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set set is an unordered and non-repeating set of elements
1.set creation
2 ways:
se = {11,22,33}
se = set([11,22,33]) #Call the __init__ method of set to create
2.Common methods of set
1.add
se = {11,22,33}se.add(44)
print(se) => {33,11,44,22} #Because it is unordered, the execution result will be different, but 44 is indeed added to the original set set
2.remove
se = {11,22,33 }
se.remove(11)
print (se) => {22,33}
se.remove(44) #Error report, prompting that the specified element cannot be found
3.discard
se = {11 ,22,33}se.discard(11)
print (se) => {22,33}se.discard(44)
print (se) => {11,22,33} #Cannot find the specified element, no deletion, no error
4.pop
se = {11,22,33}
se.pop()print (se) => {11,22} # Randomly pop an element from the stack , the execution results may be different ret = se.pop()print (ret) => {33} #Print the result of popping out the stack
5.difference
se1 = {11, 22, 33, 44}
se2 = {22, 33, 44, 55}
print(se1.difference(se2)) = > 11 # Print elements that exist in se1 but do not exist in se2 print(se2.difference(se1)) = > ; 55 #Print elements that exist in se2 but do not exist in se1
6.difference_update
se1 = {11,22,33,44}
se2 = {22,33,44,55}
se1.difference_update (se2)print (se1) => 11 #Overwrite the elements that exist in se1 and do not exist in se2 to se1, and update the set collection
7.intersection
se1 = {11,22,33,44 }
se2 = {22,33,44,55}
print (se1.intersection(se2)) => {22,33,44} #Se1, se2 intersection
8.intersection_update
se1 = { 11,22,33,44}
se2 = {22,33,44,55}
se1.intersection_update(se2)
print (se1) => {33,44,22} #Overwrite the intersection of se1 and se2 Write to the set of se1
9.union
se1 = {11,22,33,44}
se2 = {22,33,44,55}
print (se1.union(se2)) => {11,22,33,44,55} #se1, the union of se2