Detailed explanation of python3 set (set)-Python Tutorial-php.cn

Detailed explanation of python3 set (set)

迷茫

Release： 2017-03-25 13:18:45

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add(add elements)

name = set([&#39;Tom&#39;,&#39;Lucy&#39;,&#39;Ben&#39;])
name.add(&#39;Juny&#39;)print(name)#输出：{&#39;Lucy&#39;, &#39;Juny&#39;, &#39;Ben&#39;, &#39;Tom&#39;}

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clear(clear all elements)

name = set([&#39;Tom&#39;,&#39;Lucy&#39;,&#39;Ben&#39;])
name.clear()print(name)#输出：set()

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copy(copy set collection)

name = set([&#39;Tom&#39;,&#39;Lucy&#39;,&#39;Ben&#39;])
new_name = name.copy()print(new_name)#输出：{&#39;Tom&#39;, &#39;Lucy&#39;, &#39;Ben&#39;}

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difference(returns different elements in two or more sets and generates a new set)

A = set([2,3,4,5])
B = set([3,4])
C = set([2])
n = n1.difference(n2,n3)
print(n)
#输出：{5}
#返回A集合里面，在B和C集合中没有的元素，并生成新的集合

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difference_update（delete In the A set, the elements that exist in the B set)

A = set([2,3,4,5])
B = set([4,5])
A.difference_update(B)
print(A)

#输出：{2, 3}

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discard(remove the element)

n = set([2,3,4])
n.discard(3)print(n)#输出：{2, 4}

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intersection(take the intersection and generate a new set)

n1 = set([2,3,4,5])
n2 = set([4,5,6,7])
n = n1.intersection(n2)print(n)#输出：{4, 5}

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intersection_update(Get the intersection and modify the original set)

n1 = set([2,3,4,5])
n2 = set([4,5,6,7])
n1.intersection_update(n2)print(n1)#输出：{4, 5}

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isdisjoint(Judge the intersection, return False if it is, True if not)

n1 = set([2,3,4,5])
n2 = set([4,5,6,7])print(n1.isdisjoint(n2))#输出：False

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issubset(Judge the subset)

A = set([2,3])
B = set([2,3,4,5])print(A.issubset(B))#输出：True#A是B的子集

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issuperset(Judge the parent set)

A = set([2,3])
B = set([2,3,4,5])print(B.issuperset(A))# 输出：True#B是A的父集

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pop(Randomly remove an element)

n = set([2,3,4,5])
n1 = n.pop()print(n,n1)# 输出：{3, 4, 5} 2

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remove(Remove the specified element)

n = set([2,3,4,5])
n.remove(2)print(n)# 输出：{3, 4, 5}

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symmetric_difference(Get the intersection , and generate a new set)

A = set([2,3,4,5])
B = set([4,5,6,7])print(A.symmetric_difference(B))# 输出：{2, 3, 6, 7}

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symmetric_difference_update(take the intersection, change the original set)

A = set([2,3,4,5])
B = set([4,5,6,7])
A.symmetric_difference_update(B)print(A)# 输出：{2, 3, 6, 7}

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union(take the union, and generate a new set)

A = set([2,3,4,5])
B = set([4,5,6,7])print(A.union(B))# 输出：{2, 3, 4, 5, 6, 7}

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update(take the union and change the original set)

A = set([2,3,4,5])
B = set([4,5,6,7])
A.update(B)print(A)# 输出：{2, 3, 4, 5, 6, 7}

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