python - 根据元组中的值拆分元组列表

Question

我有一个如下所示的列表: 每个元组都有一个名称、起点、终点和方向:

    major_list = [('a',20,30,-1),('b',31,40,-1),('c',41,50,-1),('d',51,60,+1),('z',90,100,-1),('e',61,70,+1),('f',71,80,+1)]

需要像这样分割:

    [[('a',20,30,-1),('b',31,40,-1),('c',41,50,-1)],[('d',51,60,+1)],[('z',90,100,-1)],[('e',61,70,+1),('f',71,80,+1)]]

拆分列表有两种规则: 1) 如果相邻元组之间的 abs(start - stop) > 20，则创建一个新列表 [OR] 2) 如果相邻元组具有相反的方向，例如 ('c',41,50,-1),('d',51,60,+1)，则在 'c' 之后创建一个新列表

这是我到目前为止所拥有的:

    SplitList = []
    for i,tup in enumerate(major_List):
        if i != len(major_List)-1:
           if i == 0:
              tmp_list = []
           next_tup = major_List[i+1]
           if (abs(int(next_tup[1]) - int(tup[2])) > 20) or (next_tup[3] != tup[3]):
              tmp_list.append(tup)
              if tmp_list:                  
                 SplitList.append(tmp_list)
                 tmp_list = []
        else:
           tmp_list.append(tup)

由于某种原因，SplitList 在末尾插入了一个空列表，我无法弄清楚我做错了什么。有没有更Pythonic的方法来做同样的事情？

Answer 1

如果它是列表中的第一个元素，则将该元素添加到列表内的 final 中，然后只需检查 final 列表的最后一个元素中所需的子元素当前:

 major_list = [('a',20,30,-1),('b',31,40,-1),('c',41,50,-1),('d',51,60,+1),('z',90,100,-1),('e',61,70,+1),('f',71,80,+1)]

final = []  

for ele in major_list:
    if not final:
        final.append([ele]) # final is empty so add the first element in a list
    # check the last item of the last sublist added to final and compare to our current element
    elif abs(final[-1][-1][1] - ele[2]) > 20 or final[-1][-1][3] != ele[3]:
    # if it does not meet the requirement, add it to final in a new list
        final.append([ele])
    else:
        # else add it to the last sublist
        final[-1].append(ele)
print(final)

[[('a', 20, 30, -1), ('b', 31, 40, -1), ('c', 41, 50, -1)], [('d', 51, 60, 1)], [('z', 90, 100, -1)], [('e', 61, 70, 1), ('f', 71, 80, 1)]]