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Now I am using MATLAB, and found one of its function named crossvalind, which provides lots of methods for cross validation, including some that I wasn't able to find on this site.

[TRAIN,TEST] = crossvalind('LeaveMOut',N,M), where M is an integer,
returns logical index vectors for cross-validation of N observations by
randomly selecting M of the observations to hold out for the evaluation
set. M defaults to 1 when omitted. Using LeaveMOut cross-validation
within a loop does not guarantee disjointed evaluation sets. Use K-fold
instead.

It seems that this kind of cross-validation can be used in a loop, but it cannot guarantee disjointed evaluation. I think that this kind of cross-validation has the advantage that its partition could be updated in every loop, rather than being fixed before the loop as in k-fold cross-validation. This means that more 'dynamics' could be tested by this kind of cross validation. But I am not sure about this leave-m-out cross-validation: Could anyone states the difference between leave-m-out cross validation and k-fold validation, what are their advantages or disadvantages, and when should we choose one rather than the other?


To @a.desantos

In k-fold cross-validation, from the set of loops one sample can be in the test set for only one time, but in leave-M-out cross-validation one sample could appear in more than one test sets.

Edit 1

Some new issues inspired by @a.desantos. What m and n should I consider? Considering (by convention) that 10-fold cross-validation is nice, I should choose m as 4 when n would be 40; the combination of C(40,4) is 91,390. Should I generate all 91,390 different combinations, then run 91,390 tests, and average the performance of the specific learning algorithm?

To @Frank Harrell

Edit 2

you mean the train-test routine will be runned for 10 times since each time the data is split into ten disjoint parts in a 10 fold cross validation, this will generate one average test error E1, and this error is not reliable, then i need to run this 10-loop-test-routine for about 5-10 times and average all E1s to get another test error named E2, this E2 stands for the ultimate average error, and is more reliable than E1, right?

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  • $\begingroup$ Welcome to this site! I have revised your post, to make it more comfortable to read as any other threads on this site. Please check that I didn't introduce any typo or error. $\endgroup$ – chl Sep 25 '13 at 12:41
  • $\begingroup$ Answering your question: Obviously not, you cannot perform 91390. It is not feasible and I don't think it could be interesting. Instead, you use a representative proportion. In my opinion, I think that 100 would be more than enough. Why 100? Because I consider it highly enough compared with the $k$ in $k$-CV. $\endgroup$ – a.desantos Sep 26 '13 at 9:20
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The X-validation method LeaveMOut it is a common strategy. In fact, when modelling a specific classifier, LeaveMOut allows you to create training and testing data easily. As this procedure is repeated several times randomly, you average the performance.

However, LeaveMOut is a kind of k-fold cross validation where (k-1) folds are used for training and the remaining fold is used for testing.

There is no such a big difference. Maybe the only difference is that LeaveMOut does not allow validation data, as it only leaves M samples out of the training data. When it said that using LMO on a loop does not guarantee disjointed evaluation sets, it means that samples may be in both places every loop and this maybe conflictive for some applications (not in general, in my opinion).

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  • $\begingroup$ hi, i have add one additional question, it's related to how to choose m given n, and how to run leavemout as kind of cross validation method, details are in edit 1 section which is now part of the original question:) $\endgroup$ – crazyminer Sep 25 '13 at 11:46
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    $\begingroup$ The question is how many time to run each method. 10-fold cv should be repeated 50-100 times to obtain adequate precision of estimates. The more random method may need to be repeated more than $10 \times 100$ times. $\endgroup$ – Frank Harrell Sep 25 '13 at 11:53
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    $\begingroup$ P.S. The need for a very high number of iterations to achieve satisfactory precision in the measure of predictive accuracy makes the bootstrap even more attractive. It can achieve the same precision with fewer iterations, and it properly penalizes for variable selection (if any selection was done) assuming the analyst does the right thing and restarts the feature selection fresh at each iteration for any of these approaches. $\endgroup$ – Frank Harrell Sep 25 '13 at 13:05
  • $\begingroup$ @FrankHarrell i think 10fold only runs for ten times, since the data is divided into 10 disjoint parts, each part act as test set for one time, so the total time is 10... $\endgroup$ – crazyminer Sep 25 '13 at 15:39
  • $\begingroup$ That is inadequate. Your results will vary too much if you split again into tenths. You need to repeat 50-100 times averaging the 10-fold cv results. Or better use the bootstrap for a total of perhaps 400 model fits and evaluations. $\endgroup$ – Frank Harrell Sep 25 '13 at 21:20

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