Issue 128 rank mahalanobis

[benthestatistician]

Performance improvements for rank Mahalanobis calculations with a within argument to match_on. Another option for how to do the rank Mahalanobis calculation. #128

[benthestatistician]

Not as ready as I thought it was... @nullsatz to look into test failures.

[nullsatz]

Not as ready as I thought it was... @nullsatz to look into test failures.

This one testthat test is failing. I'm pasting the output below.

Failure (test.rank.mahal.R:136:5): compute_pooled_cov_rank_mahalanobis results match ordinary Mahalanobis's
match_on(z ~ ., data = df, method = "pooled_cov") not equivalent to match_on(z ~ ., data = df, method = "mahalanobis").
Mean relative difference: 0.29986

[ FAIL 1 | WARN 0 | SKIP 0 | PASS 2042 ]

[benthestatistician]

That test still needs to be modified in accord with this comment to #128.

[nullsatz]

That test still needs to be modified in accord with this comment to #128.

Sorry @benthestatistician I missed that comment.

I have tried to implement your suggestion as follows and in the test, but I'm still getting the same error message.

    ## nr number of samples
    nr <- 10L
    z <- integer(nr)
    ## two outcomes: 0 (from initialization), and 1 (assigned below randomly)
    z[sample(1:nr, nr / 2L)] <- 1L 
        
    ## Goal: two groups with the same within-group variance and no rank ties
    df <- data.frame(z = z, X = integer(nr))
    df[df$z == 0, 'X'] <- seq(1, by=2, len = nr / 2)  # odds
    df[df$z == 1, 'X'] <- seq(2, by=2, len = nr / 2)  # evens

[benthestatistician]

See comment here

[benthestatistician]

Looks good to me!