go further in the process, in the last step I get: no query Note that in "source" it was suppose to show the query that originated the result set, but instead it says just "select. As a result it does't select my result or anything (besides being "successful"). Normally my query would show up there automatically and I couldn't find any option that corrects this problem in the menus. Please help me! Thanks. Aft...
Read more

Opening a url is failing in Swift

Opening a url is failing in Swift breakpointing inside the callback and logging success shows false. I added itms-apps to my plist: LSApplicationQueriesSchemes and put this in my app delegate: func application(_ app: UIApplication, open url: URL, options: [UIApplicationOpenURLOptionsKey : Any] = [:]) -> Bool { return true } func moreTapped() { let url = NSURL(string: "itms-apps://itunes.com/developer/quantum-productions/id979315877") if #available(iOS 10.0, *) { UIApplication.shared.open(url! as URL, options: [:], completionHandler: {(success: Bool) in }) } else { UIApplication.shared.openURL(url! as URL) } } Are you testing on a real device? – rmaddy Jun 21 at 23:57 I try on a device, get "Cannot Connect to App Store". Device is online ...
Read more