Deleting parts of groups that do not have equal dates

Question

Find the intersection of dates within the pair and remove the other dates & corresponding prices. Also keep the list in same format as it currently is in.

Let me explain: Each stock has closing dates and prices. There are many groups of 2 stocks in the list.(all unique pairs) Within each stock pair(group) I want to delete the dates and corresponding prices if the dates does not match both stocks(finding the intersection of dates within the pairs.)

Input: Subsets[Partition[tt, 2, 3], {2}]
Output: {{{{"AAD AT Equity", 
"Date", $Failed, {2013, 8, 1, 0, 0, 0.}, {2013, 8, 2, 0, 0, 
     0.}, {2013, 8, 5, 0, 0, 0.}, {2013, 8, 6, 0, 0, 0.}}, {"", 
    "PX_LAST", 0.82, 1.77, 1.75, 1.72, 1.71}}, {{"ABC AT Equity", 
    "Date", $Failed, {2013, 8, 5, 0, 0, 0.}, {2013, 8, 6, 0, 0, 0.}, 
    "", ""}, {"", "PX_LAST", 0.6808, 3.33, 3.35, "", 
    ""}}}, {{{"AAD AT Equity", 
    "Date", $Failed, {2013, 8, 1, 0, 0, 0.}, {2013, 8, 2, 0, 0, 
     0.}, {2013, 8, 5, 0, 0, 0.}, {2013, 8, 6, 0, 0, 0.}}, {"", 
    "PX_LAST", 0.82, 1.77, 1.75, 1.72, 1.71}}, {{"ABP AT Equity", 
    "Date", $Failed, {2013, 8, 2, 0, 0, 0.}, {2013, 8, 5, 0, 0, 
     0.}, {2013, 8, 6, 0, 0, 0.}, ""}, {"", "PX_LAST", 5.3942, 2.22, 
    2.2, 2.2, ""}}}, {{{"AAD AT Equity", 
    "Date", $Failed, {2013, 8, 1, 0, 0, 0.}, {2013, 8, 2, 0, 0, 
     0.}, {2013, 8, 5, 0, 0, 0.}, {2013, 8, 6, 0, 0, 0.}}, {"", 
    "PX_LAST", 0.82, 1.77, 1.75, 1.72, 1.71}}, {{"ACR AT Equity", 
    "Date", $Failed, {2013, 8, 6, 0, 0, 0.}, "", "", ""}, {"", 
    "PX_LAST", 1., 3.32, "", "", ""}}}, {{{"ABC AT Equity", 
    "Date", $Failed, {2013, 8, 5, 0, 0, 0.}, {2013, 8, 6, 0, 0, 0.}, 
    "", ""}, {"", "PX_LAST", 0.6808, 3.33, 3.35, "", 
    ""}}, {{"ABP AT Equity", 
    "Date", $Failed, {2013, 8, 2, 0, 0, 0.}, {2013, 8, 5, 0, 0, 
     0.}, {2013, 8, 6, 0, 0, 0.}, ""}, {"", "PX_LAST", 5.3942, 2.22, 
    2.2, 2.2, ""}}}, {{{"ABC AT Equity", 
    "Date", $Failed, {2013, 8, 5, 0, 0, 0.}, {2013, 8, 6, 0, 0, 0.}, 
    "", ""}, {"", "PX_LAST", 0.6808, 3.33, 3.35, "", 
    ""}}, {{"ACR AT Equity", "Date", $Failed, {2013, 8, 6, 0, 0, 0.}, 
    "", "", ""}, {"", "PX_LAST", 1., 3.32, "", "", 
    ""}}}, {{{"ABP AT Equity", 
    "Date", $Failed, {2013, 8, 2, 0, 0, 0.}, {2013, 8, 5, 0, 0, 
     0.}, {2013, 8, 6, 0, 0, 0.}, ""}, {"", "PX_LAST", 5.3942, 2.22, 
    2.2, 2.2, ""}}, {{"ACR AT Equity", 
    "Date", $Failed, {2013, 8, 6, 0, 0, 0.}, "", "", ""}, {"", 
"PX_LAST", 1., 3.32, "", "", ""}}}}

For the first pair of stocks I want to end up with this correct result:

{{{"AAD AT Equity", 
"Date", $Failed,, {2013, 8, 5, 0, 0, 0.}, {2013, 8, 6, 0, 0, 0.}}, {"", 
    "PX_LAST", 0.82, 1.72, 1.71}}, {{"ABC AT Equity", 
    "Date", $Failed, {2013, 8, 5, 0, 0, 0.}, {2013, 8, 6, 0, 0, 0.}, 
"", ""}, {"", "PX_LAST", 0.6808, 3.33, 3.35, "", 
""}}}

As you can see, what has happened is that AAD equity have lost it two first trading dates in the list and its corresponding close price, and we end up with a pair of stocks with equal number of corresponding dates, and each stock have same amount of closing prices.

Note: I want this to be done for each pair.

Answer 1

I answered your prior question relating to manipulating this data. I showed there how transforming the data made it easier to work with. I think perhaps you should consider a new format for your data as you apparently have a number of manipulation in mind and you may have to fight the existing format to get them done conveniently.

Also, if your data is large you may more (space) efficiently process it by not constructing all pairs (subsets) in advance, but instead iterating through them. This goes against normal recommendations for Mathematica programming but it can make sense in such cases.

These things notwithstanding I will still show you a way to manipulate this data.

filter[d1_] :=
  Module[{pos, keep},
    keep = DeleteCases[Intersection @@ d1[[All, 1, 2 ;;]], ""];
    pos = Position[#[[1]], Alternatives @@ keep, {1}] &;
    #[[All, Flatten @ {1, pos @ #}]] & /@ d1
  ]

newdat = filter /@ data;

A look at the result:

MatrixForm /@ newdat

---

Here is a different formulation that may be faster on long columns:

filter2[d1_] :=
  Module[{keep},
    Scan[
      keep[#] = True; &,
      Intersection @@ d1[[All, 1]]
    ];
    keep[""] = False;
    keep[Except[_List]] = True;
    Select[#\[Transpose], keep @ #[[1]] &]\[Transpose] & /@ d1
  ]

Answer 2

Based on the chat, here's an answer, sort of. I didn't test for stability, but it seems to work. Please confirm if this is what you want:

I define a function func, that works on each pair, identifying dates in common and then dropping out the other dates and respective prices:

func[object_, preFields_: 3] := 
  Module[{goodDates, posGood, posBad, lengthRecord},
    lengthRecord = Length[First[First@object]];
    goodDates = DeleteCases[
                  Intersection[Sequence @@ object[[All, 1, 1 + preFields ;;]]],""];
    posGood = Function[pair, 
       Flatten[Position[object[[pair, 1, 1 + preFields ;;]], #] & /@ goodDates]] 
       /@ Range[Length@object];
    posBad = Complement[Range[preFields + 1, lengthRecord], #] & /@ (posGood + preFields);
    posBad = Map[List, posBad, {2}];
    Transpose[
       Function[obj, MapIndexed[Delete[#, posBad[[First@#2]]] &, obj]] /@ Transpose[object]]
 ]

I tried to make it easy to read, but wasn't too successful in this regard. Please confirm that it's doing the right thing and I'll add some background.

To use it:

func /@ data

... long output