高深簡潔的語法我不會，這裡提供一個土法煉鋼的方法 : 用多層迴圈來處理。

多層迴圈在學校教學裡很常見的是寫九九乘法表，這就是個類似的應用，以下是範例:

A=[1,2,3,4,5],[3,4,5,7,8],[10,20,45,50,51]
B=[3,5],[2,3],[3,4],[10,50],[4,5],[7,8]

result=[]
for i in A:
    totla_match_num=0
    for j in B:
        match_len=0
        for k in j:
            if k in i:
                match_len=match_len+1
        if match_len==len(j):
            totla_match_num=totla_match_num+1
    if totla_match_num>3:
        result.append(i)
print(result)

執行結果:

[[1, 2, 3, 4, 5], [3, 4, 5, 7, 8]]

只要你邏輯清晰條件定義明確，用最基本的迴圈加條件判斷就能解決許多問題。

補充一個，如果List內元素不重複可以用的方法

A = [1,2,3,4,5],[3,4,5,7,8],[10,20,45,50,51]
B = [3,5],[2,3],[3,4],[10,50],[4,5],[7,8]

result = []
for i in A:
    count = 0
    for j in B:
        if set(j).issubset(i):
            if (count := count+1) > 3:
                result.append(i)
                break
print(result)

好孩子不要學的拚最短行數訓練。

A = [1,2,3,4,5],[3,4,5,7,8],[10,20,45,50,51]
B = [3,5],[2,3],[3,4],[10,50],[4,5],[7,8]

def countLstIn(allIn, outer):
    return sum(tuple(all([e in outer for e in i]) for i in allIn))

result = [a for a in A if countLstIn(B, a) >= 3]

• 將B裡頭的"子list"全部合起來之後，list轉set去除重複的，。
• 將setB與A裡頭的集合做「交集」，檢查結果的長度，如果>3就記錄下來

# _*_ coding:utf-8 _*_
A=[1,2,3,4,5],[3,4,5,7,8],[10,20,45,50,51]
B=[3,5],[2,3],[3,4],[10,50],[4,5],[7,8]

listB = [element for subB in B for element in subB]
setB = set(listB)
print('setB=', setB)

result = []
for data in A:
    data = set(data)
    tmp = setB.intersection(data)
    if(len(tmp)>3):
        result.append(list(data))

print(result)