找到元组匹配的INDEX为1对1和2对2等等......但不是2对1或1对2使用python

Question

R=[(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6)]

For the above set i need to get the INDEX where the first value should match to the ONLY to the first value of the set and second should match to the ONLY the second value which means when i match a[0] to R[tuples] none of the elements in a and R shouldn't be equals to each other other than 1 to 1 or 2 to 2

in simple first should match only to the first and second should match ONLY to the second element in a resulted set. and it cannot match any other element of the result sets

Expected Answer:

a=(2,5) ANS INDEX: (5,7) 
b=(3,6) ANS INDEX: (9,13) 

Here is my code is going too detail and too lengthy and taking too long to run for my actual project. so i gave a sample above pls help me to drive this in a optimistic way.., to achieve the speed.

 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

Thanks in advance.

Answer 1

Question : Find for R[0] and R[1] and the answer would be for 0: 4,5,6,7 for 1: 2,5,6,7

Answer for R[ 0 ] : (3, 4, 5, 6, 7) => R[3] has 1
Answer for R[ 1 ] : (4, 6, 7, 8, 9) => R[2] and R[5] no Match, R[4] has 2 ; R[8] has 19 ; R[9] has 21

Data :

 # 0 1 2 3 4 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), (1,5,15,20), (3,26,19,25), (4,17,18,21)] # 5 6 7 8 9 

result = []
for i, T in enumerate(R):
    r = []
    for Ii, Tt in enumerate(R[i + 1:], i + 1):
        if set(Tt).intersection(set(T)):
            r.append(Ii)

    if len(r):
        result.append(tuple(r))
    else:
        result.append(None)

print('result:{}'.format(result))

Output :

 # 0 1 2 3 4 result:[(3, 4, 5, 6, 7), (4, 6, 7, 8, 9), (3, 5, 8), (7,), None, (9,), (7, 8), None, None, None] # 5 6 7 8 9 

Tested with Python: 3.4.2