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在一个巨大的数据帧中计算p值需要很长时间

dplyr performance r

drmariod · 技术社区 · 6 年前

p.values 学生在一个非常大的数据框架内进行t检验,数据格式很长。由于我的原始数据帧在数据帧内有大约行,计算p.values需要很长时间(大约需要100分钟)。

我试图加快这个过程,但我不确定数据帧是否是提高速度的最佳格式,或者我是否应该重塑数据,并可能使用 matrix

这里是一些可重复的例子,有一个小的数据框架和一个基准在最后。

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

因此,由于我的t.test函数非常精简(没有测试等),我已经节省了一些时间。但我想知道是否可以通过矢量化来进一步改进。

2 回复 | 直到 6 年前

Zheyuan Li 6 年前

首先,替换 mean(x) 具有 sum(x) / length(x) mean is slow .

my.t.test ,我发现它80%的执行时间都花在 var 使用Rcpp实现。

library(Rcpp)

cppFunction("double var_cpp (NumericVector x, double xc) {
  size_t n = (size_t)x.size();
  double z1 = 0.0, z2 = 0.0, *p = &x[0], *q = &x[n];
  if (n & 2) {z1 = (*p - xc) * (*p - xc); p++;}
  for (; p < q; p += 2) {
    z1 += (p[0] - xc) * (p[0] - xc);
    z2 += (p[1] - xc) * (p[1] - xc);
    }
  z1 = (z1 + z2) / (double)(n - 1);
  return z1;
  }")

library(microbenchmark)
x <- runif(1e+7)
xc <- sum(x) / length(x)
microbenchmark(var_cpp(x, xc), var(x))
#Unit: milliseconds
#           expr       min        lq      mean    median        uq       max
# var_cpp(x, xc)  20.71985  20.76298  21.00832  20.80576  20.87323  25.85723
#         var(x) 109.61120 109.78513 111.92657 109.89077 114.21301 121.98907

sum 也可以提升。

cppFunction("double sum_cpp (NumericVector x) {
  size_t n = (size_t)x.size();
  double z1 = 0.0, z2 = 0.0, *p = &x[0], *q = &x[n];
  if (n & 2) z1 = *p++;
  for (; p < q; p += 2) {z1 += p[0]; z2 += p[1];}
  z1 = (z1 + z2);
  return z1;
  }")

microbenchmark(sum_cpp(x), sum(x))
#Unit: milliseconds
#       expr      min       lq     mean   median       uq      max neval
# sum_cpp(x) 15.58856 15.63613 15.70195 15.67847 15.69998 18.14852   100
#     sum(x) 30.13504 30.20687 30.23993 30.23877 30.26721 30.40525   100

my.t.test.cpp <- function (x, y = NULL) {
  nx <- length(x)
  mx <- sum_cpp(x) / nx
  vx <- var_cpp(x, mx)
  ny <- length(y)
  my <- sum_cpp(y) / ny
  vy <- var_cpp(y, my)
  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)
  }

打开 Martin Morgan's answer

谢谢马丁把 dplyr

也谢谢马丁的补充 fcpp fcpp公司 我自己(接近他的)。我对不同大小数据集的基准测试表明 fcpp公司 和 f2.2 有同样的表现(他的基准显示)。

factor 一开始就起作用。OP的数据 df1 其中分组变量 id 1:1000 ,我们可以 class(id) <- "factor"; levels(id) <- 1:1000 . 总的来说我们可以用 as.factor 如果分组变量已经是数据帧中的一个因素,这将很有帮助。见 R: Why use as.factor() instead of just factor()

Martin Morgan 6 年前

均值和方差计算需要分组进行,但t检验和p值计算可以矢量化。

my.t.test.2 <- function(grp, x, y) {
    grp <- factor(grp)

    x_g <- split(x, grp)
    x_n <- lengths(x_g)
    x_mean <- vapply(x_g, mean, numeric(1))
    x_var <- vapply(x_g, var, numeric(1))

    y_g <- split(y, grp)
    y_n <- lengths(y_g)
    y_mean <- vapply(y_g, mean, numeric(1))
    y_var <- vapply(y_g, var, numeric(1))

    x_se2 <- x_var / x_n
    y_se2 <- y_var / y_n
    se <- sqrt(x_se2 + y_se2)
    tstat <- (x_mean - y_mean) / se
    df <- se^4 / (x_se2^2 / (x_n - 1L) + (y_se2^2) / (y_n - 1L))

    2 * pt(-abs(tstat), df)
}

一个人可以试着通过避免发送来变得超级聪明(这就是为什么 mean()

my.t.test.2.1 <- compiler::cmpfun(function(grp, x, y) {
    grp <- factor(grp)

    x_g <- split.default(x, grp)
    n <- lengths(x_g)
    n1 <- n - 1L
    x_mean <- vapply(x_g, mean.default, numeric(1), USE.NAMES = FALSE)
    x_var <- vapply(x_g, var, numeric(1), USE.NAMES = FALSE)

    y_g <- split.default(y, grp)
    y_mean <- vapply(y_g, mean.default, numeric(1), USE.NAMES = FALSE)
    y_var <- vapply(y_g, var, numeric(1), USE.NAMES = FALSE)

    x_se2 <- x_var / n
    y_se2 <- y_var / n
    se <- sqrt(x_se2 + y_se2)
    tstat <- (x_mean - y_mean) / se
    df <- se^4 / ((x_se2^2 + y_se2^2) / n1)

    2 * pt(-abs(tstat), df)
})

规范解决方案和其他解决方案可以打包以提供相同的输出

f0 <- function(df)
    df %>% group_by(id) %>% summarize(p.value = t.test(A, B)$p.value)

f1 <- function(df)
    df %>% group_by(id) %>% summarize(p.value = my.t.test(A, B))

f2 <- function(df)
    tibble(id = unique(df$id), p.value = my.t.test.2(df$id, df$A, df$B))

f2.1 <- function(df)
    tibble(id = unique(df$id), p.value = my.t.test.2.1(df$id, df$A, df$B))

f2.1() 产生与规范实现相同的结果,速度大约是规范实现的两倍;担心 平均值() 等( f2() 与。 二层一()

> all.equal.default(f0(df1), f2.1(df1))
[1] TRUE
> microbenchmark(f0(df1), f1(df1), f2(df1), f2.1(df1), times = 5)
Unit: milliseconds
      expr      min       lq     mean   median       uq      max neval
   f0(df1) 374.2819 379.7749 380.8365 380.0094 381.2368 388.8794     5
   f1(df1) 249.6502 250.2525 251.8813 252.1965 253.3444 253.9630     5
   f2(df1) 154.1152 158.3243 159.8277 159.1076 162.7602 164.8311     5
 f2.1(df1) 151.0032 151.0149 152.3900 152.8105 153.2840 153.8373     5

对我来说C++实现

my.t.test.cpp <- function (x, y = NULL) {
    nx <- length(x)
    mx <- sum_cpp(x) / nx
    vx <- var_cpp(x, mx)
    ny <- length(y)
    my <- sum_cpp(y) / ny
    vy <- var_cpp(y, my)
    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)
}

fcpp <- function(df)
    df %>% group_by(id) %>% summarize(p.value = my.t.test.cpp(A, B))

分析2.1解决方案表明,大部分时间都花在 var() ,如果有人打电话给 stopifnot()

> var
function (x, y = NULL, na.rm = FALSE, use) 
{
    ...
    na.method <- pmatch(use, c("all.obs", "complete.obs", "pairwise.complete.obs", 
        "everything", "na.or.complete"))
    ...
    if (is.data.frame(x)) 
        x <- as.matrix(x)
    else stopifnot(is.atomic(x))
    ... 
    .Call(C_cov, x, y, na.method, FALSE)
}
<bytecode: 0x5e1a440>
<environment: namespace:stats>

> Rprof(); x <- my.t.test.2.1(df1$id, df1$A, df1$B); Rprof(NULL); summaryRprof()
$by.self
                      self.time self.pct total.time total.pct
"withCallingHandlers"      0.04    28.57       0.08     57.14
"tryCatchList"             0.04    28.57       0.04     28.57
"vapply"                   0.02    14.29       0.14    100.00
"stopifnot"                0.02    14.29       0.12     85.71
"match.call"               0.02    14.29       0.02     14.29

$by.total
                      total.time total.pct self.time self.pct
"vapply"                    0.14    100.00      0.02    14.29
"my.t.test.2.1"             0.14    100.00      0.00     0.00
"stopifnot"                 0.12     85.71      0.02    14.29
"FUN"                       0.12     85.71      0.00     0.00
"withCallingHandlers"       0.08     57.14      0.04    28.57
"tryCatchList"              0.04     28.57      0.04    28.57
"tryCatch"                  0.04     28.57      0.00     0.00
"match.call"                0.02     14.29      0.02    14.29

$sample.interval
[1] 0.02

$sampling.time
[1] 0.14

因此在追求速度时,可以避免参数检查,直接调用C函数

my.t.test.2.2 <- compiler::cmpfun(function(grp, x, y) {
    var <- function(x)
        .Call(stats:::C_cov, x, NULL, 4L, FALSE)
    grp <- factor(grp)

    x_g <- split.default(x, grp)
    n <- lengths(x_g)
    n1 <- n - 1L
    x_mean <- vapply(x_g, mean.default, numeric(1), USE.NAMES = FALSE)
    x_var <- vapply(x_g, var, numeric(1), USE.NAMES = FALSE)

    y_g <- split.default(y, grp)
    y_mean <- vapply(y_g, mean.default, numeric(1), USE.NAMES = FALSE)
    y_var <- vapply(y_g, var, numeric(1), USE.NAMES = FALSE)

    x_se2 <- x_var / n
    y_se2 <- y_var / n
    se <- sqrt(x_se2 + y_se2)
    tstat <- (x_mean - y_mean) / se
    df <- se^4 / ((x_se2^2 + y_se2^2) / n1)

    2 * pt(-abs(tstat), df)
})

f2.2 <- function(df)
    tibble(id = unique(df$id), p.value = my.t.test.2.2(df$id, df$A, df$B))

结果证明这是很有表现力的。

> all.equal.default(f0(df1), f2.2(df1))
[1] TRUE
> microbenchmark(
+     f0(df1), f1(df1), f2(df1), f2.1(df1), f2.2(df1), fcpp(df1),
+     times = 5
+ )
Unit: milliseconds
      expr       min        lq      mean    median       uq       max neval
   f0(df1) 378.61985 379.25525 393.38371 379.56797 386.2806 443.19488     5
   f1(df1) 250.99802 252.45281 253.55140 253.34249 255.2801 255.68362     5
   f2(df1) 156.76073 158.63126 159.63693 160.33446 161.2260 161.23216     5
 f2.1(df1) 146.64555 148.28773 151.17250 151.38536 153.9363 155.60751     5
 f2.2(df1)  25.24441  25.62982  27.50898  26.11755  30.0836  30.46951     5
 fcpp(df1) 104.20851 104.50396 105.19383 104.62905 104.7876 107.84006     5

my.t.test.2.2.cpp <- compiler::cmpfun(function(grp, x, y) {
    grp <- factor(grp)

    x_g <- split.default(x, grp)
    n <- lengths(x_g)
    n1 <- n - 1L
    x_mean <- vapply(x_g, mean.default, numeric(1), USE.NAMES = FALSE)
    x_var <- unlist(Map(var_cpp, x_g, x_mean))

    y_g <- split.default(y, grp)
    y_mean <- vapply(y_g, mean.default, numeric(1), USE.NAMES = FALSE)
    y_var <- unlist(Map(var_cpp, y_g, y_mean))

    x_se2 <- x_var / n
    y_se2 <- y_var / n
    se <- sqrt(x_se2 + y_se2)
    tstat <- (x_mean - y_mean) / se
    df <- se^4 / ((x_se2^2 + y_se2^2) / n1)

    2 * pt(-abs(tstat), df)
})

f2.2.cpp <- function(df)
    tibble(id = unique(df$id), p.value = my.t.test.2.2.cpp(df$id, df$A, df$B))

可比性能

> microbenchmark(f2.2(df1), f2.2.cpp(df1), times = 20)
Unit: milliseconds
          expr      min       lq     mean   median       uq      max neval
     f2.2(df1) 25.11237 25.69622 30.27956 26.35570 29.81884 87.34955    20
 f2.2.cpp(df1) 24.88787 25.25171 26.80836 25.43498 29.06338 30.80012    20

我不确定哪一个是黑客攻击——编写你自己的C++代码来改变,或者直接调用R的C代码。

一个更快的C++解决方案在一个调用中计算组均值和方差

cppFunction('List doit(IntegerVector group, NumericVector x) {
  int n_grp = 0;
  for (int i = 0; i < group.size(); ++i)
      n_grp = group[i] > n_grp ? group[i] : n_grp;

  std::vector<int> n(n_grp);
  std::vector<double> sum(n_grp), sumsq(n_grp);
  for (int i = 0; i < group.size(); ++i) {
      n[ group[i] - 1 ] += 1;
      sum[ group[i] - 1 ] += x[i];
      sumsq[ group[i] - 1 ] += x[i] * x[i];
  }
  NumericVector mean(n_grp), var(n_grp);
  for (size_t i = 0; i < n.size(); ++i) {
      mean[i] = sum[i] / n[i];
      var[i] = (sumsq[i] - sum[i] * mean[i]) / (n[i] - 1);
  }
  return List::create(_["n"]=n[0], _["mean"]=mean, _["var"]=var);
}')

my.t.test.2.3.cpp <- compiler::cmpfun(function(grp, x, y) {
    x <- doit(grp, x)
    y <- doit(grp, y)

    x_se2 <- x$var / x$n
    y_se2 <- y$var / y$n
    se <- sqrt(x_se2 + y_se2)
    tstat <- (x$mean - y$mean) / se
    df <- se^4 / ((x_se2^2 + y_se2^2) / (x$n - 1L))

    2 * pt(-abs(tstat), df)
})

f2.3.cpp <- function(df)
    tibble(
        id = unique(df$id),
        p.value = my.t.test.2.3.cpp(df$id, df$A, df$B)
    )

而且这很快

> all.equal.default(f0(df1), f2.3.cpp(df1))
[1] TRUE
> microbenchmark(f2.2(df1), f2.2.cpp(df1), f2.3.cpp(df1), times = 50)
Unit: milliseconds
          expr       min        lq      mean    median        uq       max
     f2.2(df1) 24.743364 25.445833 28.032135 25.873117 29.191020 88.642771
 f2.2.cpp(df1) 24.122380 24.867212 26.012985 25.369963 25.897866 30.783544
 f2.3.cpp(df1)  2.831635  2.946094  3.101408  2.992049  3.073788  7.191572
 neval
    50
    50
    50
>

另一种选择是生物导体封装 genefilter ::rowttests() ,这需要一个矩阵

set.seed(1)
m1 <- cbind(
    matrix(rnorm(8000, mean = 26, sd = 5), ncol=8, byrow = TRUE),
    matrix(rnorm(8000, mean = 25, sd = 7), ncol=8, byrow = TRUE)
)

f4 <- function(m1)
    genefilter::rowttests(m1, factor(rep(1:2, each=8)))

而且速度也很快

> microbenchmark(f2.3.cpp(df1), f4(m1), times=50)
Unit: milliseconds
          expr      min       lq     mean   median       uq      max neval
 f2.3.cpp(df1) 2.760877 2.796542 2.877030 2.845795 2.895441 3.286143    50
        f4(m1) 1.335288 1.359007 1.397601 1.377544 1.412606 1.693340    50