1.Sets!

(sets are like formal math sets)

#sets DO NOT have duplicate values -

#elements in sets aren't ordered

#you access items in a set by index --- NO ORDER!

2.create and access

{} (not a key: value) / set function

# can't use s[0]

s = set({1,2,3,4,5,5,5,6})

print(s) #{1, 2, 3, 4, 5, 6}

s2 = {1,4,5,5, 'a' 'b', 234.5, 7/5}

# 注意和dic区别是没有key：value pair

#注意上面的例子我在a，b之间忘记加 ，了，成为了一个合并的array

print(s2) #{1, 'ab', 1.4, 4, 5, 234.5}

print(type(s2)) #<class 'set'>

print(4 in s) #True

#还是可以loop through it

for stuff in s2:

print(stuff)

'''

注意顺序是乱的，且是已经完成计算的

1

1.4

4

ab

5

234.5

'''

print(len(s)) #6 -- give you the number of unique items

print(list(s)) # by this way you can convert a duplicate list to a de-dup list

# ---list(set(cities))

3.set methods

add() # add an element to a set, if the element is already in the set, the set doesn't change

s.add(7)

print(s) #{1, 2, 3, 4, 5, 6, 7}

s.add(1)

print(s) #{1, 2, 3, 4, 5, 6, 7}

#s.add(10,9) #TypeError: add() takes exactly one argument (2 given)

remove(): # removes a value from the set-- returns a KeyError if the value is not found

s.remove(7)

print(s) #{1, 2, 3, 4, 5, 6}

#s.remove(100) # KeyError: 100

#use .discard() to avoid KeyErrors

s.discard(6)

print(s) #{1, 2, 3, 4, 5} can also delete

s.discard(8)

print(s) #{1, 2, 3, 4, 5} no error reported

clear() : # remove all the contents of the set

another_s.clear()

print(another_s) #set()

print(type(another_s)) #<class 'set'>

4.copy() creates a copy of the set, a new one!

another_s = s.copy()

print(another_s) #{1, 2, 3, 4, 5}

print(another_s is s) #False

print(another_s == s) #True

5.set Math

#union -- pipe

print(math_students | biology_students) #{'Jane', 'Oliver', 'Matthew', 'Apa', 'Chard', 'Helen', 'Mesut', 'Pat', 'James'}

#intersection -- &

print(math_students & biology_students) #{'Matthew', 'James'}

6.set Comrehension --also return a set

print({ke ** 2 for ke in range(10)})

#{0, 1, 64, 4, 36, 9, 16, 49, 81, 25} check there is no order of it

# if there is : then it will be a dict

print({ke: ke ** 2 for ke in range(10)})

#{0: 0, 1: 1, 2: 4, 3: 9, 4: 16, 5: 25, 6: 36, 7: 49, 8: 64, 9: 81}

print({char.upper() for char in "hellor"})

#{'R', 'L', 'O', 'E', 'H'}

def are_all_vowels_in_string(string):

return len({char for char in string if char in "aeiou"}) ==5

string = "aeiouaaa"

string2 = "sequoia"

string3 = "saqw"

print( are_all_vowels_in_string(string)) #True

print( are_all_vowels_in_string(string2)) #True

print( are_all_vowels_in_string(string3)) #False