dict comprehension

1square_dict = {num: num*num for num in range(1, 11)}

1# dict comprehension we use same logic, with a difference of key:value pair
2# {key:value for i in list}
3
4fruits = ["apple", "banana", "cherry"]
5print({f: len(f) for f in fruits})
6
7#output
8{'apple': 5, 'banana': 6, 'cherry': 6}

1dict1 = {'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5}
2# Double each value in the dictionary
3double_dict1 = {k:v*2 for (k,v) in dict1.items()}
4# double_dict1 = {'e': 10, 'a': 2, 'c': 6, 'b': 4, 'd': 8} <-- new dict

1# Dictionary Comprehension Types
2
3{j:j*2 for i in range(10)}
4
5#using if 
6{j:j*2 for j in range(10) if j%2==0}
7
8#using if and 
9{j:j*2 for j in range(10) if j%2==0 and j%3==0}
10
11#using if else
12{j:(j*2 if j%2==0 and j%3==0 else 'invalid' )for j in range(10) }

1# dictionary comprehension
2# these are a little bit tougher ones than list comprehension
3
4sample_dict = {
5    'a': 1,
6    'b': 2,
7    'c': 3,
8    'd': 4,
9    'e': 5
10}
11
12# making squares of the numbers using  dict comprehension
13square_dict = {key:value**2 for key, value in sample_dict.items()}
14print(square_dict)
15
16square_dict_even = {key:value**2 for key, value in sample_dict.items() if value % 2 == 0}
17print(square_dict_even)
18
19# if you don't have a dictionary and you wanna create a dictionary of a number:number**2
20square_without_dict = {num:num**2 for num in range(11)}
21print(square_without_dict)
22
23

1# Dictionary comprehension
2
3{i:i*2 for i in range(10)}
4