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rstatisticsdata.tablevectorizationt-test

R - fast two sample t test


I would like to perform a two sample t test in R using separate groupings. The t.test must be "unbiased", meaning that for all transactions in the outer group (group 2 below), the T test must be run for each inner group (group 1 below) like: "inner group A" vs. "inner group not A". The for loop code shown below is probably clearer than a verbal explanation...

My current code is below. Does anyone know a faster/better way to do this? Open to using any package, but currently using data.table.

For context, I have ~1 million rows of transaction data. Group 1 indicates a person (if there are multiple rows they have multiple transactions) and contains ~30k unique values. Group 2 indicates a zip code and contains ~500 unique values

Thanks!

library(data.table)

# fake data
grp1 <- c('A','A','A','B','B','C','C','D','D','D','D','E','E','E','F','F')
grp2 <- c(1,1,1,1,1,1,1,  2,2,2,2,2,2,2,  2,2)
vals <- c(10,20,30, 40,15, 25,60, 70,100,200,300, 400,1000,2000, 3000,5000)
DT <- data.table(grp1 = grp1, grp2 = grp2, vals = vals)

# "two sample t.test" --------------------------------------------------

# non vectorized, in-place
# runtime is ~50 mins for real data
for (z in DT[,unique(grp2)]){
  for (c in DT[grp2 == z, unique(grp1)]) {

    res = t.test(
      DT[grp2 == z & grp1 == c, vals],
      DT[grp2 == z & grp1 != c, vals],
      alternative = 'greater'
    )

    DT[grp2 == z & grp1 == c, pval := res$p.value]
    DT[grp2 == z & grp1 == c, tstat := res$statistic]

  }
}

# vectorized, creates new summarized data.table
# runtime is 1-2 mins on real data
vec <- DT[,{
  grp2_vector = vals
  .SD[,.(tstat = t.test(vals, setdiff(grp2_vector, vals), alternative = 'g')$statistic,
         pval = t.test(vals, setdiff(grp2_vector, vals), alternative = 'g')$p.value), by=grp1]
} , by=grp2]


Solution

  • stats::t.test is generalized and does a number of checks. You can just calculate what you need, i.e. t-statistic and p-value and also make use of the optimization in data.table to calculate length, mean and variance. Here is a possible approach:

    #combinations of grp1 and grp2 and those not in grp1 for each grp2
    comb <- unique(DT[, .(grp1, grp2)])[,
        rbindlist(lapply(1:.N, function(n) .(g1=rep(grp1[n], .N-1L), notIn=grp1[-n]))),
        .(g2=grp2)]
    
    #this is optimized, switch on verbose to see the output
    X <- DT[, .(nx=.N, mx=mean(vals), vx=var(vals)), .(grp1, grp2)] #, verbose=TRUE]
    
    #calculate length, mean, var for values not in grp1
    Y <- DT[comb, on=.(grp2=g2, grp1=notIn), allow.cartesian=TRUE][,
        .(ny=.N, my=mean(vals), vy=var(vals)), by=.(grp1=g1, grp2=grp2)]
    
    #calculate outputs based on stats:::t.test.default
    ans <- X[Y, on=.(grp1, grp2)][, c("tstat", "pval") := {
        stderrx <- sqrt(vx/nx)
        stderry <- sqrt(vy/ny)
        stderr <- sqrt(stderrx^2 + stderry^2)
        df <- stderr^4/(stderrx^4/(nx - 1) + stderry^4/(ny - 1))
        tstat <- (mx - my)/stderr
        .(tstat, pt(tstat, df, lower.tail = FALSE))
    }, by=1:Y[,.N]]
    

    output:

       grp1 grp2 nx       mx         vx ny        my           vy       tstat       pval
    1:    C    1  2   42.500     612.50  5   23.0000     145.0000  1.06500150 0.22800432
    2:    B    1  2   27.500     312.50  5   29.0000     355.0000 -0.09950372 0.53511601
    3:    A    1  3   20.000     100.00  4   35.0000     383.3333 -1.31982404 0.87570431
    4:    F    2  2 4000.000 2000000.00  7  581.4286  489747.6190  3.30491342 0.08072148
    5:    E    2  3 1133.333  653333.33  6 1445.0000 4323350.0000 -0.32174451 0.62141500
    6:    D    2  4  167.500   10891.67  5 2280.0000 3292000.0000 -2.59809850 0.97016160
    

    timing code:

    library(data.table) #data.table_1.12.4
    set.seed(0L)
    np <- 4.2e5
    nzc <- 4.2e3
    DT <- data.table(grp1=rep(1:np, each=5), grp2=rep(1:nzc, each=np/nzc*5),
        vals=abs(rnorm(np*5, 5000, 2000)), key=c("grp1", "grp2"))
    
    mtd0 <- function() {
        DT[, {
            grp2_vector <- vals
            .SD[,{
                    tres <- t.test(vals, setdiff(grp2_vector, vals), alternative = 'g')
                    .(tstat=tres$statistic, pval=tres$p.value)
                }, by=grp1]
        } , by=grp2]
    }
    
    mtd1 <- function() {
        comb <- unique(DT[, .(grp1, grp2)])[,
            rbindlist(lapply(1:.N, function(n) .(g1=rep(grp1[n], .N-1L), notIn=grp1[-n]))),
            .(g2=grp2)]
    
        X <- DT[, .(nx=.N, mx=mean(vals), vx=var(vals)), .(grp1, grp2)] #, verbose=TRUE]
    
        Y <- DT[comb, on=.(grp2=g2, grp1=notIn), allow.cartesian=TRUE][,
            .(ny=.N, my=mean(vals), vy=var(vals)), by=.(grp1=g1, grp2=grp2)]
    
        ans <- X[Y, on=.(grp1, grp2)][, c("tstat", "pval") := {
            stderrx <- sqrt(vx/nx)
            stderry <- sqrt(vy/ny)
            stderr <- sqrt(stderrx^2 + stderry^2)
            df <- stderr^4/(stderrx^4/(nx - 1) + stderry^4/(ny - 1))
            tstat <- (mx - my)/stderr
            .(tstat, pt(tstat, df, lower.tail = FALSE))
        }, by=1:Y[,.N]]
    }
    
    microbenchmark::microbenchmark(mtd0(), mtd1(), times=1L)
    

    timings:

    Unit: seconds
       expr      min       lq     mean   median       uq      max neval
     mtd0() 65.76456 65.76456 65.76456 65.76456 65.76456 65.76456     1
     mtd1() 18.29710 18.29710 18.29710 18.29710 18.29710 18.29710     1