[英]Bootstrap confidence interval with “Inf” in final estimates - boot/dplyr package
[英]view values used by function boot to bootstrap estimates
我已经写了下面的代码来获得均值的引导估计。 我的目标是通过功能查看从数据集中选择的号码,最好是在他们的顺序进行选择, boot
在boot
包。
数据集仅包含三个数字:1、10和100,而我仅使用两个引导程序样本。
估计的平均值为23.5
,下面的R
代码指示六个数字包括一个'1',四个'10'和一个'100'。 但是,这些数字有30种可能的组合,平均得出23.5。
我是否有办法确定这30种可能的组合中的哪一种是实际出现在两个引导程序样本中的组合?
library(boot)
set.seed(1234)
dat <- c(1, 10, 100)
av <- function(dat, i) { sum(dat[i])/length(dat[i]) }
av.boot <- boot(dat, av, R = 2)
av.boot
#
# ORDINARY NONPARAMETRIC BOOTSTRAP
#
#
# Call:
# boot(data = dat, statistic = av, R = 2)
#
#
# Bootstrap Statistics :
# original bias std. error
# t1* 37 -13.5 19.09188
#
mean(dat) + -13.5
# [1] 23.5
# The two samples must have contained one '1', four '10' and one '100',
# but there are 30 possibilities.
# Which of these 30 possible sequences actual occurred?
# This code shows there must have been one '1', four '10' and one '100'
# and shows the 30 possible combinations
my.combos <- expand.grid(V1 = c(1, 10, 100),
V2 = c(1, 10, 100),
V3 = c(1, 10, 100),
V4 = c(1, 10, 100),
V5 = c(1, 10, 100),
V6 = c(1, 10, 100))
my.means <- apply(my.combos, 1, function(x) {( (x[1] + x[2] + x[3])/3 + (x[4] + x[5] + x[6])/3 ) / 2 })
possible.samples <- my.combos[my.means == 23.5,]
dim(possible.samples)
n.1 <- rowSums(possible.samples == 1)
n.10 <- rowSums(possible.samples == 10)
n.100 <- rowSums(possible.samples == 100)
n.1[1]
n.10[1]
n.100[1]
length(unique(n.1)) == 1
length(unique(n.10)) == 1
length(unique(n.100)) == 1
我认为您可以使用以下代码确定采样的数量和采样的顺序。 您必须从boot
包中提取函数ordinary.array
并将其粘贴到R
代码中。 然后指定n
, R
和strata
的值,其中n
是数据集中的观察数, R
是所需的重复样本数。
我不知道这种方法的通用性,但是它与我尝试过的几个简单示例(包括以下示例)一起使用。
library(boot)
set.seed(1234)
dat <- c(1, 10, 100, 1000)
av <- function(dat, i) { sum(dat[i])/length(dat[i]) }
av.boot <- boot(dat, av, R = 3)
av.boot
#
# ORDINARY NONPARAMETRIC BOOTSTRAP
#
#
# Call:
# boot(data = dat, statistic = av, R = 3)
#
#
# Bootstrap Statistics :
# original bias std. error
# t1* 277.75 -127.5 132.2405
#
#
mean(dat) + -127.5
# [1] 150.25
# boot:::ordinary.array
ordinary.array <- function (n, R, strata)
{
inds <- as.integer(names(table(strata)))
if (length(inds) == 1L) {
output <- sample.int(n, n * R, replace = TRUE)
dim(output) <- c(R, n)
}
else {
output <- matrix(as.integer(0L), R, n)
for (is in inds) {
gp <- seq_len(n)[strata == is]
output[, gp] <- if (length(gp) == 1)
rep(gp, R)
else bsample(gp, R * length(gp))
}
}
output
}
# I think the function ordinary.array determines which elements
# of the data are sampled in each of the R samples
set.seed(1234)
ordinary.array(n=4,R=3,1)
# [,1] [,2] [,3] [,4]
# [1,] 1 3 1 3
# [2,] 3 4 1 3
# [3,] 3 3 3 3
#
# which equals:
((1+100+1+100) / 4 + (100+1000+1+100) / 4 + (100+100+100+100) / 4) / 3
# [1] 150.25
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