CODEWITHSUNDEEP
pythonpython-2.7sortingdictionary
Nitin Kantak
Nitin Kantak

Reputation: 168

How to sort keys in dictionary if values are same ?

Sort the dictionary by score. if the score is same then sort them by name 

import operator
dt={ 
'sudha'  : {'score' : 75} ,
'Amruta' : {'score' : 95} ,
'Ramesh' : {'score' : 56} ,
'Shashi' : {'score' : 78} ,
'Manoj'  : {'score' : 69} ,
'Resham'  : {'score' : 95} 
} 

sorted_x1 = sorted(dt.items(), key=operator.itemgetter(1))
print sorted_x1

output is :

[('Ramesh', {'score': 56}), ('Manoj', {'score': 69}), 
 ('sudha', {'score':           75}), 
 ('Shashi', {'score': 78}),   ('Resham', {'score': 95}), ('Amruta',{'score': 95})]

Now I want to Sort Last two elements by its name because its scores are same .

Upvotes: 2

Views: 107

Answers (3)

John La Rooy
John La Rooy

Reputation: 304355

Careful, your code is actually comparing the dicts themselves rather than the scores. ie

{'score': 56} < {'score': 69}

May cause surprises if there are extra keys in the dictionaries. You can arrange the key func to return a tuple of the score and the key. 

sorted_x1 = sorted(dt.items(), key=lambda (k, v): (v['score'], k))

If someone needs to do this in Python3, you will have to avoid the tuple unpacking in the lambda as that is no longer allowed. 

sorted_x1 = sorted(dt.items(), key=lambda k_v: (k_v[1]['score'], k_v[0]))

Upvotes: 5

Joe T. Boka
Joe T. Boka

Reputation: 6581

This is one way to do it:

k = sorted(dt, key=dt.__getitem__)
v = sorted(dt.values())
sorted_dt_scores = zip(k,v)

output when you print sorted_dt_scores:

[('Ramesh', {'score': 56}), ('Manoj', {'score': 69}), ('sudha', {'score': 75}), ('Shashi', {'score': 78}), ('Resham', {'score': 95}), ('Amruta', {'score': 95})]

Upvotes: 1

Anand S Kumar
Anand S Kumar

Reputation: 90979

You can send the elements you want to sort on as a tuple or list, and sort will happen first on the first element, if first elements are same it will move on to the next element.

Example -

>>> dt={
... 'sudha'  : {'score' : 75} ,
... 'Amruta' : {'score' : 95} ,
... 'Ramesh' : {'score' : 56} ,
... 'Shashi' : {'score' : 78} ,
... 'Manoj'  : {'score' : 69} ,
... 'Resham'  : {'score' : 95}
... }
>>>
>>> sorted_x1 = sorted(dt.items(), key=lambda x: (x[1]['score'], x[0]))
>>> sorted_x1
[('Ramesh', {'score': 56}), ('Manoj', {'score': 69}), ('sudha', {'score': 75}), ('Shashi', {'score': 78}), ('Amruta', {'score': 95}), ('Resham', {'score': 95})]

Upvotes: 3

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