扩展项目,然后在python的嵌套列表中删除项目

时间:2020-09-27 19:53:56

标签: python list nested extend remove

我正在尝试在python中对嵌套列表进行分类,我的列表很大,由于项目索引错误,我还没有成功。我的目标是,如果列表中的两个项目具有相同的成员,则用item2扩展item1并删除item2。我在python上没有足够的经验。希望你能帮忙

My pseudo code

L = [[0, 1], [2, 3], [4, 5, 13], [6, 7], [2, 8],[3, 10, 11], [12, 13]]

for i in range(len(L)-1):
    for j in range(i+1,len(L)):
        if i!=j and set(L[i]) & set(L[j]) != set():
            L[i].extend(L[j])
            L.remove(L[j])

expected L = [[0,1], [6, 7], [2, 3, 2, 8, 3, 10, 11], [4, 5, 13, 12, 13]]

1 个答案:

答案 0 :(得分:2)

L = [[0, 1], [2, 3], [4, 5, 13], [6, 7], [2, 8],[3, 10, 11], [12, 13]]

out = []
while L:
    current = L.pop(0)
    out.append(current)
    tmp = []    
    for v in L:
        if set(v).intersection(current):
            current.extend(v)
        else:
            tmp.append(v)
    L = tmp

print(out)

打印:

[[0, 1], [2, 3, 2, 8, 3, 10, 11], [4, 5, 13, 12, 13], [6, 7]]

编辑:版本2:

L = [[0, 1], [2, 3], [4, 5], [6, 7], [8, 9],[10, 11], [1,3,5,7,9,11]] 
    
out = []
while L:
    current = L[0]
    while True:
        tmp = []
        for i, v in enumerate(L[1:], 1):
            if set(v).intersection(current):
                current.extend(L.pop(i))
                break
            else:
                tmp.append(v)
        else:
            break
    out.append(current)
    L = tmp

print(out)

打印:

[[0, 1, 1, 3, 5, 7, 9, 11, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]]

对于L = [[0, 1], [2, 3], [4, 5, 13], [6, 7], [2, 8],[3, 10, 11], [12, 13]]打印:

[[0, 1], [2, 3, 2, 8, 3, 10, 11], [4, 5, 13, 12, 13], [6, 7]]