Sorting dictionary keys based on their values

Question

I have a python dictionary setup like so

mydict = { 'a1': ['g',6],            'a2': ['e',2],            'a3': ['h',3],            'a4': ['s',2],            'a5': ['j',9],            'a6': ['y',7] } 

I need to write a function which returns the ordered keys in a list, depending on which column your sorting on so for example if we're sorting on mydict[key][1] (ascending)

I should receive a list back like so

['a2', 'a4', 'a3', 'a1', 'a6', 'a5'] 

It mostly works, apart from when you have columns of the same value for multiple keys, eg. 'a2': ['e',2] and 'a4': ['s',2]. In this instance it returns the list like so

['a4', 'a4', 'a3', 'a1', 'a6', 'a5'] 

Here's the function I've defined

def itlist(table_dict,column_nb,order="A"):     try:         keys = table_dict.keys()         values = [i[column_nb-1] for i in table_dict.values()]         combo = zip(values,keys)         valkeys = dict(combo)         sortedCols = sorted(values) if order=="A" else sorted(values,reverse=True)         sortedKeys = [valkeys[i] for i in sortedCols]     except (KeyError, IndexError), e:         pass     return sortedKeys 

And if I want to sort on the numbers column for example it is called like so

sortedkeysasc = itmethods.itlist(table,2) 

So any suggestions?

Paul

Answer 1

Wouldn't it be much easier to use

sorted(d, key=lambda k: d[k][1]) 

(with d being the dictionary)?

Answer 2

>>> L = sorted(d.items(), key=lambda (k, v): v[1]) >>> L [('a2', ['e', 2]), ('a4', ['s', 2]), ('a3', ['h', 3]), ('a1', ['g', 6]), ('a6', ['y', 7]), ('a5', ['j', 9])]  >>> map(lambda (k,v): k, L) ['a2', 'a4', 'a3', 'a1', 'a6', 'a5'] 

Here you sort the dictionary items (key-value pairs) using a key - callable which establishes a total order on the items.

Then, you just filter out needed values using a map with a lambda which just selects the key. So you get the needed list of keys.

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EDIT: see this answer for a much better solution.