Message Boards Message Boards

0
|
6562 Views
|
10 Replies
|
2 Total Likes
View groups...
Share
Share this post:

Replicate Dave Giles' permutation test using WL?

Dear community members, I'm trying to replicate Dave Giles' permutation test examples using Mathematica but I'm having difficulties randomly selecting 50,000 samples from all the possible permutations of a list of 20 prices. Mathematica runs out of memory in my laptop (8 GB RAM).

prices = {5.0, 4.8, 4.7, 4.0, 5.3, 4.1, 5.5, 4.7, 3.3, 4.0, 4.0, 4.6, 
   5.3, 3.0, 3.5, 3.9, 4.7, 5.0, 5.2, 4.6};
In[4]:= RandomSample[Permutations[prices], 50000]

During evaluation of In[4]:= General::nomem: The current computation was aborted because there was insufficient memory available to complete the computation.

During evaluation of In[4]:= Throw::sysexc: Uncaught SystemException returned to top level. Can be caught with Catch[\[Ellipsis], _SystemException].

Out[4]= SystemException["MemoryAllocationFailure"]

Is there an alternative way that I could use to generate those samples?
POSTED BY: Ruben Garcia Berasategui
10 Replies
Sort By:
Hans Milton
Hans Milton
Posted 5 years ago

Just a simple comment: Your method is obviously more efficient than the OP's. However, the resulting collection of permutations will probably contain duplicates. While Rubens method will not.
POSTED BY: Hans Milton
Jim Baldwin
Jim Baldwin, Retired
Posted 5 years ago

Point well taken. However, here's my (maybe feeble) defense:

  • The R code used in "Dave Giles' permutation test examples" allows for potential duplicates. The probability that there will be any duplicates (let alone even a few) with 50,000 samples out of 20! elements is extremely small. (Although to have a proper defense I'll need to quantify that.)

  • The number of duplicates in any particular sample using the method I proposed will almost always be zero. One could execute Length[samples] - Length[DeleteDuplicates[Sort[samples]]] to provide the number of duplicates.

  • If one looked at the distribution of the 20! test statistics and the distribution of an infinite number of samples with replacement from those 20! test statistics, those distributions would be identical as each of the 20! test statistics will be represented the same proportion of the time.

There will be no practical difference in the result with or without duplicates for this particular permutation test and dataset. (I am reminded of the Hans Christian Andersen story "The Princess and the Pea".) Or rather the only practical difference will be the ability to perform the test (allowing duplicates) or not (not allowing duplicates).
POSTED BY: Jim Baldwin
Jim Baldwin
Jim Baldwin, Retired
Posted 5 years ago

Just drop the use of Permutations.

prices = {5.0, 4.8, 4.7, 4.0, 5.3, 4.1, 5.5, 4.7, 3.3, 4.0, 4.0, 4.6, 5.3, 3.0, 
  3.5, 3.9, 4.7, 5.0, 5.2, 4.6};
samples = Table[RandomSample[prices, Length[prices]], {i, 50000}];
POSTED BY: Jim Baldwin

How about

Permute[prices, RandomPermutation[Length[prices]]]

This gives only one permutation and no duplicates.

Regards,

Neil
POSTED BY: Neil Singer
Jim Baldwin
Jim Baldwin, Retired
Posted 5 years ago

But it is apparently desired to have 50,000 permutations with none of those arrangements duplicated. With a sample of size 1 there are by definition no duplicates. What am I missing?
POSTED BY: Jim Baldwin

Jim,

Actually, after rereading what you posted more carefully, I agree with you. You can obviously expand the approach I posted above to include 50000 samples:

Map[Permute[prices, #] & , RandomPermutation[Length[prices], 50000]]

but this would have the same duplicates as your approach and would be slower than what you posted.

I like your post best. I suppose you can always guarrantee no duplicates with:

DeleteDuplicates[
 Table[RandomSample[prices, Length[prices]], {i, 50000}]]

and it is still very fast (although duplicates are unlikely).

Regards,

Neil
POSTED BY: Neil Singer
Jim Baldwin
Jim Baldwin, Retired
Posted 5 years ago

Good idea, Neil. If one needed exactly 50,000 samples with no duplicates, just sample somewhat more than 50,000, delete duplicates, and then just use the first 50,000.
POSTED BY: Jim Baldwin

Re that small probability: It appears to be a birthday paradox problem, and if I set it up correctly the probability is around 5*10^(-10).
POSTED BY: Daniel Lichtblau
Jim Baldwin
Jim Baldwin, Retired
Posted 5 years ago

Not that you need my confirmation but, yes, that is a good way to set up the problem:

1 - 50000! Binomial[20!, 50000]/(20!)^50000 // N
(* 5.13779*10^-10 *)

However, if you are seeing lots (or even one duplicate) when running the code, please buy a lotto ticket for me.
POSTED BY: Jim Baldwin

Dear all, thank your for your answers. I followed Jim's initial suggestion and I got a result (0.00184) pretty close to Dave's figures ( 0.0017 & 0.0016). Regards, Ruben
POSTED BY: Ruben Garcia Berasategui
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Tag limit exceeded
Note: Only the first five people you tag will receive an email notification; the other tagged names will appear as links to their profiles.
Reply Preview
Attachments
Remove
or Discard

Group Abstract Group Abstract