我有一个很大的2D列表，在一些子列表中有重复的内容.我想退还2个或更多的副本，如果不可能的话，请先退还副本.例如,
Df =[[2,3,5,20],[5,10,20,10],[4,13,15,15,17,34,17],[33,34,15,21],[12,16,24,32,12,33,24]]
我想要这样的结果:
Df2 = [[15,17,4,13,34],[12,24,16,32,33]]
我已经try 了下面的代码，但我希望它只返回有2个或更多副本的副本.
res = [t for t in Df if len(t) > len(set(t))]
您可以使用collections模块中的Counter来计算子列表中每个元素的出现次数.然后，过滤掉那些有重复项的子列表
from collections import Counter
Df = [[2, 3, 5, 20], [5, 10, 20, 10], [4, 13, 15, 15, 17, 34, 17], [33, 34, 15, 21], [12, 16, 24, 32, 12, 33, 24]]
# Count occurrences of elements in each sublist
counts = [Counter(sublist) for sublist in Df]
# Find sublists with duplicates and at least 2 duplicates
duplicates = [[elem for elem, count in sorted(count.items(), key=lambda x: x[1], reverse=True)]
for count in counts if sum(val > 1 for val in count.values()) >= 2]
print(duplicates)
-你在做什么？
[[15, 17, 4, 13, 34], [12, 24, 16, 32, 33]]