To create an empty collection, use s=set()

d={}, default dictionary s=set() is an empty collection

The elements in a collection are out of order and cannot be obtained from a corner label

#Collection creation and traversal
s={'Lin Qingxia','High Circle','Zhang Manyu'}
for ele in s:
    print(ele)
#Add to Collection
s = {'Lin Qingxia','High Circle','Chinese Dream'}
# s.add('zhang')
# print(s)
'''update# Traverse elements from a container to add to the current collection
         # Multiple containers''can be passed in update
s.update('efgh','xczbc',(1,2,3),[4,5])
print(s)
#Output {1, 2, 3,'Liang Wancong','Chinese Dream', 4, 5, 6,'Lin Qingxia','Chen Qianhui','Gaoyuan'}
'''Delete Collection'''
# s.remove(5)         #If there is a direct deletion if there is no program error eg:s.remove(6)
# s.clear()           #empty
# s.pop()             #Randomly deletes an element in the set and has a return value. The set is empty
s.discard(6)          #If there is a direct deletion, if there is no program no error
print(s)

Set, Subset, Superset, Union, Intersection, Difference

s1 = {1,2,3,4,5}
s2 = {3,4,5,6,7}
# s=s1.issubset(s2) is a subset, returns bool value False
# s=s1.issuperset(s2) is a superset, returns bool value False
# s=s1.union(s2) Union
# s=s1.intersection(s2) intersection
#s1.intersection_update(s2)        #Intersection, but no return value, intersection updated to s1
# s=s1.difference(s2) difference set, the union of s1-s1 and S2
# s1.difference_update(s2) difference set, but no return value, update s1
# s=s1.symmetric_difference(s2) XOR, union-intersection of S1 and S2
s1.symmetric_difference_update(s2)      # XOR, update s1
print(s1)

The role of sets

# The main purpose of a collection is to de-duplicate a list or element
# l = ['Zhang San','Li Si','Wang Wu','Zhang San']
t = ('Zhang San','Li Si','King Five','Zhang San')
s = set(t)
print(s)

str='00Da'
a=str.isalpha()
print(a)
a=str.isdecimal()
print(a)
a=str.islower()
print(a)
a=str.isupper()
print(a)
a=str.startswith('00D')
print(a)
a=str.endswith('Da')
print(a)
'''False False False False True True'''

String Operation - Find and Replace

str = 'hellohl'
#Find left-to-right ends as soon as it is found
#a=str.find('l') 
#You can specify where to start and end looking in this range
a=str.find('l',4,len(str))     #Values to the right and left of an interval can be written very large
print(a)      #6
a=str.rfind('l',2,5)
print(a)        #3
# Error occurs when there is no difference between index and find
a=str.index('e')
print(a)   #1
a=str.replace('l','hah',4)       #4 represents the maximum number of substitutions
print(a)     #hehahhahohhah, change 4 to 2 and output hehahhahohl

Split and connect

str = '''hello'''
a=str.partition('e')   #('h', 'e', 'llo')
print(a)
a=str.split('l',1)
print(a)      #['he', 'lo']
a=str.splitlines()
print(a)
# Pass Container Parameter Return String
# Segmentation of string elements
a=str.join('tkjght')
print(a)   #thellokhellojhelloghellohhellot
s = str.join(['1','2','3'])
print(s)   #1hello2hello3

toggle case

str = '''Dell2'''
a=str.lower()
print(a)    #qell2
a=str.upper()
print(a)     #QELL2

#align text
str = 'hello'
# Default space to current length
s = str.ljust(10,'-')
print(s)                #hello-----
s = str.rjust(10,'-')
print(s)   #-----hello
s = str.center(10,'-')
print(s)   #--hello---

Remove whitespace from left and right of string

str = '---hello---'
s = str.strip('-')
print(s)    #hello

Dictionaries

d = {'china':'China','england':'Britain','france':'France'}
a=d['china']
print(a)   #China
a=d.get('france')
print(a)   #France
# View objects can be converted to lists
# View objects cannot be manipulated
print(d.keys())   #dict_keys(['china', 'england', 'france'])
print(list(d.values()))  #['China','Britain','France']
# Returns a nested tuple in the list of view objects for an entry
print(d.items())      #dict_items([('china','China', ('england','UK'), ('france','France')))
for a in d:
    print(a)   '''#china
                 england
                 france '''
print(d)      #{'china':'china','england':'Britain','france':'France'}

Dictionary Add, Delete, Modify

Dictionary traversal

d = {'china':'China','england':'Britain','france':'France'}
# View object list nested tuples [(key,value),(key,value)]
 items = d.items()   #items is a list, list elements are tuples, and tuples contain keys and values

 for key,value in items:
     print(key,'  ',value)
    '''China
//Britain
//France''

Section

Format: String [Start Index: End Index: Step]

Slice note:

* 1. The specified interval is left closed right open and contains the start index but not the end index

* 2. Step size defaults to 1 and can be omitted

* 3. From the beginning, index numbers can be omitted and colons cannot be omitted

* 4. By the end, you can add directly after the start of the index:

str = "Welcome to the Republic of China"
# Print out the first to last two (not included) strings
# newStr = str[0:-2:1]
# print(newStr)
# From the fifth to the first (not included)
# newStr = str[-5:-1:1]     #Left Close Right Open
# print(newStr)

# From the fourth to the last (including the last)
print(str[-4:10])

Inverse index when step size is negative

str = "Welcome to the Republic of China"
# Print out the corner labels 2 to 7 (inclusive) in reverse order
# Name Republic of China
newStr = str[2:8:-1]                #Error, because the 8 is at the beginning
# newStr = str[-3:1:-1]
# newStr = str[7:1:-1]
print(newStr)

List tuple set dictionary derivation

# 2. Create a list of all the even numbers from 1 to 100

# l = [ele for ele in range(1,101) if ele%2==0]

Comprehensive Practice

'''
//Requirements:
//Determine whether the date entered by the user is a valid date

//Users can enter "20170229"
//Determine whether it is a valid date, such as "20170229" is not a valid date and "20171345" is not a valid date

//Analysis:
1.Enter Date
2.Judge Length
3.Are they all numbers
4.Is the month satisfied:>=1  <=12
5.Date of each month in normal or leap years
'''
# Number of days per month for the average year
p = [0,31,28,31,30,31,30,31,31,30,31,30,31]
# Number of days per month for leap years
r = [0,31,29,31,30,31,30,31,31,30,31,30,31]

while True:
    # 1. Enter date
    date = input('Please enter a date:')
    # 2. Judging Length
    if len(date) != 8:
        print('Illegal date,Re-enter')
        continue
    # 3. Are they all numbers
    if not date.isdecimal():
        print('Date must be all numbers')
        continue
    # Specific date
    year = int(date[:4])
    month = int(date[4:6])
    day = int(date[6:])
    # 4.Month Satisfaction: >=1 <=12
    if month<1 or month>12:
        print('Month not satisfied')
        continue
    # Judging whether a normal year is a leap year
    monthDay = 0
    if (year%4==0 and year%100!=0) or year%400==0:
        monthDay = r[month]
    else:
        monthDay = p[month]
    # 5. Dates are monthly according to the average or leap year
    if day>monthDay:
        print('Date not satisfied')
        continue
    break
print('Date Satisfaction')

Indexes

Dictionaries are key-value pairs to store complex data

eval becomes the original type