/Python/Take first n row of dictionary python

Take first n row of dictionary python

a_dictionary = {"name" : "John", "age" : 35, "height" : 65}

dict_items = a_dictionary.items()

first_two = list(dict_items)[:2]
print(first_two)

OUTPUT
[('name', 'John'), ('age', 35)]

first_three = list(dict_items)[:3]
print(first_three)

OUTPUT
[('name', 'John'), ('age', 35), ('height', 65)]

Similar codes

4 votes378 views4 answers
How to change django admin text
Python Django
2 votes342 views1 answer
Python numpy array replace nan with string
Python
0 vote326 views1 answer
Add zero in front of numbers lower than 10 python
Python
9 votes378 views2 answers
Keras model save
Python
2 votes373 views1 answer
Python how to make an array of ones
Python
0 vote371 views1 answer
Convert list of lists to numpy array matrix python
Python
16 votes349 views2 answers
Python to excel
Python
1 vote390 views1 answer
Import * meaning in python
Python
1 vote401 views1 answer
How to parse dicts in reqparse in flask
Python Flask
0 vote398 views2 answers
Python time modülü
Python