How to add an element to the beginning of an OrderedDict?
Asked Answered
A

11

118

I have this:

d1 = OrderedDict([('a', '1'), ('b', '2')])

If I do this:

d1.update({'c':'3'})

Then I get this:

OrderedDict([('a', '1'), ('b', '2'), ('c', '3')])

but I want this:

[('c', '3'), ('a', '1'), ('b', '2')]

without creating new dictionary.

Altimetry answered 21/5, 2013 at 7:56 Comment(3)
i think you should redesign your programmLippizaner
"An OrderedDict is a dict that remembers the order that keys were first inserted." @ZagorulkinDmitry is right :) (docs.python.org/2/library/…)Wardmote
#39487Zugzwang
G
109

There's no built-in method for doing this in Python 2. If you need this, you need to write a prepend() method/function that operates on the OrderedDict internals with O(1) complexity.

For Python 3.2 and later, you should use the move_to_end method. The method accepts a last argument which indicates whether the element will be moved to the bottom (last=True) or the top (last=False) of the OrderedDict.

Finally, if you want a quick, dirty and slow solution, you can just create a new OrderedDict from scratch.

Details for the four different solutions:


Extend OrderedDict and add a new instance method

from collections import OrderedDict

class MyOrderedDict(OrderedDict):

    def prepend(self, key, value, dict_setitem=dict.__setitem__):

        root = self._OrderedDict__root
        first = root[1]

        if key in self:
            link = self._OrderedDict__map[key]
            link_prev, link_next, _ = link
            link_prev[1] = link_next
            link_next[0] = link_prev
            link[0] = root
            link[1] = first
            root[1] = first[0] = link
        else:
            root[1] = first[0] = self._OrderedDict__map[key] = [root, first, key]
            dict_setitem(self, key, value)

Demo:

>>> d = MyOrderedDict([('a', '1'), ('b', '2')])
>>> d
MyOrderedDict([('a', '1'), ('b', '2')])
>>> d.prepend('c', 100)
>>> d
MyOrderedDict([('c', 100), ('a', '1'), ('b', '2')])
>>> d.prepend('a', d['a'])
>>> d
MyOrderedDict([('a', '1'), ('c', 100), ('b', '2')])
>>> d.prepend('d', 200)
>>> d
MyOrderedDict([('d', 200), ('a', '1'), ('c', 100), ('b', '2')])

Standalone function that manipulates OrderedDict objects

This function does the same thing by accepting the dict object, key and value. I personally prefer the class:

from collections import OrderedDict

def ordered_dict_prepend(dct, key, value, dict_setitem=dict.__setitem__):
    root = dct._OrderedDict__root
    first = root[1]

    if key in dct:
        link = dct._OrderedDict__map[key]
        link_prev, link_next, _ = link
        link_prev[1] = link_next
        link_next[0] = link_prev
        link[0] = root
        link[1] = first
        root[1] = first[0] = link
    else:
        root[1] = first[0] = dct._OrderedDict__map[key] = [root, first, key]
        dict_setitem(dct, key, value)

Demo:

>>> d = OrderedDict([('a', '1'), ('b', '2')])
>>> ordered_dict_prepend(d, 'c', 100)
>>> d
OrderedDict([('c', 100), ('a', '1'), ('b', '2')])
>>> ordered_dict_prepend(d, 'a', d['a'])
>>> d
OrderedDict([('a', '1'), ('c', 100), ('b', '2')])
>>> ordered_dict_prepend(d, 'd', 500)
>>> d
OrderedDict([('d', 500), ('a', '1'), ('c', 100), ('b', '2')])

Use OrderedDict.move_to_end() (Python >= 3.2)

Python 3.2 introduced the OrderedDict.move_to_end() method. Using it, we can move an existing key to either end of the dictionary in O(1) time.

>>> d1 = OrderedDict([('a', '1'), ('b', '2')])
>>> d1.update({'c':'3'})
>>> d1.move_to_end('c', last=False)
>>> d1
OrderedDict([('c', '3'), ('a', '1'), ('b', '2')])

If we need to insert an element and move it to the top, all in one step, we can directly use it to create a prepend() wrapper (not presented here).


Create a new OrderedDict - slow!!!

If you don't want to do that and performance is not an issue then easiest way is to create a new dict:

from itertools import chain, ifilterfalse
from collections import OrderedDict


def unique_everseen(iterable, key=None):
    "List unique elements, preserving order. Remember all elements ever seen."
    # unique_everseen('AAAABBBCCDAABBB') --> A B C D
    # unique_everseen('ABBCcAD', str.lower) --> A B C D
    seen = set()
    seen_add = seen.add
    if key is None:
        for element in ifilterfalse(seen.__contains__, iterable):
            seen_add(element)
            yield element
    else:
        for element in iterable:
            k = key(element)
            if k not in seen:
                seen_add(k)
                yield element

d1 = OrderedDict([('a', '1'), ('b', '2'),('c', 4)])
d2 = OrderedDict([('c', 3), ('e', 5)])   #dict containing items to be added at the front
new_dic = OrderedDict((k, d2.get(k, d1.get(k))) for k in \
                                           unique_everseen(chain(d2, d1)))
print new_dic

output:

OrderedDict([('c', 3), ('e', 5), ('a', '1'), ('b', '2')])

Graeae answered 21/5, 2013 at 7:59 Comment(23)
Note if c already exists, this will not update the old valueCredits
this answer is simply wrong. you can, see answer below.Quintal
@Quintal If you are referring to move_to_end then there's no Python 3 tag on the question, move_to_end works only in Python 3.2+. I will update my answer to include the Python 3 based solution. Much thanks for update though!Graeae
@AshwiniChaudhary correct me if I'm wrong, but move_to_end can be simulated by deleting and readding a key - so technically it's not even required.Quintal
@Quintal Yeah, but in this case we are adding it as the first key not last. move_to_end covers both part with the optional last argument.Graeae
"Deleting an entry and reinserting it will move it to the end." - from docs.python.org (for any python version)Quintal
@Quintal Question is about adding it as the top not at the end.Graeae
Oh right, sorry - but this can be simulated, by moving everything else to the end. My point is: if you find this question, and do not at all care about performance, but just want to move something to the top and keep the old reference and read this question, then the answer "you can't" simply does not seem correct to me. Sorry for having been rather imprecise before.Quintal
I don't understand the appeal of ifilterfalse(seen.__contains__, iterable) while you can just do (x not in seen for x in iterable) or event just an if in the for loop. It's shorter, easier to read and does not require an additional import.Domineca
@AshwiniChaudhary You are right. Looking at the code of prepend, it is O(1) indeed. Thanks for noticing my oversight. I misinterpreted your first sentence and assumed complexity was O(n). "There's no built-in way to do this in Python 2 in O(1) time." could also be interpreted as implying a fundamental shortcoming with the Python 2 implementation. I rephrased the sentence to make it more clear.Farnham
@AshwiniChaudhary Since Python 3 has already added move_to_front, maybe it is better to implement a move_to_front method instead of a separate prepend method? This will make your code more portable if you ever need to support both Python 2 and Python 3 from the same code-base.Farnham
What's the reason behind dict_setitem=dict.__setitem__ as a param to prepend? Why would/should one pass a different setter?Surrejoinder
There must be a bug in ordered_dict_prepend above. Calling ordered_dict_prepend(d, 'c', 100) twice and trying to print the resulting dict (by simply entering d in Python's console) results in Python process keep grabbing memory. Tested with Python 2.7.10Licastro
@Surrejoinder dict_setitem=dict.__setitem__ is a micro-optimisation done to prevent extra lookups, in this case dict is a built-in variable.Graeae
@Piotr Dobrogost Thanks for pointing out the bug. I have edited (peer preview pending) the answer to fix that bug. I would have copied the code directly if not for you ;)Homerhomere
The edit was rejected since bugfixes are apparently not valid edit. stackoverflow.com/review/suggested-edits/19321924Homerhomere
I don't think it was a good idea to make the first thing suggested be to operate on an OrderedDict's internals—especially without some sort of warning about why it's generally considered a poor programming practice.Hayner
@Hayner Feel free to edit it, I will accept the changes if they look fine. :)Graeae
Ashwini: Well, maybe, if I get around to it. Seems like you really ought to do it though—assuming you know the reasons doing so is generally discouraged. A concrete reason: Under @Jared's answer there's a comment indicating that OrderedDict was rewritten in C for Python 3.5.Hayner
@Hayner I am aware of the re-write in 3, but the first section itself mentions the fact that 3.2+ has a dedicated method to do this now. The other version(s) that relies on internals is for Python 2, and Python 2 doesn't receive such rewrites anymore.Graeae
Ashwini: Agree it's probably safe to break the "rule" for Python 2 code since it very stable nowadays. It's unclear what version the OP was asking about (because you're the one who added both py2.x and 3.x tags to their question. Regardless, my original criticism still applies—about the first thing you suggest is accessing the internals of the built-in class without mentioning any caveats. You did mention that in 3.2+ it can be done via the new move_to_end() method, but needing to write something portable was criteria you effectively added by adding those tags (instead of asking the OP).Hayner
@Hayner While you checked the edit history you should've also checked their last seen time of May 27 2013. Good luck asking them for clarifications and also thank you for the downvote(no wonder I don't contribute here anymore).Graeae
I did note that—and what I meant was ask them back on May 21 '13 when the question was asked (and answered by yourself).Hayner
B
15

EDIT (2019-02-03) Note that the following answer only works on older versions of Python. More recently, OrderedDict has been rewritten in C. In addition this does touch double-underscore attributes which is frowned upon.

I just wrote a subclass of OrderedDict in a project of mine for a similar purpose. Here's the gist.

Insertion operations are also constant time O(1) (they don't require you to rebuild the data structure), unlike most of these solutions.

>>> d1 = ListDict([('a', '1'), ('b', '2')])
>>> d1.insert_before('a', ('c', 3))
>>> d1
ListDict([('c', 3), ('a', '1'), ('b', '2')])
Bora answered 20/8, 2013 at 4:20 Comment(3)
I get TypeError: '_Link' object does not support indexing when using this on Python 3.4.Fissionable
It also doesnt work with python 3.5: AttributeError: 'ListDict' object has no attribute '_OrderedDict__map'Volz
This no longer works because OrderedDict has been rewritten in C as of Python 3.5, and this subclass committed the taboo of mucking about with internals (actually reversing name mangling to access __ properties).Unopened
S
14

You have to make a new instance of OrderedDict. If your keys are unique:

d1=OrderedDict([("a",1),("b",2)])
d2=OrderedDict([("c",3),("d",99)])
both=OrderedDict(list(d2.items()) + list(d1.items()))
print(both)

#OrderedDict([('c', 3), ('d', 99), ('a', 1), ('b', 2)])

But if not, beware as this behavior may or may not be desired for you:

d1=OrderedDict([("a",1),("b",2)])
d2=OrderedDict([("c",3),("b",99)])
both=OrderedDict(list(d2.items()) + list(d1.items()))
print(both)

#OrderedDict([('c', 3), ('b', 2), ('a', 1)])
Spermary answered 6/8, 2013 at 12:44 Comment(3)
In python3, the items method no longer returns a list, but rather a view, which acts like a set. In this case you'll need to take the set union since concatenating with + won't work: dict(x.items() | y.items())Zugzwang
@TheDemz I thought set union would not preserve the order, thus making the final order of the items in the resulting OrderedDict unstable?Aricaarick
@Aricaarick Yes its unstable. The objects returned from dict.keys(), dict.values(), and dict.items() are called dictionary views. They are lazy sequences that will see changes in the underlying dictionary. To force the dictionary view to become a full list use list(dictview). See Dictionary view objects. docs.python.org/3.4/library/…Zugzwang
K
6

This is now possible with move_to_end(key, last=True)

>>> d = OrderedDict.fromkeys('abcde')
>>> d.move_to_end('b')
>>> ''.join(d.keys())
'acdeb'
>>> d.move_to_end('b', last=False)
>>> ''.join(d.keys())
'bacde'

https://docs.python.org/3/library/collections.html#collections.OrderedDict.move_to_end

Komsomolsk answered 22/1, 2015 at 11:28 Comment(0)
T
6

If you know you will want a 'c' key, but do not know the value, insert 'c' with a dummy value when you create the dict.

d1 = OrderedDict([('c', None), ('a', '1'), ('b', '2')])

and change the value later.

d1['c'] = 3
Tenderfoot answered 25/3, 2015 at 16:45 Comment(2)
Is there way to insert into an ordereddict such that the elements are still sorted(increasing/decreasing).Panga
An ordered dict is not sorted by any property of the items. It is ordered by insertion order, which has nothing to do with the items themselves. In CPthon 3.6 and in Python 3.7, all dicts are so ordered and there is little reason to use an OrderedDict.Tenderfoot
V
4

FWIW Here is a quick-n-dirty code I wrote for inserting to an arbitrary index position. Not necessarily efficient but it works in-place.

class OrderedDictInsert(OrderedDict):
    def insert(self, index, key, value):
        self[key] = value
        for ii, k in enumerate(list(self.keys())):
            if ii >= index and k != key:
                self.move_to_end(k)
Vendee answered 18/5, 2018 at 16:39 Comment(0)
G
4

You may want to use a different structure altogether, but there are ways to do it in python 2.7.

d1 = OrderedDict([('a', '1'), ('b', '2')])
d2 = OrderedDict(c='3')
d2.update(d1)

d2 will then contain

>>> d2
OrderedDict([('c', '3'), ('a', '1'), ('b', '2')])

As mentioned by others, in python 3.2 you can use OrderedDict.move_to_end('c', last=False) to move a given key after insertion.

Note: Take into consideration that the first option is slower for large datasets due to creation of a new OrderedDict and copying of old values.

Gerger answered 8/6, 2018 at 13:19 Comment(0)
K
2

If you need functionality that isn't there, just extend the class with whatever you want:

from collections import OrderedDict

class OrderedDictWithPrepend(OrderedDict):
    def prepend(self, other):
        ins = []
        if hasattr(other, 'viewitems'):
            other = other.viewitems()
        for key, val in other:
            if key in self:
                self[key] = val
            else:
                ins.append((key, val))
        if ins:
            items = self.items()
            self.clear()
            self.update(ins)
            self.update(items)

Not terribly efficient, but works:

o = OrderedDictWithPrepend()

o['a'] = 1
o['b'] = 2
print o
# OrderedDictWithPrepend([('a', 1), ('b', 2)])

o.prepend({'c': 3})
print o
# OrderedDictWithPrepend([('c', 3), ('a', 1), ('b', 2)])

o.prepend([('a',11),('d',55),('e',66)])
print o
# OrderedDictWithPrepend([('d', 55), ('e', 66), ('c', 3), ('a', 11), ('b', 2)])
Kadi answered 21/5, 2013 at 8:55 Comment(0)
H
1

I would suggest adding a prepend() method to this pure Python ActiveState recipe or deriving a subclass from it. The code to do so could be a fairly efficient given that the underlying data structure for ordering is a linked-list.

Update

To prove this approach is feasible, below is code that does what's suggested. As a bonus, I also made a few additional minor changes to get to work in both Python 2.7.15 and 3.7.1.

A prepend() method has been added to the class in the recipe and has been implemented in terms of another method that's been added named move_to_end(), which was added to OrderedDict in Python 3.2.

prepend() can also be implemented directly, almost exactly as shown at the beginning of @Ashwini Chaudhary's answer—and doing so would likely result in it being slightly faster, but that's been left as an exercise for the motivated reader...

# Ordered Dictionary for Py2.4 from https://code.activestate.com/recipes/576693

# Backport of OrderedDict() class that runs on Python 2.4, 2.5, 2.6, 2.7 and pypy.
# Passes Python2.7's test suite and incorporates all the latest updates.

try:
    from thread import get_ident as _get_ident
except ImportError:  # Python 3
#    from dummy_thread import get_ident as _get_ident
    from _thread import get_ident as _get_ident  # Changed - martineau

try:
    from _abcoll import KeysView, ValuesView, ItemsView
except ImportError:
    pass

class MyOrderedDict(dict):
    'Dictionary that remembers insertion order'
    # An inherited dict maps keys to values.
    # The inherited dict provides __getitem__, __len__, __contains__, and get.
    # The remaining methods are order-aware.
    # Big-O running times for all methods are the same as for regular dictionaries.

    # The internal self.__map dictionary maps keys to links in a doubly linked list.
    # The circular doubly linked list starts and ends with a sentinel element.
    # The sentinel element never gets deleted (this simplifies the algorithm).
    # Each link is stored as a list of length three:  [PREV, NEXT, KEY].

    def __init__(self, *args, **kwds):
        '''Initialize an ordered dictionary.  Signature is the same as for
        regular dictionaries, but keyword arguments are not recommended
        because their insertion order is arbitrary.

        '''
        if len(args) > 1:
            raise TypeError('expected at most 1 arguments, got %d' % len(args))
        try:
            self.__root
        except AttributeError:
            self.__root = root = []  # sentinel node
            root[:] = [root, root, None]
            self.__map = {}
        self.__update(*args, **kwds)

    def prepend(self, key, value):  # Added to recipe.
        self.update({key: value})
        self.move_to_end(key, last=False)

    #### Derived from cpython 3.2 source code.
    def move_to_end(self, key, last=True):  # Added to recipe.
        '''Move an existing element to the end (or beginning if last==False).

        Raises KeyError if the element does not exist.
        When last=True, acts like a fast version of self[key]=self.pop(key).
        '''
        PREV, NEXT, KEY = 0, 1, 2

        link = self.__map[key]
        link_prev = link[PREV]
        link_next = link[NEXT]
        link_prev[NEXT] = link_next
        link_next[PREV] = link_prev
        root = self.__root

        if last:
            last = root[PREV]
            link[PREV] = last
            link[NEXT] = root
            last[NEXT] = root[PREV] = link
        else:
            first = root[NEXT]
            link[PREV] = root
            link[NEXT] = first
            root[NEXT] = first[PREV] = link
    ####

    def __setitem__(self, key, value, dict_setitem=dict.__setitem__):
        'od.__setitem__(i, y) <==> od[i]=y'
        # Setting a new item creates a new link which goes at the end of the linked
        # list, and the inherited dictionary is updated with the new key/value pair.
        if key not in self:
            root = self.__root
            last = root[0]
            last[1] = root[0] = self.__map[key] = [last, root, key]
        dict_setitem(self, key, value)

    def __delitem__(self, key, dict_delitem=dict.__delitem__):
        'od.__delitem__(y) <==> del od[y]'
        # Deleting an existing item uses self.__map to find the link which is
        # then removed by updating the links in the predecessor and successor nodes.
        dict_delitem(self, key)
        link_prev, link_next, key = self.__map.pop(key)
        link_prev[1] = link_next
        link_next[0] = link_prev

    def __iter__(self):
        'od.__iter__() <==> iter(od)'
        root = self.__root
        curr = root[1]
        while curr is not root:
            yield curr[2]
            curr = curr[1]

    def __reversed__(self):
        'od.__reversed__() <==> reversed(od)'
        root = self.__root
        curr = root[0]
        while curr is not root:
            yield curr[2]
            curr = curr[0]

    def clear(self):
        'od.clear() -> None.  Remove all items from od.'
        try:
            for node in self.__map.itervalues():
                del node[:]
            root = self.__root
            root[:] = [root, root, None]
            self.__map.clear()
        except AttributeError:
            pass
        dict.clear(self)

    def popitem(self, last=True):
        '''od.popitem() -> (k, v), return and remove a (key, value) pair.
        Pairs are returned in LIFO order if last is true or FIFO order if false.

        '''
        if not self:
            raise KeyError('dictionary is empty')
        root = self.__root
        if last:
            link = root[0]
            link_prev = link[0]
            link_prev[1] = root
            root[0] = link_prev
        else:
            link = root[1]
            link_next = link[1]
            root[1] = link_next
            link_next[0] = root
        key = link[2]
        del self.__map[key]
        value = dict.pop(self, key)
        return key, value

    # -- the following methods do not depend on the internal structure --

    def keys(self):
        'od.keys() -> list of keys in od'
        return list(self)

    def values(self):
        'od.values() -> list of values in od'
        return [self[key] for key in self]

    def items(self):
        'od.items() -> list of (key, value) pairs in od'
        return [(key, self[key]) for key in self]

    def iterkeys(self):
        'od.iterkeys() -> an iterator over the keys in od'
        return iter(self)

    def itervalues(self):
        'od.itervalues -> an iterator over the values in od'
        for k in self:
            yield self[k]

    def iteritems(self):
        'od.iteritems -> an iterator over the (key, value) items in od'
        for k in self:
            yield (k, self[k])

    def update(*args, **kwds):
        '''od.update(E, **F) -> None.  Update od from dict/iterable E and F.

        If E is a dict instance, does:           for k in E: od[k] = E[k]
        If E has a .keys() method, does:         for k in E.keys(): od[k] = E[k]
        Or if E is an iterable of items, does:   for k, v in E: od[k] = v
        In either case, this is followed by:     for k, v in F.items(): od[k] = v

        '''
        if len(args) > 2:
            raise TypeError('update() takes at most 2 positional '
                            'arguments (%d given)' % (len(args),))
        elif not args:
            raise TypeError('update() takes at least 1 argument (0 given)')
        self = args[0]
        # Make progressively weaker assumptions about "other"
        other = ()
        if len(args) == 2:
            other = args[1]
        if isinstance(other, dict):
            for key in other:
                self[key] = other[key]
        elif hasattr(other, 'keys'):
            for key in other.keys():
                self[key] = other[key]
        else:
            for key, value in other:
                self[key] = value
        for key, value in kwds.items():
            self[key] = value

    __update = update  # let subclasses override update without breaking __init__

    __marker = object()

    def pop(self, key, default=__marker):
        '''od.pop(k[,d]) -> v, remove specified key and return the corresponding value.
        If key is not found, d is returned if given, otherwise KeyError is raised.

        '''
        if key in self:
            result = self[key]
            del self[key]
            return result
        if default is self.__marker:
            raise KeyError(key)
        return default

    def setdefault(self, key, default=None):
        'od.setdefault(k[,d]) -> od.get(k,d), also set od[k]=d if k not in od'
        if key in self:
            return self[key]
        self[key] = default
        return default

    def __repr__(self, _repr_running={}):
        'od.__repr__() <==> repr(od)'
        call_key = id(self), _get_ident()
        if call_key in _repr_running:
            return '...'
        _repr_running[call_key] = 1
        try:
            if not self:
                return '%s()' % (self.__class__.__name__,)
            return '%s(%r)' % (self.__class__.__name__, self.items())
        finally:
            del _repr_running[call_key]

    def __reduce__(self):
        'Return state information for pickling'
        items = [[k, self[k]] for k in self]
        inst_dict = vars(self).copy()
        for k in vars(MyOrderedDict()):
            inst_dict.pop(k, None)
        if inst_dict:
            return (self.__class__, (items,), inst_dict)
        return self.__class__, (items,)

    def copy(self):
        'od.copy() -> a shallow copy of od'
        return self.__class__(self)

    @classmethod
    def fromkeys(cls, iterable, value=None):
        '''OD.fromkeys(S[, v]) -> New ordered dictionary with keys from S
        and values equal to v (which defaults to None).

        '''
        d = cls()
        for key in iterable:
            d[key] = value
        return d

    def __eq__(self, other):
        '''od.__eq__(y) <==> od==y.  Comparison to another OD is order-sensitive
        while comparison to a regular mapping is order-insensitive.

        '''
        if isinstance(other, MyOrderedDict):
            return len(self)==len(other) and self.items() == other.items()
        return dict.__eq__(self, other)

    def __ne__(self, other):
        return not self == other

    # -- the following methods are only used in Python 2.7 --

    def viewkeys(self):
        "od.viewkeys() -> a set-like object providing a view on od's keys"
        return KeysView(self)

    def viewvalues(self):
        "od.viewvalues() -> an object providing a view on od's values"
        return ValuesView(self)

    def viewitems(self):
        "od.viewitems() -> a set-like object providing a view on od's items"
        return ItemsView(self)

if __name__ == '__main__':

    d1 = MyOrderedDict([('a', '1'), ('b', '2')])
    d1.update({'c':'3'})
    print(d1)  # -> MyOrderedDict([('a', '1'), ('b', '2'), ('c', '3')])

    d2 = MyOrderedDict([('a', '1'), ('b', '2')])
    d2.prepend('c', 100)
    print(d2)  # -> MyOrderedDict([('c', 100), ('a', '1'), ('b', '2')])
Hayner answered 21/5, 2013 at 9:39 Comment(1)
@LazyLeopard: It's not thread-safe. You may find How to make built-in containers (sets, dicts, lists) thread safe? of interest.Hayner
G
1

I got an infinity loop while trying to print or save the dictionary using @Ashwini Chaudhary answer with Python 2.7. But I managed to reduce his code a little, and got it working here:

def move_to_dict_beginning(dictionary, key):
    """
        Move a OrderedDict item to its beginning, or add it to its beginning.
        Compatible with Python 2.7
    """

    if sys.version_info[0] < 3:
        value = dictionary[key]
        del dictionary[key]
        root = dictionary._OrderedDict__root

        first = root[1]
        root[1] = first[0] = dictionary._OrderedDict__map[key] = [root, first, key]
        dict.__setitem__(dictionary, key, value)

    else:
        dictionary.move_to_end( key, last=False )
Gametocyte answered 11/1, 2019 at 11:56 Comment(1)
this is brilliant! Worked for me!Feuilleton
D
0

This is a default, ordered dict which allows to insert items in any position and use the . operator to create keys:

from collections import OrderedDict

class defdict(OrderedDict):

    _protected = ["_OrderedDict__root", "_OrderedDict__map", "_cb"]    
    _cb = None

    def __init__(self, cb=None):
        super(defdict, self).__init__()
        self._cb = cb

    def __setattr__(self, name, value):
        # if the attr is not in self._protected set a key
        if name in self._protected:
            OrderedDict.__setattr__(self, name, value)
        else:
            OrderedDict.__setitem__(self, name, value)

    def __getattr__(self, name):
        if name in self._protected:
            return OrderedDict.__getattr__(self, name)
        else:
            # implements missing keys
            # if there is a callable _cb, create a key with its value
            try:
                return OrderedDict.__getitem__(self, name)
            except KeyError as e:
                if callable(self._cb):
                    value = self[name] = self._cb()
                    return value
                raise e

    def insert(self, index, name, value):
        items = [(k, v) for k, v in self.items()]
        items.insert(index, (name, value))
        self.clear()
        for k, v in items:
            self[k] = v


asd = defdict(lambda: 10)
asd.k1 = "Hey"
asd.k3 = "Bye"
asd.k4 = "Hello"
asd.insert(1, "k2", "New item")
print asd.k5 # access a missing key will create one when there is a callback
# 10
asd.k6 += 5  # adding to a missing key
print asd.k6
# 15
print asd.keys()
# ['k1', 'k2', 'k3', 'k4', 'k5', 'k6']
print asd.values()
# ['Hey', 'New item', 'Bye', 'Hello', 10, 15]
Dabster answered 4/7, 2019 at 18:31 Comment(0)

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