R - 快速二样本 t 检验 [英] R - fast two sample t test

问题描述

我想使用单独的分组在 R 中执行两个样本 t 检验.t.test 必须是无偏的"，这意味着对于外部组(下面的第 2 组)中的所有事务，必须为每个内部组(下面的第 1 组)运行 T 测试，例如:内部组 A"与内部组不是 A".下面显示的 for 循环代码可能比口头解释更清楚...

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...

我当前的代码如下.有谁知道更快/更好的方法来做到这一点?可以使用任何包，但目前使用的是 data.table.

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.

就上下文而言，我有大约 100 万行交易数据.第 1 组表示一个人(如果有多个行，他们有多个事务)并包含约 30k 个唯一值.第 2 组表示邮政编码，包含约 500 个唯一值

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

谢谢！

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]

推荐答案

stats::t.test 是通用的，并进行了许多检查.您可以只计算您需要的，即 t-statistic 和 p-value，还可以利用 data.table 中的优化来计算长度、均值和方差.这是一种可能的方法:

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]]

输出:

   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

计时码:

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)

时间安排:

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

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