从两个列表中删除两个集合列表的交集，并将其附加到python中的新列表中

我想取集合列表的交点

x1，y2

，将它们从两个列表中删除，然后将其附加到一个新的列表“res”。
这是我的代码：（我认为这是可行的）

我也试过这样做：

但当它是一个集合列表时，会出现错误：

x1=[{'A'},{'B'},{'c'},{'D'}]
y1=[{'A'},{'B'},{'C'},{'d'}]
res=[]
res.append(list(set(x1) & set(x2)))
print(x1,y1)
print(res)

>TypeError: unhashable type: 'set'

有没有更好的方法写这篇文章？
如有任何帮助或建议，将不胜感激

更新：

x1=[{'A'},{'B'},{'c'},{'D'}]
y1=[{'A'},{'B'},{'C'},{'d'}]
x2= {e for i in x1 for e in i}
y2= {e for i in y1 for e in i}
z1 = x2.intersection(y2)
res = [{e} for e in z1]
x1=[{e} for e in x2-z1]
y1=[{e} for e in y2-z1]
print(x1,y1)
>[{'c'}, {'D'}] [{'C'}, {'d'}]
print(res)
>[{'A'}, {'B'}]

但当集合的大小大于1时：

x1=[{'A','B'},{'B'},{'c'},{'D'}]
y1=[{'A'},{'B'},{'C'},{'d'}]
x2= {e for i in x1 for e in i}
y2= {e for i in y1 for e in i}
z1 = x2.intersection(y2)
res = [{e} for e in z1]
x1=[{e} for e in x2-z1]
y1=[{e} for e in y2-z1]
print(x1,y1)
>[{'D'}, {'c'}] [{'C'}, {'d'}]
print(res)
>[{'B'}, {'A'}]

假设输出：

x1=[{'A','B'},{'B'},{'c'},{'D'}]
y1=[{'A'},{'B'},{'C'},{'d'}]
res=[]
for i in x1:
    if i in y1:
        res.append(i)
        x1.remove(i)
        y1.remove(i)
for i in y1:
    if i in x1:
        res.append(i)
        x1.remove(i)
        y1.remove(i)
print(x1,y1)
>[{'B', 'A'}, {'c'}, {'D'}] [{'A'}, {'C'}, {'d'}]
print(res)
[{'B'}]

---

如何修复此问题？

使用

.intersection（）

方法尝试此方法

>>> x1=[{'A'},{'B'},{'c'},{'D'}]
>>> _x1= {e for i in x1 for e in i}     
>>> _x1     
{'c', 'B', 'D', 'A'}
>>> y1=[{'A'},{'B'},{'C'},{'d'}]        
>>> _y1= {e for i in y1 for e in i} 
>>> _y1 
{'d', 'B', 'A', 'C'}
>>> z1 = _x1.intersection(_y1) # intersection   

---

输出：

>>> [{e} for e in z1]   
[{'B'}, {'A'}]

>>> print(x1,y1)
[{'c'}, {'D'}] [{'C'}, {'d'}]

说明：

>>> [{e} for e in z1]   
[{'B'}, {'A'}]

>>> print(x1,y1)
[{'c'}, {'D'}] [{'C'}, {'d'}]

• 将集合列表转换为集合并将其存储到临时变量中，并使用由提供的

  .intersection（）

  内置函数

  set（）

• 最后，转换为所需格式（即集合列表）

---

使用

.intersection（）

方法尝试此操作

>>> x1=[{'A'},{'B'},{'c'},{'D'}]
>>> _x1= {e for i in x1 for e in i}     
>>> _x1     
{'c', 'B', 'D', 'A'}
>>> y1=[{'A'},{'B'},{'C'},{'d'}]        
>>> _y1= {e for i in y1 for e in i} 
>>> _y1 
{'d', 'B', 'A', 'C'}
>>> z1 = _x1.intersection(_y1) # intersection   

---

输出：

>>> [{e} for e in z1]   
[{'B'}, {'A'}]

>>> print(x1,y1)
[{'c'}, {'D'}] [{'C'}, {'d'}]

说明：

>>> [{e} for e in z1]   
[{'B'}, {'A'}]

>>> print(x1,y1)
[{'c'}, {'D'}] [{'C'}, {'d'}]

• 将集合列表转换为集合并将其存储到临时变量中，并使用由提供的

  .intersection（）

  内置函数

  set（）

• 最后，转换为所需格式（即集合列表）

将这些集合列表转换为冻结集合集合，取交叉点，然后再转换回：

x1=[{'A'},{'B'},{'c'},{'D'}]
y1=[{'A'},{'B'},{'C'},{'d'}]

def list_of_sets_to_set_of_frozensets(l):
    return set(map(frozenset, l))

def set_of_frozensets_to_list_of_sets(s):
    return list(map(set, s))

res = set_of_frozensets_to_list_of_sets(list_of_sets_to_set_of_frozensets(x1) & list_of_sets_to_set_of_frozensets(y1))

print(x1,y1)
print(res)

将这些集合列表转换为冻结集合集合，获取交点，然后再转换回：

x1=[{'A'},{'B'},{'c'},{'D'}]
y1=[{'A'},{'B'},{'C'},{'d'}]

def list_of_sets_to_set_of_frozensets(l):
    return set(map(frozenset, l))

def set_of_frozensets_to_list_of_sets(s):
    return list(map(set, s))

res = set_of_frozensets_to_list_of_sets(list_of_sets_to_set_of_frozensets(x1) & list_of_sets_to_set_of_frozensets(y1))

print(x1,y1)
print(res)

---

我终于明白了，谢谢！我认为使用这种方法似乎还有一些缺陷，请检查update@Manx我已经在

\ux1

，

\uy1

中存储了，我想你没有注意到前面的编辑。我终于找到了，谢谢！我认为使用这种方法似乎还有一些缺陷，请检查update@Manx我已经在

\ux1

，

\uy1

中存储了，我想你没有注意到前面的编辑。