列表重叠查找器

Question

我有两个列表：

a = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89]
b = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]

需要找到重叠的元素

我正在做功课。

if (a[i]==b[i]):
print(i)```
But I got this error:
IndexError: list index out of range,

I expect result:
1 , 2 , 3 , 5 , 8 , 13

Answer 1

您收到此错误是因为一个list比另一个长。要解决此问题，您只需将lists转换为sets以消除重复项，然后使用intersection：

a = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89]
b = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
list(set(a) & set(b))
# >> [1, 2, 3, 5, 8, 13]

Answer 2

a = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89]
b = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
print(set(a).intersection(set(b)))
# output {1, 2, 3, 5, 8, 13}

当列表中出现 1 个以上的元素时

a = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89]
b = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,2]
from collections import defaultdict
dic=defaultdict(int)
dic2=defaultdict(int)
for i in a:
dic[i]+=1
for i in b:
dic2[i]+=1
l=list(set(a) & set(b))
solution = [j for i in l for j in [i]*min(dic[i],dic2[i])]
print(solution)
# output [1, 2, 3, 5, 8, 13]