比较元组并找到它在python中的相同位置的下一个索引

Question

R=[(1,10,14,34),(2,5,19,21),(3,7,31,32),(1,9,12,31),(2,10,11‌​,22),(4,8,14,32),(13‌​,15,19,34),(1,5,15,2‌​0),(3,26,19,25),(4,1‌​7,19,21),(4,1‌​7,20,21)]

对于R中的每个元组，找到具有相同元素位置的每个元素的下一个索引，其中两个元组中不应存在其他元素。

FOR索引0和1这里是预期的样本结果

0: 4,5,6,7  
1: 2,5,6,8
.
.
.
.
.
so on

这是我的代码太详细，太冗长，需要太长时间才能运行我的实际项目。所以我给了一个上面的样本请帮助我以乐观的方式驾驶它......，以达到速度。

for j in range(i+1,i+10):
        b=set(Results[j])
        if (len(a&b)==0):

            for k in range(i+10, i+200):
                c=set(Results[k])
                if ( (len(a&c)==0) and (len(b&c)==0) ):

                    for l in range(i+200, i+600):
                        d=set(Results[l])
                        if ( (len(a&d)==0) and (len(b&d)==0)  and (len(c&d)==0) ):

                            for m in range(i+500, i+1000):
                                e=set(Results[m])
                                if ( (len(a&e)==0) and (len(b&e)==0)  and (len(c&e)==0)   and (len(d&e)==0) ):

                                    for n in range(i+1000, i+2000):
                                        f=set(Results[n])
                                        if ( (len(a&f)==0) and (len(b&f)==0)  and (len(c&f)==0)   and (len(d&f)==0)   and (len(e&f)==0) ):

                                            for o in range(i+2000, i+3000):
                                                g=set(Results[o])
                                                if ( (len(a&g)==0) and (len(b&g)==0)  and (len(c&g)==0)   and (len(d&g)==0)   and (len(e&g)==0)  and (len(f&g)==0) ):              

                                                        for p in range(i+3000, XRUN):
                                                            h=set(Results[p])
                                                            if ( (len(a&h)==0) and (len(b&h)==0)  and (len(c&h)==0)   and (len(d&h)==0)   and (len(e&h)==0)  and (len(f&h)==0)  and (len(g&h)==0)):              

                                                                CN=CN+1

提前致谢。