this is quite complex, so please allow me some space to make the problem clear.
I have several groups which has 3 types T1,T2 and T1*T2.
The T1 (and T2) values are always in doubles ( aa, cc, gg , tt , dd ,ii ).
The T1*T2 values can be either equal to T1 , or equal to T2 or equal to a mixture of T1 and T2.
As an example if within a group, T1 is aa , T2 is tt, then T1*T2 can be aa, tt or at .
Since T1*T2 values are derivatives of T1 and/or T2 values , the values of T1 or T2 if missing, can be imputed from a value of T1*T2 and T1 (or T2).
So if value of T1 is aa, T2 is missing and one of the T1*T2 values is gg (different from T1 and a double ), the value of T2 has to be gg.
If a double does not exist in T1*T2, then the missing value is to be determined from a mixed value (like ag, gt, at)
as another example , in the following data, t2 value is missing in Group5, but from a mixed value of gt , T2 can be inferred as tt since T1 is gg.
Similarly, in the following example , where T1 is missing , it can be inferred as aa.
So the idea is, if either one of T1 and T2 is missing for a group, then the missing value is imputed for that group. If both T1 and T2 are missing, but there exists two different doubles like aa and tt in T1*T2 ( or there exists a mixed value of T1*T2 like at ) then T1 and T2 both can be inferred for the group.
If both T1 and T2 are missing and can be inferred from different doubles, the order doesnt matter, meaning if there T1 and T2 are inferred to have values aa and tt, but if it cant be determined which one is which, then any value can be assigned to T1 and the other one can be assigned to T2.
In Group7, since there is at least a mixed value ga, T1 is gg and T2 is aa.
If T1 and T2 both exist for a group, then we let that group be since there is nothing to impute.
Example input
This is how the imputed output looks like
Well, this again is far from elegant, but try
As you can see, the "imputed" values are printed at the end of each group. Pipe it through a sort step if that's not acceptable.
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