怎样遍历集合里面的所有元素并且做运算

Question

问题遇到的现象和发生背景

问题相关代码，请勿粘贴截图

运行结果及报错内容

我的解答思路和尝试过的方法

我想要达到的结果

比如有个由同样维度的集合组成的列表
list =[('01', '02', '13', '18', '34'), ('05', '22', '23', '25', '35'), ('01', '05', '07', '17', '20'), ('02', '13', '17', '19', '34'), ('07', '20', '27', '30', '33'), ('02', '18', '24', '25', '34'), ('04', '06', '07', '09', '24'), ('08', '11', '19', '22', '29'), ('04', '10', '11', '23', '26'), ('20', '26', '28', '30', '35'), ('01', '04', '12', '16', '29'), ('01', '02', '08', '12', '26'), ('03', '05', '08', '18', '20'), ('03', '08', '25', '29', '30'), ('03', '20', '25', '30', '34'), ('09', '17', '19', '29', '34'), ('11', '13', '17', '19', '25'), ('02', '08', '16', '29', '34'), ('05', '24', '26', '33', '35'), ('05', '13', '18', '23', '34'), ('09', '11', '12', '15', '31'), ('02', '07', '13', '14', '19'), ('08', '12', '19', '27', '33'), ('04', '07', '17', '29', '31'), ('02', '06', '14', '19', '28'), ('08', '09', '17', '26', '28'), ('04', '21', '28', '29', '33'), ('06', '08', '16', '34', '35'), ('02', '03', '14', '27', '28'), ('02', '16', '17', '18', '34'), ('14', '17', '18', '23', '27'), ('01', '02', '15', '17', '26'), ('01', '11', '12', '34', '35'), ('17', '19', '21', '27', '31'), ('10', '17', '22', '25', '35'), ('08', '23', '26', '27', '33'), ('14', '15', '26', '27', '29'), ('02', '15', '19', '29', '30'), ('05', '10', '21', '25', '31'), ('01', '02', '09', '16', '30'), ('14', '17', '18', '28', '34'), ('11', '18', '21', '22', '33'), ('04', '25', '28', '29', '35'), ('08', '13', '19', '20', '29'), ('08', '09', '13', '29', '30'), ('17', '18', '25', '27', '28'), ('18', '21', '22', '23', '35'), ('07', '08', '15', '19', '28'), ('11', '14', '29', '31', '32'), ('03', '04', '07', '27', '31')]

我想要遍历里面的集合，并且互相做交集，得到只有两个共同元素的子集

for i in list:
for j in list:
s = set(i)&set(j)
if len(s) ==2 and i != j:
print(s)

但是这就会有很多的重复，请问怎样修改就不会有重复 ，谢谢

Answer 1

a=set()
for i in list:
    for j in list:
        s = set(i)&set(j)
        if len(s) ==2 and i != j:
            a.add(tuple(s))
print(a)