Question

问题：我如何使用boostrap来获取集合的置信区间统计量的协方差矩阵的特征值，分别为数据框中的每个组（因子级别）？

问题：我无法完成数据计算结构我需要包含适用于boot函数的这些结果，或者一种在组中“映射”引导程序并以适合绘图的形式获得置信区间的方法。

上下文：在heplots包中，boxM计算协方差矩阵相等的Box M检验。有一种绘图方法可以生成一个有用的日志决定因素图测试。该图中的置信区间基于渐近理论近似。

> library(heplots)
> iris.boxm <- boxM(iris[, 1:4], iris[, "Species"])
> iris.boxm

        Box's M-test for Homogeneity of Covariance Matrices

data:  iris[, 1:4]
Chi-Sq (approx.) = 140.94, df = 20, p-value < 2.2e-16

> plot(iris.boxm, gplabel="Species")

绘图方法还可以显示特征值的其他函数，但没有理论值在这种情况下可以获得置信区间。

op <- par(mfrow=c(2,2), mar=c(5,4,1,1))
plot(iris.boxm, gplabel="Species", which="product")
plot(iris.boxm, gplabel="Species", which="sum")
plot(iris.boxm, gplabel="Species", which="precision")
plot(iris.boxm, gplabel="Species", which="max")
par(op)

因此，我希望能够使用boostrap计算这些CI，并将其显示在相应的图中。

我尝试了什么：

以下是提升这些统计数据的功能，但总数样本，不考虑群组（Species）。

cov_stat_fun <- function(data, indices, 
            stats=c("logdet", "prod", "sum", "precision", "max")
            ) {
    dat <- data[indices,]
    cov <- cov(dat, use="complete.obs")
    eigs <- eigen(cov)$values

    res <- c(
        "logdet" = log(det(cov)),
        "prod" = prod(eigs),
        "sum" = sum(eigs),
        "precision" = 1/ sum(1/eigs),
        "max" = max(eigs)
        )
}

boot_cov_stat <- function(data, R=500,  ...) {
    boot(data, cov_stat_fun, R=R,  ...)
}

这样可行，但我需要按组（以及总样本）的结果

> iris.boot <- boot_cov_stat(iris[,1:4])
>
> iris.ci <- boot.ci(iris.boot)
> iris.ci
BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS
Based on 500 bootstrap replicates

CALL : 
boot.ci(boot.out = iris.boot)

Intervals : 
Level      Normal              Basic             Studentized     
95%   (-6.622, -5.702 )   (-6.593, -5.653 )   (-6.542, -5.438 )  

Level     Percentile            BCa          
95%   (-6.865, -5.926 )   (-6.613, -5.678 )  
Calculations and Intervals on Original Scale
Some BCa intervals may be unstable
>

我还编写了一个函数来计算每个组的单独协方差矩阵，但我无法看到如何在我的引导函数中使用它。有人可以帮忙吗？

# calculate covariance matrices by group and pooled
covs <- function(Y, group) {
   Y <- as.matrix(Y)
   gname <- deparse(substitute(group))
   if (!is.factor(group)) group <- as.factor(as.character(group))

   valid <- complete.cases(Y, group)
   if (nrow(Y) > sum(valid)) 
      warning(paste(nrow(Y) - sum(valid)), " cases with missing data have been removed.")
   Y <- Y[valid,]
   group <- group[valid]
   nlev <- nlevels(group)
   lev <- levels(group)
   mats <- aux <- list()
   for(i in 1:nlev) {
      mats[[i]] <- cov(Y[group == lev[i], ])
   }
   names(mats) <- lev
   pooled <- cov(Y)
   c(mats, "pooled"=pooled)
}

修改：在一个看似相关的问题Bootstrap by groups中，建议通过使用strata boot()参数来提供答案，但没有示例。 [啊：strata参数只是确保在bootstrap样本中表示层与数据中的频率相关。]

为我的问题尝试这个，我没有进一步开悟，因为我想得到的是每个Species的单独的置信区间。

> iris.boot.strat <- boot_cov_stat(iris[,1:4], strata=iris$Species)
> 
> boot.ci(iris.boot.strat, conf=0.95, type=c("basic", "bca"))
BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS
Based on 500 bootstrap replicates

CALL : 
boot.ci(boot.out = iris.boot.strat, conf = 0.95, type = c("basic", 
    "bca"))

Intervals : 
Level      Basic                BCa          
95%   (-6.587, -5.743 )   (-6.559, -5.841 )  
Calculations and Intervals on Original Scale
Some BCa intervals may be unstable
>

Answer 1

如果我理解你的问题，你可以按组运行你的引导功能，如下所示：

library(boot)
library(tidyverse)

# Pooled
iris.boot <- boot_cov_stat(iris[,1:4])
iris.ci <- boot.ci(iris.boot)

# By Species
boot.list = setNames(unique(iris$Species), unique(iris$Species)) %>% 
  map(function(group) {
    iris.boot = boot_cov_stat(iris[iris$Species==group, 1:4])
    boot.ci(iris.boot)
  })

# Combine pooled and by-Species results
boot.list = c(boot.list, list(Pooled=iris.ci))

boot.list

$setosa
BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS
Based on 500 bootstrap replicates

CALL : 
boot.ci(boot.out = iris.boot)

Intervals : 
Level      Normal              Basic             Studentized     
95%   (-13.69, -11.86 )   (-13.69, -11.79 )   (-13.52, -10.65 )  

Level     Percentile            BCa          
95%   (-14.34, -12.44 )   (-13.65, -11.99 )  
Calculations and Intervals on Original Scale
Warning : BCa Intervals used Extreme Quantiles
Some BCa intervals may be unstable

$versicolor
BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS
Based on 500 bootstrap replicates

CALL : 
boot.ci(boot.out = iris.boot)

Intervals : 
Level      Normal              Basic             Studentized     
95%   (-11.37,  -9.81 )   (-11.36,  -9.78 )   (-11.25,  -8.97 )  

Level     Percentile            BCa          
95%   (-11.97, -10.39 )   (-11.35, -10.09 )  
Calculations and Intervals on Original Scale
Warning : BCa Intervals used Extreme Quantiles
Some BCa intervals may be unstable

$virginica
BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS
Based on 500 bootstrap replicates

CALL : 
boot.ci(boot.out = iris.boot)

Intervals : 
Level      Normal              Basic             Studentized     
95%   (-9.467, -7.784 )   (-9.447, -7.804 )   (-9.328, -6.959 )  

Level     Percentile            BCa          
95%   (-10.050,  -8.407 )   ( -9.456,  -8.075 )  
Calculations and Intervals on Original Scale
Warning : BCa Intervals used Extreme Quantiles
Some BCa intervals may be unstable

$Pooled
BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS
Based on 500 bootstrap replicates

CALL : 
boot.ci(boot.out = iris.boot)

Intervals : 
Level      Normal              Basic             Studentized     
95%   (-6.620, -5.714 )   (-6.613, -5.715 )   (-6.556, -5.545 )  

Level     Percentile            BCa          
95%   (-6.803, -5.906 )   (-6.624, -5.779 )  
Calculations and Intervals on Original Scale
Some BCa intervals may be unstable

Answer 2

我认为最好的一般答案是@ eipi10提出的扩展，使用某种方法从bootci对象中提取所需的置信区间。这是broom包中缺少的。

作为一种有益的替代方案，我尝试直接在引导程序的结果上使用broom::tidy()。它不是（通常是不对称的）置信区间，而是将引导程序估计值设为statistic，bias和std.error。但是，根据我得到的结果（见下文），我怀疑broom::tidy()在这种情况下是否给出了正确的结果。

# try just using tidy on the bootstrap results

## pooled
iris.boot <- boot_cov_stat(iris[,1:4])
iris.pooled <- tidy(iris.boot)

，并提供：

> iris.pooled
       term   statistic          bias    std.error
1    logdet -6.25922391 -0.0906294902 0.2469587430
2      prod  0.00191273 -0.0001120317 0.0004485317
3       sum  4.57295705 -0.0382145128 0.2861790776
4 precision  0.01692092 -0.0005047993 0.0016818910
5       max  4.22824171 -0.0329408193 0.2815648589
>

现在，使用map对群组的另一个答案中描述的方法，并结合：

## individual groups
boot.list2 = setNames(unique(iris$Species), unique(iris$Species)) %>% 
  map(function(group) {
    iris.boot = boot_cov_stat(iris[iris$Species==group, 1:4])
    tidy(iris.boot)
  })

# Combine pooled and by-Species results
boot.list <- c(boot.list2, list(Pooled=iris.pooled))

转换为数据框：

## transform this list to a data frame, with a group variable
result <- bind_rows(boot.list) %>% 
    mutate(group = rep(c( levels(iris$Species), "Pooled"), 5)) %>%
    arrange(term)

> result
        term     statistic          bias    std.error      group
1     logdet -1.306736e+01 -3.240621e-01 4.660334e-01     setosa
2     logdet -1.087433e+01 -2.872073e-01 3.949917e-01 versicolor
3     logdet -8.927058e+00 -2.925485e-01 4.424367e-01  virginica
4     logdet -6.259224e+00 -9.062949e-02 2.469587e-01     Pooled
5        max  2.364557e-01 -6.696719e-03 4.426305e-02     setosa
6        max  4.878739e-01 -6.798321e-03 8.662880e-02 versicolor
7        max  6.952548e-01 -6.517223e-03 1.355433e-01  virginica
8        max  4.228242e+00 -3.294082e-02 2.815649e-01     Pooled
9  precision  5.576122e-03 -5.928678e-04 8.533907e-04     Pooled
10 precision  7.338788e-03 -6.894908e-04 1.184594e-03     setosa
11 precision  1.691212e-02 -1.821494e-03 2.000718e-03 versicolor
12 precision  1.692092e-02 -5.047993e-04 1.681891e-03  virginica
13      prod  2.113088e-06 -4.158518e-07 7.850009e-07 versicolor
14      prod  1.893828e-05 -3.605691e-06 6.100376e-06  virginica
15      prod  1.327479e-04 -2.381536e-05 4.792428e-05     Pooled
16      prod  1.912730e-03 -1.120317e-04 4.485317e-04     setosa
17       sum  3.092041e-01 -1.005543e-02 4.623437e-02  virginica
18       sum  6.248245e-01 -1.238896e-02 8.536621e-02     Pooled
19       sum  8.883673e-01 -1.500578e-02 1.409230e-01     setosa
20       sum  4.572957e+00 -3.821451e-02 2.861791e-01 versicolor
>

这给出了一些现在可以绘制的东西，据说对应于原始问题中没有误差条的情节：

result %>% mutate(Pooled = group == "Pooled") %>%
    filter (term != "logdet") %>%
    ggplot(aes(y=statistic, x=group, color=Pooled)) +
    geom_point(size=2.5) +
    geom_errorbar(aes(ymin=statistic-2*std.error, 
                      ymax=statistic+2*std.error), width=0.4) +
    facet_wrap( ~ term, scales="free") +
    coord_flip() + guides(color=FALSE)

然而，这个“整洁的情节”似乎明显错误。理论上说，轮询样本的结果必须在每种情况下介于不同组之间，因为它在某种意义上是对群体的“凸组合”。将下面的图表与原始问题中给出的图表进行比较。（这可能是我在这里做错了，但我看不出有什么缺陷。）

在R

2 个答案: