Is there a faster way to create a matrix of indices from ragged data?

Question

I have data that is given as a list of ordered pairs mixed with scalars. The pairs can contain infinite bounds. My goal is to convert the data into an index used in future computations.

data = {{1, ∞}, {-∞, 2}, 3, {2, 2}, {2, 3}};

This gives me all of the unique values present in data.

udata = Sort[DeleteDuplicates[Flatten@data], Less]

==> {-∞, 1, 2, 3, ∞}

Now I use Dispatch to create replacement rules based on the unique values.

dsptch = Dispatch[Thread[udata -> Range[Length[udata]]]];

Finally I replace the values with their indices and expand scalars a such that they are also pairs {a,a}. This results in a matrix of indices which is what I'm after.

Replace[data /. dsptch, a_Integer :> {a, a}, 1]

==> {{2, 5}, {1, 3}, {4, 4}, {3, 3}, {3, 4}}

NOTES:

1. The number of unique values is generally small compared to the length of data but this doesn't have to be the case.

2. Any real numbers are possible. The data I've shown simply gives a sense of the structural possibilities.

Question: Is there a way to create the final matrix of indices that is much faster than what I'm doing here?

Edit: To test the how potential solutions scale I recommend using the following data. It is fairly representative of a true-to-life case.

inf = {#, ∞} & /@ RandomChoice[Range[1000], 3*10^5];
neginf = {-∞, #} & /@ RandomChoice[Range[1000], 10^5];
int = Sort /@ RandomChoice[Range[1000], {10^5, 2}];
num = RandomChoice[Range[1000], 5*10^5];

testData = RandomSample[Join[inf, neginf, int, num]];

Answer 1

A modest improvement when you replace Replace[...] with Transpose@Thread:

 (udata = Sort[DeleteDuplicates[Flatten@testData], Less]; 
 dsptch = Dispatch[Thread[udata -> Range[Length[udata]]]]; 
 out1 = Replace[testData /. dsptch, a_Integer :> {a, a}, 1];) // AbsoluteTiming 
 (* {2.1282128, Null} *)

 (udata = Sort[DeleteDuplicates[Flatten@testData], Less]; 
 dsptch = Dispatch[Thread[udata -> Range[Length[udata]]]]; 
 out2 = Transpose@Thread[testData /. dsptch];) // AbsoluteTiming 
 (* {1.9421942, Null} *)
 out1==out2
 (* True  *)

Answer 2

I get about a 20% speed up by using kguler's Thread trick to transform the data at the beginning, saving the Transpose until the end. There seems to be a slight advantage to working on data with dimensions {2,10^6} over data with dimensions {10^6,2}. I'm not sure why.

twobiglists=Thread[testData];
udata=Sort[DeleteDuplicates[Flatten@twobiglists],Less];
dsptch=Dispatch[Thread[udata->Range[Length[udata]]]];
result=Transpose[twobiglists/.dsptch];

Answer 3

This is a little faster approach:

first,transform data and udata a little, represent Infinity and -Infinity by "a1" and "a0" :

 data2 = Block [{DirectedInfinity = "a" <> ToString[# + 1] &}, data]
 =>{{1, "a2"}, {"a0", 2}, 3, {2, 2}, {2, 3}}
 udata2 = Block [{DirectedInfinity = "a" <> ToString[# + 1] &}, udata]
 =>{"a0", 1, 2, 3, "a2"}

second, rebuild dispatch table:

dsptch2 = Dispatch[Thread[udata2 -> Range[Length[udata2]]]];

third, Replace and Replace:

Replace[Replace[data2, dsptch2, {-1}], a_Integer :> {a, a}, 1]
==>{{2, 5}, {1, 3}, {4, 4}, {3, 3}, {3, 4}}

the main difference is the inner Replace, make some bigger test data:

l = Join[Range[ 100], {\[Infinity] , -\[Infinity] }];
l2 = Partition [RandomChoice[l, 10^6], 2];
data = Riffle[l2, Join[{\[Infinity] , -\[Infinity] }, Range[ 100]], 5];

now timing the inner Replace part alone:

c1 = data2 /. dsptch2; // Timing (*original approach*)
c2 = Replace[data2, dsptch2, {-1}]; // Timing (*modified approach*)
c1 == c2

=>{0.749, Null}
=>{0.343, Null}
=>True

we see the speed is doubled, now timing the whole:

(udata = Sort[DeleteDuplicates[Flatten@data], Less];
  dsptch = Dispatch[Thread[udata -> Range[Length[udata]]]];
  a1 = Replace[data /. dsptch, a_Integer :> {a, a}, 1];) // Timing 

(data2 = Block [{DirectedInfinity = "a" <> ToString[# + 1] &}, data];
  udata = Sort[DeleteDuplicates[Flatten@data], Less];
  udata2 = 
   Block [{DirectedInfinity = "a" <> ToString[# + 1] &}, udata];
  dsptch2 = Dispatch[Thread[udata2 -> Range[Length[udata2]]]];
  a4 = Replace[Replace[data2, dsptch2, {-1}], a_Integer :> {a, a}, 
    1];) // Timing 
a4 == a1

=>{1.092, Null}
=>{0.889, Null}
=>True

a little faster...

Answer 4

Now incorporating Simon Woods's improvement and including his code in the timings.

Pardon the very late answer but hopefully it is useful to someone with a similar problem.

• Since the depth of the data is known it is faster to use Replace rather than ReplaceAll

• Using an Association and Lookup⁽¹⁾ in place of the Dispatch table is faster still

Test data:

SeedRandom[0]

inf = {#, ∞} & /@ RandomChoice[Range[1000], 3*10^5];
neginf = {-∞, #} & /@ RandomChoice[Range[1000], 10^5];
int = Sort /@ RandomChoice[Range[1000], {10^5, 2}];
num = RandomChoice[Range[1000], 5*10^5];

testData = RandomSample[Join[inf, neginf, int, num]];

Data and time common to each method:

udata = Sort[DeleteDuplicates[Flatten@testData], Less]; // RepeatedTiming

{0.247, Null}

Relative performance:

(* kglr *)    
(dsptch = Dispatch[Thread[udata -> Range[Length[udata]]]];
  r0 = Transpose @ Thread[testData /. dsptch];) // RepeatedTiming

(* Simon Woods *)
(dsptch = Dispatch[Thread[udata -> Range[Length[udata]]]];
  r1 = Transpose[Thread @ testData /. dsptch];) // RepeatedTiming

(* Replace *)
(dsptch = Dispatch@Thread[udata -> Range@Length@udata];
  r2 = Replace[Thread @ testData, dsptch, {2}]\[Transpose];) // RepeatedTiming

(* Association, for v10+ *)
(asc = AssociationThread[udata -> Range@Length@udata];
  r3 = (Lookup[asc, #] & /@ Thread @ testData)\[Transpose];) // RepeatedTiming

{0.673, Null}

{0.573, Null}

{0.446, Null}

{0.390, Null}

r0 === r1 === r2 === r3

True

Answer 5

Using PositionIndex (new in 10.0) and SequenceReplace (new in
11.3)

data = {{1, \[Infinity]}, {-\[Infinity], 2}, 3, {2, 2}, {2, 3}};

udata = Sort[DeleteDuplicates[Flatten @ data], Less];

SequenceReplace[{a_Integer} :> {a, a}][data /. First /@ PositionIndex @ udata]

{{2, 5}, {1, 3}, 4, {3, 3}, {3, 4}}