问在一个巨大的数据帧中计算pvalue需要很长的时间

EN

library(dplyr)

my.t.test <- function (x, y = NULL) {
  nx <- length(x)
  mx <- mean(x)
  vx <- var(x)
  ny <- length(y)
  my <- mean(y)
  vy <- var(y)
  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 - 0)/stderr
  pval <- 2 * pt(-abs(tstat), df)
  return(pval)
}

cont <- c("A", "B")
set.seed(1)
df1 <- data.frame(id=rep(1:1000, each=8),
                  replicate=1:4,
                  A=rnorm(8000, mean=26, sd=5),
                  B=rnorm(8000, mean=25, sd=7))

completeDF <- function() {
  df1 %>%
  group_by(id) %>%
  summarise(Comparison=paste(cont, collapse=' - '),
            p.value=t.test(get(cont[1]), get(cont[2]))$p.value,
            log10.p.value=-log10(p.value),
            log2.foldchange=mean(get(cont[1]), na.rm=TRUE) - mean(get(cont[2]), na.rm=TRUE)
  )}
noPvalue <- function() {
  df1 %>%
    group_by(id) %>%
    summarise(Comparison=paste(cont, collapse=' - '),
              log2.foldchange=mean(get(cont[1]), na.rm=TRUE) - mean(get(cont[2]), na.rm=TRUE)
    )}
myPvalue <- function() {
  df1 %>%
    group_by(id) %>%
    summarise(Comparison=paste(cont, collapse=' - '),
              p.value=my.t.test(get(cont[1]), get(cont[2])),
              log10.p.value=-log10(p.value),
              log2.foldchange=mean(get(cont[1]), na.rm=TRUE) - mean(get(cont[2]), na.rm=TRUE)
    )}
microbenchmark::microbenchmark(
  completeDF(), noPvalue(), myPvalue()
)

Unit: milliseconds
         expr       min        lq      mean    median        uq      max neval
 completeDF() 358.38330 365.09423 424.60255 369.20453 377.40354 655.2009   100
   noPvalue()  57.42996  58.89978  81.86222  59.66851  60.96582 337.2346   100
   myPvalue() 216.04812 220.98277 318.09568 224.19516 493.74908 609.4516   100