CODEWITHSUNDEEP
python
John Smith
John Smith

Reputation: 2738

Python dictionary with frequency

I have a dictionary with a key and an item. Basically, keys keep people name and the values keep salary. What I want is that ordering items according to their frequency with the higher value. 

d = {'name-1': 100, 'name-2':90, 'name-3': 80, 'name-1': 80}
    print OrderedDict(sorted(d.items(), key=lambda t: t[1]))

So, the output should be ('name-1':100). Because, the occurrence is two in the dict and the salary is the highest.

I can only sort this dict. Could you please help me to get that result ?

Upvotes: 1

Views: 234

Answers (3)

Joran Beasley
Joran Beasley

Reputation: 114118

as others have pointed out you cannot have duplicate keys in a dict... however you could easily use a list

#d = (('name-1', 100), ('name-2',90), ('name-3',80), ('name-1',80))
names = "name1,name2,name3,name1".split(",")
salaries = map(int,"100,90,80,80".split(","))
d = zip(names,salaries)
print max(d,key=lambda x:names.count(x[0])*1000+x[1])

#('name-1', 100)
print sorted(d,key=lambda x:names.count(x[0])*1000+x[1],reverse=True)
[('name-1', 100), ('name-1', 80), ('name-2', 90), ('name-3', 80)]

Upvotes: 3

tMC
tMC

Reputation: 19355

Have you tried creating that d dict?

>>> d = {'name-1': 100, 'name-2':90, 'name-3': 80, 'name-1': 80}
>>> d
{'name-1': 80, 'name-2': 90, 'name-3': 80} # notice 'name-1' is only in there once.
>>>

If you try to create a second key with the same name in the dict, it will over write the value that was associated with that key. Every key is unique in a dict... If they weren't unique, dictionaries would no longer be near as useful.

You could create a nested tuples/lists if you need this sort of data structure... However, it will be far slower as finding a key will require iterating over all objects in the list:

d = (
        ('name-1', 100),
        ('name-2', 90),
        ('name-3', 80),
        ('name-1', 80),
     )

Upvotes: 0

John La Rooy
John La Rooy

Reputation: 304503

One approach would be to use a dict, but store the values in lists for each key

>>> d = {'name-1': [100, 80], 'name-2':[90], 'name-3': [80]}

and then

>>> sorted(d, key=lambda x:max(d.get(x)), reverse=True)
['name-1', 'name-2', 'name-3']

Upvotes: 4

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