I am trying to get a sorted dictionary. But the order of the items between mydict and orddict doesn't seem to change.
from collections import OrderedDict
mydict = {'a': 1, 'b': 2, 'c': 3, 'd': 4}
orddict = OrderedDict(mydict)
print(mydict, orddict)
# print items in mydict:
print('mydict')
for k, v in mydict.items():
print(k, v)
print('ordereddict')
# print items in ordered dictionary
for k, v in orddict.items():
print(k, v)
# print the dictionary keys
# for key in mydict.keys():
# print(key)
# print the dictionary values
# for value in mydict.values():
# print(value)
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 some comparisons between Python 3.7+ dict and OrderedDict:
from collections import OrderedDict
d = {'b': 1, 'a': 2}
od = OrderedDict([('b', 1), ('a', 2)])
# they are equal with content and order
assert d == od
assert list(d.items()) == list(od.items())
assert repr(dict(od)) == repr(d)
Obviously, there is a difference between the string representation of the two object, with the dict object in more natural and compact form.
str(d) # {'b': 1, 'a': 2}
str(od) # OrderedDict([('b', 1), ('a', 2)])
As for different methods between the two, this question can be answered with set theory:
d_set = set(dir(d))
od_set = set(dir(od))
od_set.difference(d_set)
# {'__dict__', '__reversed__', 'move_to_end'} for Python 3.7
# {'__dict__', 'move_to_end'} for Python 3.8+
This means OrderedDict has at most two features that dict does not have built-in, but work-arounds are shown here:
Workaround for __reversed__ / reversed()
No workaround is really needed for Python 3.8+, which fixed this issue. OrderedDict can be "reversed", which simply reverses the keys (not the whole dictionary):
reversed(od) # <odict_iterator at 0x7fc03f119888>
list(reversed(od)) # ['a', 'b']
# with Python 3.7:
reversed(d) # TypeError: 'dict' object is not reversible
list(reversed(list(d.keys()))) # ['a', 'b']
# with Python 3.8+:
reversed(d) # <dict_reversekeyiterator at 0x16caf9d2a90>
list(reversed(d)) # ['a', 'b']
To properly reverse a whole dictionary using Python 3.7+:
dict(reversed(list(d.items()))) # {'a': 2, 'b': 1}
Workaround for move_to_end
OrderedDict has a move_to_end method, which is simple to implement:
od.move_to_end('b') # now it is: OrderedDict([('a', 2), ('b', 1)])
d['b'] = d.pop('b') # now it is: {'a': 2, 'b': 1}
An OrderedDict preserves the order elements were inserted:
>>> od = OrderedDict()
>>> od['c'] = 1
>>> od['b'] = 2
>>> od['a'] = 3
>>> od.items()
[('c', 1), ('b', 2), ('a', 3)]
>>> d = {}
>>> d['c'] = 1
>>> d['b'] = 2
>>> d['a'] = 3
>>> d.items()
[('a', 3), ('c', 1), ('b', 2)]
So an OrderedDict does not order the elements for you, it preserves the order you give it.
If you want to "sort" a dictionary, you probably want
>>> sorted(d.items())
[('a', 1), ('b', 2), ('c', 3)]
Starting with CPython 3.6 and all other Python implementations starting with Python 3.7, the built-in dict is ordered - you get the items out in the order you inserted them. Which makes dict and OrderedDict effectively the same.
The documentation for OrderedDict lists the remaining differences. The most important one is that
The equality operation for OrderedDict checks for matching order.
Then there's a few minor practical differences:
dict.popitem() takes no arguments, whereas OrderedDict.popitem(last=True) accepts an optional last= argument that lets you pop the first item instead of the last item.
OrderedDict has a move_to_end(key, last=True) method to efficiently reposition an element to the end or the beginning. With dicts you can move a key to the end by re-inserting it: mydict['key'] = mydict.pop('key')
Until Python 3.8, you could do reversed(OrderedDict()) but reversed({}) would raise a TypeError: 'dict' object is not reversible error because they forgot to add a __reversed__ dunder method to dict when they made it ordered. This is now fixed.
And there are a few under-the-hood differences that might mean that you could get better performance for some specific usecase with OrderedDict:
The regular dict was designed to be very good at mapping
operations. Tracking insertion order was secondary.
The OrderedDict was designed to be good at reordering operations.
Space efficiency, iteration speed, and the performance of update
operations were secondary.
Algorithmically, OrderedDict can handle frequent reordering
operations better than dict. This makes it suitable for tracking
recent accesses (for example in an LRU cache).
See this great talk from 2016 by Raymond Hettinger for details on how Python dictionaries are implemented.
Adding on to the answer by Brian, OrderedDict is really great. Here's why:
You can use it as simple dict object because it supports equality testing with other Mapping objects like collections.counter.
OrderedDict preserves the insertion order as explained by Brian. In addition to that it has a method popitem which returns (key,value) pairs in LIFO order. So, you can also use it as a mapped 'stack'.
You not only get the full features of a dict but also, some cool tricks.
Ordered dictionaries are just like regular dictionaries but they remember the order that items were inserted. When iterating over an ordered dictionary, the items are returned in the order their keys were first added.
So it only sorts by order of adding into the dict
You can build an OrderedDict order by key as follow,
orddict = OrderedDict(sorted(mydict.items(), key = lambda t: t[0]))
or simply as #ShadowRanger mentioned in comment
orddict = OrderedDict(sorted(d.items()))
If you want to order by value,
orddict = OrderedDict(sorted(mydict.items(), key = lambda t: t[1]))
More information in 8.3.5.1. OrderedDict Examples and Recipes
Related
how to merge two dictionaries in python 2.7 without changing the order of keys in it. as I have to make a CSV file in the required order. I just want to add B dictionary after A dictionary values.
def Merge(A,B):
m=A.copy()
m.update(B)
return m
I am using this method. I also try with +. but the same result. ** is not working in python 2.7
Dictionaries are unordered containers. So, what you say is not correct. That is, if you simply have two dictionaries, then you have no order.
BUT! You have can order the keys of a dictionary, using the OrderedDict container.
You can consider something like:
>>> from collections import OrderedDict
>>> a = OrderedDict({'a': 1, 'b': 2})
>>> b = OrderedDict({'c': 3, 'd': 4})
>>> c = OrderedDict()
>>>for d in [a, b]:
... for k, v in d.items():
... c.update({k :v})
>>> print(c)
OrderedDict([('a', 1), ('b', 2), ('c', 3), ('d', 4)])
# Or
>>> d = OrderedDict()
>>> d.update(a.copy())
>>> d.update(b.copy())
>>> print(d)
OrderedDict([('a', 1), ('b', 2), ('c', 3), ('d', 4)])
Or, if you want a function:
from collections import OrderedDict
def merge(a, b):
c = OrderedDict()
c.update(a.copy())
c.update(b.copy())
return c
You can read more about OrderedDict here.
Dictionaries are unordered in Python 2. If you need to preserve the order of keys, you'll have to use something else, like a list of pairs, or an OrderedDict. Here's one approach.
from collections import OrderedDict
def merge(a, b):
return OrderedDict((k, v) for d in [a, b] for k, v in d.items())
Dictionaries is considered as an unordered collection of values mapped to certain keys. First, make an ordered dictionary by using OrderedDict from the collections library. Then using the method update() you can create the new dictionary. Here's an example.
# A Python program to demonstrate working of OrderedDict
from collections import OrderedDict
#Form OrderedDicts
dict1 = OrderedDict({"a": 1, "b": 2})
dict2 = OrderedDict({"c": 3, "d": 4})
print "These are two ordered dictionaries:\n"
print dict1 , dict2
#Using update method in python
def Merge(dict1, dict2):
return(dict1.update(dict2))
#Make a duplicate of the first dictionary
dict3 = dict1
#Call function
Merge(dict3, dict2)
print "Combined dictionary:", dict3
If you are just going to make a csv you can create a list of items from both ensuring that the items from one are after the items of the other. Then use the list of tuples to make the csv.
a = {1:1,2:2}
b = {3:3,4:4}
combined = []
combined.extend(a.items())
combined.extend(b.items())
In [8]: combined
Out[8]: [(1, 1), (2, 2), (3, 3), (4, 4)]
I am new to Python, and I am familiar with implementations of Multimaps in other languages. Does Python have such a data structure built-in, or available in a commonly-used library?
To illustrate what I mean by "multimap":
a = multidict()
a[1] = 'a'
a[1] = 'b'
a[2] = 'c'
print(a[1]) # prints: ['a', 'b']
print(a[2]) # prints: ['c']
Such a thing is not present in the standard library. You can use a defaultdict though:
>>> from collections import defaultdict
>>> md = defaultdict(list)
>>> md[1].append('a')
>>> md[1].append('b')
>>> md[2].append('c')
>>> md[1]
['a', 'b']
>>> md[2]
['c']
(Instead of list you may want to use set, in which case you'd call .add instead of .append.)
As an aside: look at these two lines you wrote:
a[1] = 'a'
a[1] = 'b'
This seems to indicate that you want the expression a[1] to be equal to two distinct values. This is not possible with dictionaries because their keys are unique and each of them is associated with a single value. What you can do, however, is extract all values inside the list associated with a given key, one by one. You can use iter followed by successive calls to next for that. Or you can just use two loops:
>>> for k, v in md.items():
... for w in v:
... print("md[%d] = '%s'" % (k, w))
...
md[1] = 'a'
md[1] = 'b'
md[2] = 'c'
Just for future visitors. Currently there is a python implementation of Multimap. It's available via pypi
Stephan202 has the right answer, use defaultdict. But if you want something with the interface of C++ STL multimap and much worse performance, you can do this:
multimap = []
multimap.append( (3,'a') )
multimap.append( (2,'x') )
multimap.append( (3,'b') )
multimap.sort()
Now when you iterate through multimap, you'll get pairs like you would in a std::multimap. Unfortunately, that means your loop code will start to look as ugly as C++.
def multimap_iter(multimap,minkey,maxkey=None):
maxkey = minkey if (maxkey is None) else maxkey
for k,v in multimap:
if k<minkey: continue
if k>maxkey: break
yield k,v
# this will print 'a','b'
for k,v in multimap_iter(multimap,3,3):
print v
In summary, defaultdict is really cool and leverages the power of python and you should use it.
You can take list of tuples and than can sort them as if it was a multimap.
listAsMultimap=[]
Let's append some elements (tuples):
listAsMultimap.append((1,'a'))
listAsMultimap.append((2,'c'))
listAsMultimap.append((3,'d'))
listAsMultimap.append((2,'b'))
listAsMultimap.append((5,'e'))
listAsMultimap.append((4,'d'))
Now sort it.
listAsMultimap=sorted(listAsMultimap)
After printing it you will get:
[(1, 'a'), (2, 'b'), (2, 'c'), (3, 'd'), (4, 'd'), (5, 'e')]
That means it is working as a Multimap!
Please note that like multimap here values are also sorted in ascending order if the keys are the same (for key=2, 'b' comes before 'c' although we didn't append them in this order.)
If you want to get them in descending order just change the sorted() function like this:
listAsMultimap=sorted(listAsMultimap,reverse=True)
And after you will get output like this:
[(5, 'e'), (4, 'd'), (3, 'd'), (2, 'c'), (2, 'b'), (1, 'a')]
Similarly here values are in descending order if the keys are the same.
The standard way to write this in Python is with a dict whose elements are each a list or set. As stephan202 says, you can somewhat automate this with a defaultdict, but you don't have to.
In other words I would translate your code to
a = dict()
a[1] = ['a', 'b']
a[2] = ['c']
print(a[1]) # prints: ['a', 'b']
print(a[2]) # prints: ['c']
Or subclass dict:
class Multimap(dict):
def __setitem__(self, key, value):
if key not in self:
dict.__setitem__(self, key, [value]) # call super method to avoid recursion
else
self[key].append(value)
There is no multi-map in the Python standard libs currently.
WebOb has a MultiDict class used to represent HTML form values, and it is used by a few Python Web frameworks, so the implementation is battle tested.
Werkzeug also has a MultiDict class, and for the same reason.
It is a bit hard for me to explain it in words, so I'll show an example:
What I have (data is a dict instance):
data = {'a':[4,5,3], 'b':[1,0,2], 'c':[6,7,8]}
What I need (ordered_data is an OrderedDict instance):
ordered_data = {'b':[0,1,2], 'a':[3,4,5], 'b':[6,7,8]}
The order of keys should be changed with respect to order of items in nested lists
tmp = {k:sorted(v) for k,v in data.items()}
ordered_data = OrderedDict((k,v) for k,v in sorted(tmp.items(), key=lambda i: i[1]))
First sort the values. If you don't need the original data, it's OK to do this in place, but I made a temporary variable.
key is a function that returns a key to be sorted on. In this case, the key is the second element of the item tuple (the list), and since lists are comparable, that's good enough.
You can use OrderedDict by sorting your items and the values :
>>> from operator import itemgetter
>>> from collections import OrderedDict
>>> d = OrderedDict(sorted([(k, sorted(v)) for k, v in data.items()], key=itemgetter(1)))
>>> d
OrderedDict([('b', [0, 1, 2]), ('a', [3, 4, 5]), ('c', [6, 7, 8])])
Usually, you should not worry about the data order in the dictionary itself, and instead, jsut order it when you retrieve the dictionary's contents (i.e.: iterate over it):
data = {'a':[4,5,3], 'b':[1,0,2], 'c':[6,7,8]}
for datum in sorted(data.items(), key=lambda item: item[1]):
...
I have a defaultdict with unsorted by key values:
d = defaultdict(int)
...
defaultdict(<type 'int'>, {'2014-06-30': 2, '2013-04-18': 7, '2013-04-19': 9})
Sort list of keys:
tmp = sorted(d.keys())
And create new dict:
nd = {}
for i in tmp:
nd[i] = d[i]
But as a result I am getting exact copy of source dict.
Although I added elements in new order.
dict instances are arbitrarily ordered. If you want a specific order then use collections.OrderedDict instead.
from collections import OrderedDict
ord_dict = OrderedDict()
for i in tmp:
ord_dict[i] = d[i]
print(ord_dict)
OrderedDict([('2013-04-18', 7), ('2013-04-19', 9), ('2014-06-30', 2)])
Or as #jon Clements suggests simply:
OrderedDict(sorted(d.iteritems()))
Or OrderedDict(sorted(d.items())) using python 3
python dictionaries does not work that way. Dictionary is actually a hash map which means values are put in a spot in dict depending on key hash, not the order you put your item and when iterating, you will not get "ordered dict". Use OrderedDict class instead
Consider the following dictionary, d:
d = {'a': 3, 'b': 2, 'c': 3, 'd': 4, 'e': 5}
I want to return the first N key:value pairs from d (N <= 4 in this case). What is the most efficient method of doing this?
There's no such thing a the "first n" keys because a dict doesn't remember which keys were inserted first.
You can get any n key-value pairs though:
n_items = take(n, d.items())
This uses the implementation of take from the itertools recipes:
from itertools import islice
def take(n, iterable):
"""Return the first n items of the iterable as a list."""
return list(islice(iterable, n))
See it working online: ideone
For Python < 3.6
n_items = take(n, d.iteritems())
A very efficient way to retrieve anything is to combine list or dictionary comprehensions with slicing. If you don't need to order the items (you just want n random pairs), you can use a dictionary comprehension like this:
# Python 2
first2pairs = {k: mydict[k] for k in mydict.keys()[:2]}
# Python 3
first2pairs = {k: mydict[k] for k in list(mydict)[:2]}
Generally a comprehension like this is always faster to run than the equivalent "for x in y" loop. Also, by using .keys() to make a list of the dictionary keys and slicing that list you avoid 'touching' any unnecessary keys when you build the new dictionary.
If you don't need the keys (only the values) you can use a list comprehension:
first2vals = [v for v in mydict.values()[:2]]
If you need the values sorted based on their keys, it's not much more trouble:
first2vals = [mydict[k] for k in sorted(mydict.keys())[:2]]
or if you need the keys as well:
first2pairs = {k: mydict[k] for k in sorted(mydict.keys())[:2]}
To get the top N elements from your python dictionary one can use the following line of code:
list(dictionaryName.items())[:N]
In your case you can change it to:
list(d.items())[:4]
Python's dicts are not ordered, so it's meaningless to ask for the "first N" keys.
The collections.OrderedDict class is available if that's what you need. You could efficiently get its first four elements as
import itertools
import collections
d = collections.OrderedDict((('foo', 'bar'), (1, 'a'), (2, 'b'), (3, 'c'), (4, 'd')))
x = itertools.islice(d.items(), 0, 4)
for key, value in x:
print key, value
itertools.islice allows you to lazily take a slice of elements from any iterator. If you want the result to be reusable you'd need to convert it to a list or something, like so:
x = list(itertools.islice(d.items(), 0, 4))
foo = {'a':1, 'b':2, 'c':3, 'd':4, 'e':5, 'f':6}
iterator = iter(foo.items())
for i in range(3):
print(next(iterator))
Basically, turn the view (dict_items) into an iterator, and then iterate it with next().
in py3, this will do the trick
{A:N for (A,N) in [x for x in d.items()][:4]}
{'a': 3, 'b': 2, 'c': 3, 'd': 4}
You can get dictionary items by calling .items() on the dictionary. then convert that to a list and from there get first N items as you would on any list.
below code prints first 3 items of the dictionary object
e.g.
d = {'a': 3, 'b': 2, 'c': 3, 'd': 4, 'e': 5}
first_three_items = list(d.items())[:3]
print(first_three_items)
Outputs:
[('a', 3), ('b', 2), ('c', 3)]
For Python 3.8 the correct answer should be:
import more_itertools
d = {'a': 3, 'b': 2, 'c': 3, 'd': 4, 'e': 5}
first_n = more_itertools.take(3, d.items())
print(len(first_n))
print(first_n)
Whose output is:
3
[('a', 3), ('b', 2), ('c', 3)]
After pip install more-itertools of course.
Did not see it on here. Will not be ordered but the simplest syntactically if you need to just take some elements from a dictionary.
n = 2
{key:value for key,value in d.items()[0:n]}
Were d is your dictionary and n is the printing number:
for idx, (k, v) in enumerate(d.items()):
if idx == n: break
print(k, v)
Casting your dictionary to a list can be slow.
Your dictionary may be too large and you don't need to cast all of it just for printing a few of the first.
See PEP 0265 on sorting dictionaries. Then use the aforementioned iterable code.
If you need more efficiency in the sorted key-value pairs. Use a different data structure. That is, one that maintains sorted order and the key-value associations.
E.g.
import bisect
kvlist = [('a', 1), ('b', 2), ('c', 3), ('e', 5)]
bisect.insort_left(kvlist, ('d', 4))
print kvlist # [('a', 1), ('b', 2), ('c', 3), ('d', 4), ('e', 5)]
just add an answer using zip,
{k: d[k] for k, _ in zip(d, range(n))}
This will work for python 3.8+:
d_new = {k:v for i, (k, v) in enumerate(d.items()) if i < n}
This depends on what is 'most efficient' in your case.
If you just want a semi-random sample of a huge dictionary foo, use foo.iteritems() and take as many values from it as you need, it's a lazy operation that avoids creation of an explicit list of keys or items.
If you need to sort keys first, there's no way around using something like keys = foo.keys(); keys.sort() or sorted(foo.iterkeys()), you'll have to build an explicit list of keys. Then slice or iterate through first N keys.
BTW why do you care about the 'efficient' way? Did you profile your program? If you did not, use the obvious and easy to understand way first. Chances are it will do pretty well without becoming a bottleneck.
For Python 3 and above,To select first n Pairs
n=4
firstNpairs = {k: Diction[k] for k in list(Diction.keys())[:n]}
This might not be very elegant, but works for me:
d = {'a': 3, 'b': 2, 'c': 3, 'd': 4, 'e': 5}
x= 0
for key, val in d.items():
if x == 2:
break
else:
x += 1
# Do something with the first two key-value pairs
You can approach this a number of ways. If order is important you can do this:
for key in sorted(d.keys()):
item = d.pop(key)
If order isn't a concern you can do this:
for i in range(4):
item = d.popitem()
Dictionary maintains no order , so before picking top N key value pairs lets make it sorted.
import operator
d = {'a': 3, 'b': 2, 'c': 3, 'd': 4}
d=dict(sorted(d.items(),key=operator.itemgetter(1),reverse=True))
#itemgetter(0)=sort by keys, itemgetter(1)=sort by values
Now we can do the retrieval of top 'N' elements:, using the method structure like this:
def return_top(elements,dictionary_element):
'''Takes the dictionary and the 'N' elements needed in return
'''
topers={}
for h,i in enumerate(dictionary_element):
if h<elements:
topers.update({i:dictionary_element[i]})
return topers
to get the top 2 elements then simply use this structure:
d = {'a': 3, 'b': 2, 'c': 3, 'd': 4}
d=dict(sorted(d.items(),key=operator.itemgetter(1),reverse=True))
d=return_top(2,d)
print(d)
consider a dict
d = {'a': 3, 'b': 2, 'c': 3, 'd': 4, 'e': 5}
from itertools import islice
n = 3
list(islice(d.items(),n))
islice will do the trick :)
hope it helps !
I have tried a few of the answers above and note that some of them are version dependent and do not work in version 3.7.
I also note that since 3.6 all dictionaries are ordered by the sequence in which items are inserted.
Despite dictionaries being ordered since 3.6 some of the statements you expect to work with ordered structures don't seem to work.
The answer to the OP question that worked best for me.
itr = iter(dic.items())
lst = [next(itr) for i in range(3)]
def GetNFirstItems(self):
self.dict = {f'Item{i + 1}': round(uniform(20.40, 50.50), 2) for i in range(10)}#Example Dict
self.get_items = int(input())
for self.index,self.item in zip(range(len(self.dict)),self.dict.items()):
if self.index==self.get_items:
break
else:
print(self.item,",",end="")
Unusual approach, as it gives out intense O(N) time complexity.
I like this one because no new list needs to be created, its a one liner which does exactly what you want and it works with python >= 3.8 (where dictionaries are indeed ordered, I think from python 3.6 on?):
new_d = {kv[0]:kv[1] for i, kv in enumerate(d.items()) if i <= 4}