# Select rows from table by keys

I am looking for a way to iteratively select all sublists with the same ID (my 1st column, residual columns are AbsoluteTime entries). First, I obtained the list of unique IDs with Union. Then I want to use this list for the selection task. I assume that it must work with Sequence and ReplaceAll, but could not make it work. Here's what I have so far:

ticketList = Union[dataList[[1 ;;, 1]]];
tempList = Select[dataList, #1[] == x &] /. x ->  Sequence[ticketList];


My ultimate goal is to build up a TemporalData array with all items which each belong to the same ID.

My data is has dimensions n x m. The ticket list has dimension t. The resulting table should have dimensions t x i x m, where i is the cardinality of items with the same ID, and the i's sum to n. I assumed, that I can then map TemporalData on every t-th item in order to build the temporal list.

I searched the site but didn't find a previous question that seem to address my problem. Maybe I misunderstood the use cases of similiar questions.

Here's an attempt to address your question based on the sort of data I imagine you have in mind. First, some data. Note: the integers are understood to be ID's.

times = RandomReal[{0, 100}, {20, 5}];
ids = RandomInteger[4, 20];
data = Transpose[Prepend[Transpose[times], ids]]


{{1, 95.2049, 98.1609, 35.1946, 32.636, 81.2383}, {3, 92.8072, 36.0249, 68.3793, 54.1311, 95.1051}, {3, 4.26078, 47.9919, 31.627, 53.0064, 96.8995}, {4, 18.2792, 94.4752, 5.66575, 5.62435, 35.8945}, {4, 67.4149, 68.1577, 25.281, 17.8322, 15.3123}, {2, 77.6343, 35.3486, 81.3541, 73.6091, 51.4884}, {0, 28.5346, 29.2705, 78.9446, 32.2112, 7.05296}, {4, 91.9405, 79.4176, 75.6947, 72.1069, 5.94247}, {2, 11.0662, 24.4381, 79.5328, 20.2195, 79.4703}, {0, 61.6779, 73.8532, 61.5861, 7.79641, 38.3836}, {4, 52.3064, 95.9954, 62.3701, 34.7654, 99.8931}, {2, 92.6636, 57.8744, 23.4167, 76.3724, 44.8631}, {2, 73.7149, 84.6716, 36.1234, 89.5432, 39.8183}, {3, 50.4444, 7.98911, 69.4092, 96.3191, 79.6566}, {3, 41.2261, 58.5009, 57.7989, 58.9187, 8.87187}, {4, 53.5552, 11.3278, 8.24034, 67.3065, 19.4648}, {2, 19.9501, 30.5649, 73.9395, 13.3722, 36.5546}, {3, 96.5631, 19.1948, 23.226, 92.8776, 64.4002}, {0, 65.9084, 10.2902, 8.83754, 93.0831, 36.464}, {1, 18.2795, 12.4673, 42.0253, 39.8816, 50.8252}}

Now to gather the records that have the same id:

GatherBy[data, First]//TableForm • That's exactly what I was aiming for and the absolute elegant solution. I am sorry that I did not find GatherBy in the manual. Sometimes the power of MM is really amazing / overwhelming. – Frank Mar 10 '13 at 19:57
• Thanks. Yes, Mathematica knows many useful ways to manipulate lists. BTW, if you wanted the id's to appear in standard order, you might do a SortBy[data,First] before GatherBy. – DavidC Mar 10 '13 at 20:37
• Actually, Sort[data] is sufficient if it's inside the GatherBy. – DavidC Mar 10 '13 at 22:18

Using @David's example data

 times = RandomReal[{0, 100}, {20, 5}];
ids =   RandomInteger[4, 20];
data = Transpose[Prepend[Transpose[times], ids]];


"few" more alternatives (none as elegant as David's answer):

 dt1 = Select[data, (#1[] == x) & /. x -> #] & /@  Union[data[[All, 1]]];
dt2 = Function[{xx}, Select[data, #[] == xx &]] /@ Union@data[[All, 1]];
dt3 = Cases[data, {#, __}] & /@ Union[data[[All, 1]]];
dt4 = DeleteCases[data, Except[{#, __}]] & /@ Union[data[[All, 1]]];
dt5 = Pick[data, data[[All, 1]], #] & /@ Union@data[[All, 1]];
dt6 = Pick[data, data[[All, 1]] - #, 0] & /@ Union@data[[All, 1]];
dt7 = SplitBy[SortBy[data, {First}], First];
dt8 = SortBy[GatherBy[data, First], First];
dt9 = GatherBy[SortBy[data, {First}], First];
dt10 = Extract[data, #] & /@ Position[data, {#, __}] & /@ Union[data[[All, 1]]];
dt11 = data[[#]] & /@ Join @@ Position[data, {#, __}] & /@ Union[data[[All, 1]]];
dt12 = Reap[Sow[#, First[#]] & /@ SortBy[data, {First}]][];
dt1 == dt2 == dt3 == dt4 == dt5 == dt6 == dt7 == dt8 == dt9 == dt10 == dt11 == dt12
(* True *)


Note: If the ordering of the elements in the sublists does not matter, Last@Reap[Sow[#, First[#]] & /@ data] gives the same result as GatherBy[data,First]:

 GatherBy[data,First]==Last@Reap[Sow[#, First[#]] & /@ data]
(* True *)

• Whatever I had you've plainly caught. Get well soon! xD – Mr.Wizard Mar 11 '13 at 0:20
• Also, where the devil is Sow & Reap? – Mr.Wizard Mar 11 '13 at 0:20
• @Mr.Wizard, added Reap[Sow[...]...]:) – kglr Mar 11 '13 at 0:45
• Another one for your list: Reap[{##} ~Sow~ # & @@@ data, Union[First /@ data]][[2, All, 1]] -- no claims of efficiency. – Mr.Wizard Mar 11 '13 at 0:59
• I think in many cases, SortBy[data,{First}] may be simplified to Sort[data]. – DavidC Mar 11 '13 at 1:18