switching keys and values in a dictionary in python [duplicate]
my_dict = {2:3, 5:6, 8:9}
new_dict = {}
for k, v in my_dict.items():
new_dict[v] = kswitching keys and values in a dictionary in python [duplicate]
my_dict = {2:3, 5:6, 8:9}
new_dict = {}
for k, v in my_dict.items():
new_dict[v] = kpython sets
# You can't create a set like this in Python
my_set = {} # ---- This is a Dictionary/Hashmap
# To create a empty set you have to use the built in method:
my_set = set() # Correct!
set_example = {1,3,2,5,3,6}
print(set_example)
# OUTPUT
# {1,3,2,5,6} ---- Sets do not contain duplicates and are unorderedpython set &
>>> A = {0, 2, 4, 6, 8};
>>> B = {1, 2, 3, 4, 5};
>>> print("Union :", A | B)
Union : {0, 1, 2, 3, 4, 5, 6, 8}
>>> print("Intersection :", A & B)
Intersection : {2, 4}
>>> print("Difference :", A - B)
Difference : {0, 8, 6}
# elements not present both sets
>>> print("Symmetric difference :", A ^ B)
Symmetric difference : {0, 1, 3, 5, 6, 8}sets in python
The simplest way to create set is:
1. from list
code:
s = [1,2,3]
set = set(s)
print(set)
2. s,add() method
code:
set.add(1)
set.add(2)
set.remove(2)
print(set) // 1
3. Set conatins unique elementssets in python
basket = {'apple', 'orange', 'apple', 'pear', 'orange', 'banana'}
print(basket) # show that duplicates have been removed
# OUTPUT {'orange', 'banana', 'pear', 'apple'}
print('orange' in basket) # fast membership testing
# OUTPUT True
print('crabgrass' in basket)
# OUTPUT False
# Demonstrate set operations on unique letters from two words
print(a = set('abracadabra'))
print(b = set('alacazam'))
print(a) # unique letters in a
# OUTPUT {'a', 'r', 'b', 'c', 'd'}
print(a - b) # letters in a but not in b
# OUTPUT {'r', 'd', 'b'}
print(a | b) # letters in a or b or both
# OUTPUT {'a', 'c', 'r', 'd', 'b', 'm', 'z', 'l'}
print(a & b) # letters in both a and b
# OUTPUT {'a', 'c'}
print(a ^ b) # letters in a or b but not both
# OUTPUT {'r', 'd', 'b', 'm', 'z', 'l'}sets in python
set_of_base10_numbers = {1, 2, 3, 4, 5, 6, 7, 8, 9, 0}
set_of_base2_numbers = {1, 0}
intersection = set_of_base10_numbers.intersection(set_of_base2_numbers)
union = set_of_base10_numbers.union(set_of_base2_numbers)
'''
intersection: {0, 1}:
if the number is contained in both sets it becomes part of the intersection
union: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}:
if the number exists in at lease one of the sets it becomes part of the union
'''Copyright © 2021 Codeinu
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