Answers for "order dict by key python"

python sort dict by key

A={1:2, -1:4, 4:-20}
{k:A[k] for k in sorted(A)}

output:
{-1: 4, 1: 2, 4: -20}

sort dictionary python

l = {1: 40, 2: 60, 3: 50, 4: 30, 5: 20}
d1 = dict(sorted(l.items(),key=lambda x:x[1],reverse=True))
print(d1) #output : {2: 60, 3: 50, 1: 40, 4: 30, 5: 20}
d2 = dict(sorted(l.items(),key=lambda x:x[1],reverse=False))
print(d2) #output : {5: 20, 4: 30, 1: 40, 3: 50, 2: 60}

python get dictionary keys sorted by value

sorted(A, key=A.get)

sort dict of dicts by key

channels = {
'24': {'type': 'plain', 'table_name': 'channel.items.AuctionChannel'}, 
'26': {'type': 'plain', 'table_name': 'channel.gm.DeleteAvatarChannel'}, 
'27': {'type': 'plain', 'table_name': 'channel.gm.AvatarMoneyChannel'}, 
'20': {'type': 'plain', 'table_name': 'channel.gm.AvatarMoneyAssertChannel'}, 
'21': {'type': 'plain', 'table_name': 'channel.gm.AvatarKillMobComplexChannel'}, 
'22': {'type': 'plain', 'table_name': 'channel.gm.DistributionMarkChannel'}, 
'23': {'type': 'plain', 'table_name': 'channel.gm.MailChannel'}
}

channels = collection.OrderedDict(sorted(channels.items(), key=lambda item: item[0]))
for key,value in channels.items():
    print(key, ':', value)

sort dictionary by key

dictionary_items = a_dictionary.items()
sorted_items = sorted(dictionary_items)