Return first N key:value pairs from dict

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?

Solution 1:

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.iteritems())

This uses the implementation of take from the itertools recipes:

from itertools import islice

def take(n, iterable):
    "Return first n items of the iterable as a list"
    return list(islice(iterable, n))

See it working online: ideone

Update for Python 3.6

n_items = take(n, d.items())

Solution 2:

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]}

Solution 3:

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]

Solution 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))

Solution 5:

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().