/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

0 vote158 views1 answer
Pandas save csv list as columns
Python
5 votes132 views1 answer
Dataframe unique values in each column
Python
0 vote174 views1 answer
How to read multiple files in a loop in python
Python
61 votes140 views6 answers
Python classes
Python
1 vote133 views2 answers
Add ML
Python
1 vote126 views1 answer
Wtf forms required
Python
0 vote137 views1 answer
How to get last dimension of an array python
Python
15 votes127 views2 answers
Python re compile
Python
0 vote143 views1 answer
Set header in dataframe 2nd line
Python
0 vote123 views1 answer
Create a dictionary from index and column pandas
Python