0
$\begingroup$

I have a association list, S, whose size is about 5*10^4. I also have a function F which takes in ordered pairs of elements from S, and spits out ordered pairs of integers. There is some amount of repetition among these outputs of F.

I'd like to compute the exact cardinality of the image of F, if that's possible.

Edited to add: By modifying the function F, we might as well assume that S is just Range[1,2.5*10^9], and that F outputs a single 20 digit integer for each input. Computing F[S] is highly memory intensive. I was thinking that perhaps one way to handle this would be to print the outputs, according to their modulus, mod 1000, to different files, and then compute the number of distinct outputs in each file, separately. Are there fast ways to do that?

Improve this question
$\endgroup$
3
  • 5
    $\begingroup$ can you post F and a small part of S ? $\endgroup$
    – kglr
    Commented Oct 5, 2019 at 1:18
  • 3
    $\begingroup$ CountDistinctBy[F][S]? $\endgroup$
    – kglr
    Commented Oct 5, 2019 at 6:20
  • $\begingroup$ One can take S to be Range[1,5*10^4]. For the purposes of my computation, we can take F[i_,j_]:=i*j. $\endgroup$
    – Pace Nielsen
    Commented Oct 7, 2019 at 14:36

1 Answer 1

Reset to default
1
$\begingroup$

You can use CountDistinctBy:

CountDistinctBy[F][S]
Improve this answer
$\endgroup$
2
  • $\begingroup$ My function isn't defined on S, it is defined on ordered pairs from S. By modifying F a little, you can just define it on S=Range[1,2.5*10^9], but then creating the set S takes 20 GB of RAM, and running the function you gave definitely goes over my memory limits. $\endgroup$
    – Pace Nielsen
    Commented Oct 7, 2019 at 20:16
  • $\begingroup$ @PaceNielsen, maybe (for the example function F) Length[Union @@ ParallelTable[i j, {i, 10000}, {j, 1, i}]] or Length[DeleteDuplicates[Join @@ ParallelTable[i j, {i, 10000}, {j, 1, i}]]]? $\endgroup$
    – kglr
    Commented Oct 8, 2019 at 19:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.