/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

1 vote191 views1 answer
Droping Duplicates
Python
3 votes172 views1 answer
Starting vscode on colab
Python
4 votes171 views5 answers
'datetime' has no attribute 'now'
Python
0 vote172 views1 answer
Python string: string concatenation
Python
3 votes172 views3 answers
Convert all columns to float pandas
Python
15 votes173 views5 answers
Pandas profiling
Python
0 vote232 views1 answer
Parallel loops in python
Python
0 vote182 views1 answer
Sklearn recognising sentences
Python
1 vote152 views2 answers
Python3 shebang
Python
-1 vote160 views1 answer
Python data first column indices
Python