CODEWITHSUNDEEP
pythondictionary
Saphire
Saphire

Reputation: 1930

How to make a sorted dictionary class?

I am having a hard time writing a class, which should be able to iterate through a sorted dicitonary. My main problem is at the iter-overload. I don't how to get the dic sorted.

class SortedDict():
    def __init__(self, dic = None):
        self.dic = {}
        if len(dic) > 0: self.dic = dic;

    def __iter__(self):
        self.dic = sorted(self.dic.keys())
        self.index = 0
        return self

    def next(self):
        if self.index+1 < len(self.dic):
            self.index += 1
            return self.dic.keys()[self.index]

Upvotes: 2

Views: 950

Answers (2)

thefourtheye
thefourtheye

Reputation: 239463

You don't have to reinvent the wheel. You can simply subclass the dict and implement the SortedDict, like this

class SortedDict(dict):
    def __iter__(self):
        return iter(sorted(super(SortedDict, self).__iter__()))

    def items(self):
        return iter((k, self[k]) for k in self)

    def keys(self):
        return list(self)

    def values(self):
        return [self[k] for k in self]

Thanks Poke and Martijn Pieters, for helping me with this answer.

You can see the difference between collections.OrderedDict, dict and SortedDict.

a = OrderedDict()
a["2"], a["1"], a["3"] = 2, 1, 3
print list(a.items()), a.keys(), a.values()

b = {}
b["2"], b["1"], b["3"] = 2, 1, 3
print list(b.items()), b.keys(), b.values()

c = SortedDict()
c["2"], c["1"], c["3"] = 2, 1, 3
print list(c.items()), c.keys(), c.values()

Output

[('2', 2), ('1', 1), ('3', 3)] ['2', '1', '3'] [2, 1, 3]
[('1', 1), ('3', 3), ('2', 2)] ['1', '3', '2'] [1, 3, 2]
[('1', 1), ('2', 2), ('3', 3)] ['1', '2', '3'] [1, 2, 3]

Upvotes: 5

Steve Jessop
Steve Jessop

Reputation: 279255

Since you're willing to sort at the point you begin the iteration, all you need is:

def __iter__(self):
    return iter(sorted(self.dic))

__iter__ has to return an iterator, and the builtin function iter() gets one from the sorted list of keys. Job done, no need for a next() function.

Upvotes: 3

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