Aligning two lists

Question

I have two lists, L1 and L2, each with a key and some data. Let us say the key is a person's name, a string, and the data follows. (To respond to rasher's query:) Let us also assume the lists are sorted by key:

L1 = {
       {"Joseph O'Rourke", data1, data2, ... },
       ...
     }

I would like to "align" the two lists in the following sense. If L1 has a name A that is not in L2, then L2 is padded to include a "blank" record for A. And vice versa: If L2 has a name B that is not in L1, then L1 is modified to include a "blank" record for B. Then I will have two lists that "align":

L1' = { {A,...}, {0,...}, {C,...},  {D,...}, ...}
L2' = { {A,...}, {B,...}, {0,...},  {D,...}, ...}

where maybe 0 == {}. With the lists aligned in this fashion, I could make a two-column table (one column per list) that would directly compare one list against the other. My question is:

What is a clean method for accepting L1 and L2 as input, and returning L1' and L2' as output, with the latter two lists aligned as above?

I can accomplish this via tedious list For-loops, but I suspect the cognoscenti :-) will offer more concise and efficient methods. Thanks for your help!

Answer 1

If the lists are long (several hundred or more), Alternatives will get to be slow. Here's another way that will be faster on longer lists:

align[list1_, list2_] := Module[{base, replace1, replace2},
  base = Union[First /@ list1, First /@ list2];
  With[{arg = First[#]}, replace1[arg] = #] & /@ list1;
  replace1[_] = {0, {}};
  With[{arg = First[#]}, replace2[arg] = #] & /@ list2;
  replace2[_] = {0, {}};
  {replace1 /@ base, replace2 /@ base}

Example:

data = {"that", "natural", "cowards", "delay", "country", "himself", 
   "my", "will", "cast", "office", "native", "is", "awry", "s", "ay"};

l1 = {#, ToCharacterCode[#]} & /@ Sort@RandomSample[data, 10]
l2 = {#, {StringLength[#]}} & /@ Sort@RandomSample[data, 10]
(*
  {{"ay", {97, 121}}, {"cast", {99, 97, 115, 116}},
   {"cowards", {99, 111, 119, 97, 114, 100, 115}}, {"delay", {100, 101, 108, 97, 121}},
   {"himself", {104, 105, 109, 115, 101, 108, 102}}, {"is", {105, 115}},
   {"my", {109, 121}}, {"native", {110, 97, 116, 105, 118, 101}},
   {"that", {116, 104, 97, 116}}, {"will", {119, 105, 108, 108}}}

  {{"cast", {4}}, {"cowards", {7}}, {"delay", {5}}, {"is", {2}}, {"my", {2}},
   {"native", {6}},{"natural", {7}}, {"s", {1}}, {"that", {4}}, {"will", {4}}} 
*)

align[l1, l2];
Grid[Transpose@{newl1, newl2}]

Answer 2

I'm late to the party but I like this kind of problem so I'm going to answer anyway.

I propose this:

f1[a_List, b_List, fill_: {0, {}}] :=
  With[{all = a ⋃ b},
    Replace[
      Join[all, #] ~GatherBy~ First,
      {{_} -> fill, {__, x_} :> x},
      1
    ] & /@ {a, b}
  ]

Test:

a = {{1, 7}, {3, 7}, {5, 2}, {8, 7}};
b = {{3, 1}, {6, 6}, {8, 7}, {9, 3}};

f1[a, b] // Grid

$ \begin{array}{cccccc} \{1,7\} & \{3,7\} & \{5,2\} & \{0,\{\}\} & \{8,7\} & \{0,\{\}\} \\ \{0,\{\}\} & \{3,1\} & \{0,\{\}\} & \{6,6\} & \{8,7\} & \{9,3\} \end{array} $

Note that this sample includes keys with both identical ({8, 7}) and divergent ({3, 7}, {3, 1}) data.

Answer 3

TemporalData + ResamplingMethod

ClearAll[timeAlign]
timeAlign = Module[{td = TemporalData[{##}, ResamplingMethod -> {"Constant", {}}], times},
  times = Union @@ td["TimeList"];
  Thread[{times, #}] & /@ Through @ td["PathFunctions"] @ times]&;

Example: using Mr.Wizard's example lists

a = {{1, 7}, {3, 7}, {5, 2}, {8, 7}};
b = {{3, 1}, {6, 6}, {8, 7}, {9, 3}};
timeAlign[a, b]

{{{1, 7}, {3, 7}, {5, 2}, {6, {}}, {8, 7}, {9, {}}},
{{1, {}}, {3, 1}, {5, {}}, {6, 6}, {8, 7}, {9, 3}}}

Association + KeyUnion

KeyValueMap[List]/@KeyUnion[Association/@(Rule @@@ # & /@ {a, b})]/. _Missing -> {}

{{{1, 7}, {3, 7}, {5, 2}, {8, 7}, {6, {}}, {9, {}}},
{{1, {}}, {3, 1}, {5, {}}, {8,7}, {6, 6}, {9, 3}}}