Return first N key:value pairs from dict

Question

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 f

Answer 1

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]}

Answer 2

For Python 3 and above,To select first n Pairs

n=4
firstNpairs = {k: Diction[k] for k in list(Diction.keys())[:n]}

Answer 3

just add an answer using zip,

{k: d[k] for k, _ in zip(d, range(n))}

Answer 4

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()

Answer 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().

Answer 6

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 !