The OrderedDict will preserve any order that it has access to. The only way to pass ordered data to it to initialize is to pass a list (or, more generally, an iterable) of key-value pairs, as in your last two examples. As the documentation you linked to says, the OrderedDict does not have access to … Read more
You can use the recipe that uses collections.Counter and collections.OrderedDict: from collections import Counter, OrderedDict class OrderedCounter(Counter, OrderedDict): ‘Counter that remembers the order elements are first encountered’ def __repr__(self): return ‘%s(%r)’ % (self.__class__.__name__, OrderedDict(self)) def __reduce__(self): return self.__class__, (OrderedDict(self),) words = [“oranges”, “apples”, “apples”, “bananas”, “kiwis”, “kiwis”, “apples”] c = OrderedCounter(words) print(c) # OrderedCounter(OrderedDict([(‘oranges’, 1), … Read more
OrderedDict is a subclass of dict, and needs more memory to keep track of the order in which keys are added. This isn’t trivial. The implementation adds a second dict under the covers, and a doubly-linked list of all the keys (that’s the part that remembers the order), and a bunch of weakref proxies. It’s … Read more
OrderedCounter is given as an example in the OrderedDict documentation, and works without needing to override any methods: class OrderedCounter(Counter, OrderedDict): pass When a class method is called, Python has to find the correct method to execute. There is a defined order in which it searches the class hierarchy called the “method resolution order” or … Read more
You are doing it wrong. You need not only to insert values sequentially into dictionary, but also remove some elements and see how the order has changed after this. The next code demonstrates this: OrderedDictionary od = new OrderedDictionary(); Dictionary<String, String> d = new Dictionary<String, String>(); Random r = new Random(); for (int i = … Read more
Your problem is that you are constructing a dict to give the initial data to the OrderedDict – this dict doesn’t store any order, so the order is lost before it gets to the OrderedDict. The solution is to build from an ordered data type – the easiest being a list of tuples: >>> from … Read more
Here’s a hack that almost gives you the syntax you want: class _OrderedDictMaker(object): def __getitem__(self, keys): if not isinstance(keys, tuple): keys = (keys,) assert all(isinstance(key, slice) for key in keys) return OrderedDict([(k.start, k.stop) for k in keys]) ordereddict = _OrderedDictMaker() from nastyhacks import ordereddict menu = ordereddict[ “about” : “about”, “login” : “login”, ‘signup’: “signup” … Read more
First, I don’t understand why you are adding all the keys and values count times, Index is never used. I tried this example : var source = new BindingSource(); List<MyStruct> list = new List<MyStruct> { new MyStruct(“fff”, “b”), new MyStruct(“c”,”d”) }; source.DataSource = list; grid.DataSource = source; and that work pretty well, I get two … Read more
If its an OrderedDict() you can easily access the elements by indexing by getting the tuples of (key,value) pairs as follows >>> import collections >>> d = collections.OrderedDict() >>> d[‘foo’] = ‘python’ >>> d[‘bar’] = ‘spam’ >>> d.items() [(‘foo’, ‘python’), (‘bar’, ‘spam’)] >>> d.items()[0] (‘foo’, ‘python’) >>> d.items()[1] (‘bar’, ‘spam’) Note for Python 3.X dict.items … Read more
You’ll have to create a new one since OrderedDict is sorted by insertion order. In your case the code would look like this: foo = OrderedDict(sorted(foo.iteritems(), key=lambda x: x[1][‘depth’])) See http://docs.python.org/dev/library/collections.html#ordereddict-examples-and-recipes for more examples. Note for Python 3 you will need to use .items() instead of .iteritems().