You could try: >>> d = OrderedDict([(‘a’, 1), (‘c’, 3), (‘b’, 2)]) >>> d OrderedDict([(‘a’, 1), (‘c’, 3), (‘b’, 2)]) >>> d2 = OrderedDict([(‘__C__’, v) if k == ‘c’ else (k, v) for k, v in d.items()]) >>> d2 OrderedDict([(‘a’, 1), (‘__C__’, 3), (‘b’, 2)])
From the docs: The OrderedDict constructor and update() method both accept keyword arguments, but their order is lost because Python’s function call semantics pass-in keyword arguments using a regular unordered dictionary. So initialization loses the ordering, because it’s basically calling a constructor with **kwargs. Edit: In terms of a solution (not just an explanation)—as pointed … Read more
As of Python 3.7, a new improvement to the dict built-in is: the insertion-order preservation nature of dict objects has been declared to be an official part of the Python language spec. This means there is no real need for OrderedDict anymore 🎉. They are almost the same. Some minor details to consider… Here are … Read more
You may get a list of keys with the keys property: In [20]: d=OrderedDict(((“fruit”, “banana”), (“drinks”, ‘water’), (“animal”, “cat”))) In [21]: d.keys().index(‘animal’) Out[21]: 2 Better performance could be achieved with the use of iterkeys() though. For those using Python 3: >>> list(d.keys()).index(‘animal’) 2
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
Yes, you can use del: del dct[key] Below is a demonstration: >>> from collections import OrderedDict >>> dct = OrderedDict() >>> dct[‘a’] = 1 >>> dct[‘b’] = 2 >>> dct[‘c’] = 3 >>> dct OrderedDict([(‘a’, 1), (‘b’, 2), (‘c’, 3)]) >>> del dct[‘b’] >>> dct OrderedDict([(‘a’, 1), (‘c’, 3)]) >>> In fact, you should always … 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
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
In Python 3, dict.keys() and dict.values() return special iterable classes – respectively a collections.abc.KeysView and a collections.abc.ValuesView. The first one inherit it’s __eq__ method from set, the second uses the default object.__eq__ which tests on object identity.