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')])

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)])

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')])

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)])

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')])

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

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

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)

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.

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 )

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]