7
$\begingroup$

Specifically, the R implementation of t-SNE in the Rtsne package has a "check_duplicates" argument, and the documentation suggests that "it is best to make sure there are no duplicates present and set this option to FALSE, especially for large datasets".

Further, if you attempt to run t-SNE on a dataset in R that does have duplicates, you get the error message: "Error in [command snipped by user]: Remove duplicates before running TSNE."

So, why does this behavior occur? I have a dataset in which multiple samples coincidentally have the same measurements.

Is it simply a "duplicate datapoints, after reduction, will have the same data points anyway, so don't waste processing power"? Does the presence of duplicates affect the process' calculations?

$\endgroup$
2
$\begingroup$

The algorithm is designed to handle datasets without duplicate information, so the package mades a check before apply the technique. They suggest you to remove duplicates and set check_duplicates = FALSE for a performance improvement.

The implementation in R is this:

if (check_duplicates & !is_distance){
if (any(duplicated(X))) { stop("Remove duplicates before running TSNE.")}

Whith default values check_duplicates = TRUE and is_distance = FALSE.

The paper, for who wants to undestand more about the method, is here.

$\endgroup$
  • 7
    $\begingroup$ Your answer basically paraphrases the question. It doesn't answer why you need to remove duplicated datapoints. $\endgroup$ – Daniel Falbel Nov 21 '16 at 20:30
  • $\begingroup$ The first sentence is not correct. The method is not designed to be without time-domain duplicates. The Rtsne package checks the duplicates mostly in the time-domain. - - Also tsne package does not make such a check, only Rtsne. - - To set check_duplicates=FALSE is not because of the performance improvement. It is not the main argument. - - There are several experimental features in many implementatons, such as the check-up for the distance matrix, which - yes - is turned off default. Again, you are not answering the question. - - Your reference is not relevant to the question. $\endgroup$ – Léo Léopold Hertz 준영 Dec 29 '16 at 8:25
2
$\begingroup$

t-SNE method does not require the removal of duplicates. The fact that it is a default feature in Rtsne does not imply its requirement. It is useful for some short-term event monitoring. For characterising long-term trends and/or patterns with big data sets, I see little use. The Rtsne default setup can be more inclined for characterising events in the time-domain, without any studies in Fourier domain.

Assume you have points in the time-domain. The duplicate algorithm causes significant amount of false positives because the duplicate checking is mostly designed on the time-domain signal. Fourier space can show that those events which are considered by the algorithm duplicate are not necessary so.

So my observation is that the algorithm is greedy about duplicate points in the time-domain, which is not useful for me when considering long-term signals, long-terms trends and long-term patterns. The fact that the point is duplicate in the time-domain does not actually mean that it is duplicate also in Fourier domain. I think it will be more a coincidence if is a duplicate in a time domain in the real-life applications. So turning off the feature, should be ok. To estimate how much of the points are really duplicates in both domains is specific on the case study. I get significantly better descriptors of events and/or phenomena by considering long-term data sets without the duplicate check in many real-life applications.

I think the Rtsne documentation is not clear about the case in saying [turn off check_duplicates and] don't wast processing power. There are really other reasons as described above why the check_duplicates can be turned off as realised also by some other implementations of the method. The check_duplicates=TRUE is a personal selection of the Rtsne developer by default at the moment. I would love to hear if there is any implementation reasons for the decision.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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