python sort dictionary items by value and then key

Question

All,

I'm having difficulty sorting a dictionary by value and then printing.

My object (dataSet) looks like the following...

dict_items([(0, {'studentName': 'dan', 'Score': 80.0}), (1, {'studentName': 'rob', 'Score': 92.0})])

I would like to sort by Score and print, but I am failing miserably. I used the following method as advised to sort by StudentName, if it is of help.

entries = sorted([(dataSet[entry]['studentName'], dataSet[entry]['Score']) for entry in dataSet]) for name, score in entries: print(('Student: {} -- Score: {}%'.format(name, score)))

Sorting by score and then names probably isn't going to work unless you are going to have ranges of scores that you can sort the names in. — SirParselot
– SirParselot, Commented Dec 8, 2015 at 13:15
@SirParselot Thinking about it, I am happy by simply sorting by Score. Python is not my domain and beggars can't be choosers! — IronAces
– IronAces, Commented Dec 8, 2015 at 13:20
It is not possible to sort a dict, only to get a representation of a dict that is sorted. Dicts are inherently orderless, but other types, such as lists and tuples, are not. So you need a sorted representation, which will be a list—probably a list of tuples. — Ahmed Lotfy
– Ahmed Lotfy, Commented Dec 8, 2015 at 13:23

Ahmed Lotfy · Accepted Answer · 2015-12-08 13:35:21Z

MyDict = {0: {'Score': 80.0, 'studentName': 'dan'}, 1: {'Score': 92.0, 'studentName': 'rob'}, 2: {'Score': 10.0, 'StudentName': 'xyz'}}

This returns the list of key-value pairs in the dictionary, sorted by value from highest to lowest:

sorted(MyDict.items(), key=lambda x: x[1], reverse=True)

For the dictionary sorted by key, use the following:

sorted(MyDict.items(), reverse=True)

The return is a list of tuples because dictionaries themselves can't be sorted.

This can be both printed or sent into further computation.

Ayush · Accepted Answer · 2015-12-08 13:28:55Z

assuming your object is a list of tuples of dicts (closest interpretation I can reach via the data given), This would suffice:

>>> dict_items = [(0, {'Score': 80.0, 'studentName': 'dan'}), (1, {'Score': 92.0, 'studentName': 'rob' })] >>> sorted(dict_items, key=lambda x: x[1]['Score']) # [(0, {'Score': 80.0, 'studentName': 'dan'}), (1, {'Score': 92.0, 'studentName': 'rob' })]

liu · Accepted Answer · 2015-12-08 13:37:52Z

Another method is below, maybe someone needs :

d = {0: {'Score': 80.0, 'studentName': 'dan'}, 1: {'Score': 92.0, 'studentName': 'rob'}, 2: {'Score': 10.0, 'StudentName': 'xyz'}} sorted(d, cmp = lambda a,b: cmp(d[a]['Score'],d[b]['Score']))

Pynchia · Accepted Answer · 2015-12-08 13:45:58Z

If all you need is a list of keys ordered by Score value:

d = {0: {'Score': 80.0, 'studentName': 'dan'}, 1: {'Score': 92.0, 'studentName': 'rob'}, 2: {'Score': 10.0, 'studentName': 'xyz'}} sorted(d, key=lambda k: d[k]['Score'])

produces

[2, 0, 1]

Otherwise, you can use an ordered dictionary

from collections import OrderedDict od = OrderedDict(sorted(d.items(), key=lambda i: i[1]['Score']))

which gives you

OrderedDict([(0, {'Score': 80.0, 'studentName': 'dan'}), (1, {'Score': 92.0, 'studentName': 'rob'})])

that you can print out nicely

for v in od.values(): print(v['Score'], v['studentName']) 10.0 xyz 80.0 dan 92.0 rob

Serge Ballesta · Accepted Answer · 2015-12-08 13:57:49Z

I would just use:

 def comp(x,y): if x[1]['Score'] != y[1]['Score']: return 1 if x[1]['Score'] > y[1]['Score'] else -1 elif x[1]['studentName'] < y[1]['studentName']: return -1 elif x[1]['studentName'] > y[1]['studentName']: return 1 else: return 0

then

sorted(dict_items, comp)

For example for

dict_items = [(0, {'Score': 80.0, 'studentName': 'dan'}), (1, {'Score': 92.0, 'studentName': 'rob'}), (2, {'Score': 70.0, 'studentName': 'foo'})]

it gives:

[(2, {'Score': 70.0, 'studentName': 'foo'}), (0, {'Score': 80.0, 'studentName': 'dan'}), (1, {'Score': 92.0, 'studentName': 'rob'})]

But beware: your question title was about dictionnary items. The answer is coherent with the code in the question but only sort a list of tuples. If you really have a dict d, you should use sorted(d.items())

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