Pythonで集合を使ってみます．

• 集合の生成
• 集合内包表記
• 和集合(統合)
• 積集合(共通)
• 差集合(片側差分)
• 対称差集合(両側差分)
• 部分集合か判定
• 部分集合(上位)か判定
• 互いに素か判定(独立)
• まとめ
• 参考サイト

集合の生成

s = {1, 2, 2, 3, 1, 4}

print(s)
print(type(s))
# {1, 2, 3, 4}
# <class 'set'>

# 順序は受け継がれない
s = {1.23, 'abc', (0, 1, 2), 'abc'}

print(s)
# {(0, 1, 2), 1.23, 'abc'}

# リストから集合に変換
l = [1, 2, 2, 3, 1, 4]
s = set(_list)
# 重複削除にも使える
l_unique = list(set(l))

print(s)
print(type(s))
print(l_unique)
# {1, 2, 3, 4}
# <class 'set'>
# [1, 2, 3, 4]

集合内包表記

# リスト
l = [i**2 for i in range(5)]
# 集合
s = {i**2 for i in range(5)}

print(l, len(l))
# [0, 1, 4, 9, 16] 5
print(s, len(s))
# {0, 1, 4, 9, 16} 5

# リスト
l = [i**2 for i in range(5)]
# 集合
s = {i**2 for i in range(5)}

# リストの追加
l.append(3)
print(l, len(l))
# [0, 1, 4, 9, 16, 3] 6

# 集合の追加
s.add(3)
print(s, len(s))
# {0, 1, 3, 4, 9, 16} 6

和集合(統合)

s1 = {0, 1, 2}
s2 = {1, 2, 3}

# 和集合
s_union = s1 | s2
print(s_union)
# {0, 1, 2, 3}

s_union = s1.union(s2)
print(s_union)
# {0, 1, 2, 3}

積集合(共通)

s1 = {0, 1, 2}
s2 = {1, 2, 3}

# 積集合(共通)
s_union = s1 & s2
print(s_union)
# {1, 2}

s_union = s1.intersection(s2)
print(s_union)
# {1, 2}

差集合(片側差分)

s1 = {0, 1, 2}
s2 = {1, 2, 3}

# 差集合(片側差分)
s_union = s1 - s2
print(s_union)
# {0}

s_union = s1.difference(s2)
print(s_union)
# {0}

s1 = {0, 1, 2}
s2 = {1, 2, 3}

# 差集合(片側差分)
s_union = s2 - s1
print(s_union)
# {3}

s_union = s2.difference(s1)
print(s_union)
# {3}

対称差集合(両側差分)

s1 = {0, 1, 2}
s2 = {1, 2, 3}

# 対称差集合(両側差分)
s_union = s1 ^ s2
print(s_union)
# {0, 3}

s_union = s1.symmetric_difference(s2)
print(s_union)
# {0, 3}

部分集合か判定

s1 = {0, 1, 2}
s2 = {1, 2, 3}
s3 = {1, 2}

# 部分集合
s_union = s1 <= s2
print(s_union)
# False
s_union = s1.issubset(s2)
print(s_union)
# False

# 部分集合
s_union = s1 <= s3
print(s_union)
# False
s_union = s1.issubset(s3)
print(s_union)
# False

# 部分集合
s_union = s3 <= s1
print(s_union)
# True
s_union = s3.issubset(s1)
print(s_union)
# True

部分集合(上位)か判定

s1 = {0, 1, 2}
s2 = {1, 2, 3}
s3 = {1, 2}

# 部分集合(上位)
s_union = s1 >= s2
print(s_union)
# False
s_union = s1.issuperset(s2)
print(s_union)
# False

# 部分集合(上位)
s_union = s1 >= s3
print(s_union)
# True
s_union = s1.issuperset(s3)
print(s_union)
# True

# 部分集合(上位)
s_union = s3 >= s1
print(s_union)
# False
s_union = s3.issuperset(s1)
print(s_union)
# False

互いに素か判定(独立)

s1 = {0, 1}
s2 = {1, 2}
s3 = {2, 3}

print(s1.isdisjoint(s2))
# False
print(s1.isdisjoint(s3))
# True

まとめ

Pythonで集合を使ってみました．

参考サイト

Python, set型で集合演算（和集合、積集合や部分集合の判定など） (opens new window)