Hi Christian,
here's a clumsy solution to get you started. It's not optimized for speed, so if your lists are long and you have to do this multiple times, you may have to play around with other approaches.
First I'll change raceFriends to gather up all rows that have the same first column:
resorted = Transpose /@ GatherBy[raceFriends, #[[1]] &]
From each of the sublists, I'll use the date, join all the names together, and take the total for each of the other original columns:
newRaceFriends = {#[[1, 1]], StringJoin[Riffle[#[[2]], ","]],Total[#[[3]]], Total[#[[4]]], Total[#[[5]]]} & /@ resorted
This isn't the only way to do this, but the result now looks like this:
{{"6/5/2014", "Bob", 0, 1, 1}, {"6/25/2014","Jill,Hannah,Robert,Peter", 1, 4, -4}, {"7/11/2014", "Nancy,Fred",1, 2, -2}, {"8/16/2014", "John", 0, 1, 1}, {"11/2/2014","Bob,Jill,Hannah,Jack,Peter", 3, 5, -1}, {"11/5/2014","Robert,Greg", 0, 2, 0}}
Now that the first list is in the desired format, I can join it with the second one. If the first column of newRaceFriends and raceTimes are identical, you can just do this:
raceTable=Join[newRaceFriends[[#]], {raceTimes[[#, 2]]}] & /@Range[Length[newRaceFriends]]
However, if you need to look up the dates explicitly, you could do something like this:
raceTable =Join[#, {raceTimes[[Position[raceTimes, #[[1]]][[1, 1]], 2]]}] & /@newRaceFriends
As I said above, this is certainly not the only way to do this and not too elegant either, so play around with the functions I used a bit more.
