Question

我想使用单独的分组在R中执行两次样本t检验。 t.test必须是“无偏的”，这意味着对于外部组（下面的组2）中的所有交易，必须对每个内部组（下面的组1）进行T检验，例如：“内部组A”与“内部群组不是A”。下面显示的for循环代码可能比语言解释更清晰...

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？开放使用任何软件包，但当前使用data.table。

就上下文而言，我有约100万行交易数据。第1组表示一个人（如果有多行，他们有多个交易），并且包含约3万个唯一值。第2组指示一个邮政编码，并包含〜500个唯一值

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]

Answer 1

stats::t.test是通用的，并进行了许多检查。您可以只计算所需的量，即t统计量和p值，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：

#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

计时代码：

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：

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

Answer 2

我建议查看软件包Rfast。有一个命令，例如ttest1，ttest2和ttest用于一个样本，两个样本和多个t

R-快速两次样本t检验

2 个答案: