我正在尝试从我的数据库的2列（工作时间和访问患者的相对数量）中提取随机样本，然后我想逐步计算平均值。我的意思是，首先计算前2个样本之间的平均值，然后计算我刚刚计算出的平均值和第三个样本之间的平均值......以此类推。
这是否可能？是否有相应的函数？
感谢大家的帮助。
L.
以下是我提取样本的方法。
library(dplyr)
set.seed(2020)
obs <- rnorm(10, mean = 0, sd = 1)
time <- rnorm(10, mean = 0.5, sd = 1)
rdf <- data.frame(obs, time)
sample_n(rdf, 1)
p <- replicate(100, expr = (sample_n(rdf, 1) + sample_n(rdf, 1))/2)
set.seed(2020)
obs <- rnorm(10, mean = 0, sd = 1)
time <- rnorm(10, mean = 0.5, sd = 1)
rdf <- data.frame(obs, time)
nsamp <- 5 # number of samples
mean_vect <- numeric(nsamp) # create a vector to store the means
mean_vect[1] <- mean(sample_n(rdf, 1)$obs) # mean of first sample as starting point
# start calculations to fifth sample iteratively
for (i in 2:nsamp) {
# select the next sample
next_samp <- sample_n(rdf, 1)
# calculate the mean between the previous mean and the next sample
mean_vect[i] <- mean(c(mean_vect[i-1], next_samp$obs))
}
# print the means
print(mean_vect)
[1] -1.13040590 -0.20491620 0.04831609 0.08284144 0.40170747
f <- function(S, R, i=1, cm=NULL, res=NULL, ...) {
S <- rbind(cm, rdf[sample.int(nrow(rdf), 1), ])
cm <- colMeans(S)
res <- rbind(res, cm)
return(if (i < R) {
f(S, R=R, i=i + 1, cm=cm, res=res) ## also `Recall(.)` instead of `f(.)`
} else {
`rownames<-`(as.data.frame(res), NULL)
})
}
set.seed(42)
f(rdf[sample.int(nrow(rdf), 1), ], R=10)
# obs time
# 1 0.376972125 -0.35312282
# 2 -1.209781097 0.01180847
# 3 -0.416404486 -0.17065718
# 4 0.671363430 -0.97981606
# 5 0.394365109 -0.21075628
# 6 -0.368020398 -0.04117009
# 7 -0.033236012 0.68404454
# 8 0.042065388 0.62117402
# 9 0.209518756 0.13402560
# 10 -0.009929495 -1.20236950
你可能需要增加C语言栈大小。
但你也可以使用for循环。
R <- 10
res1 <- matrix(nrow=0, ncol=2)
set.seed(42)
for (i in seq_len(R - 1)) {
if (nrow(res1) == 0) {
res1 <- rdf[sample.int(nrow(rdf), 1), ]
}
S <- rdf[sample.int(nrow(rdf), 1), ]
res1 <- rbind(res1, colMeans(rbind(res1[nrow(res1), ], S)))
}
res1
# obs time
# 1 0.376972125 -0.35312282
# 2 -1.209781097 0.01180847
# 3 -0.416404486 -0.17065718
# 4 0.671363430 -0.97981606
# 5 0.394365109 -0.21075628
# 6 -0.368020398 -0.04117009
# 7 -0.033236012 0.68404454
# 8 0.042065388 0.62117402
# 9 0.209518756 0.13402560
# 10 -0.009929495 -1.20236950
这里是两个版本的快速基准测试（R=2K），递归似乎快了近一倍。
# Unit: milliseconds
# expr min lq mean median uq max neval cld
# recursive 577.0595 582.0189 587.3052 586.9783 592.4281 597.8778 3 a
# for-loop 991.4360 993.7170 997.2436 995.9980 1000.1473 1004.2966 3 b
数据：
rdf <- structure(list(obs = c(0.376972124936433, 0.301548373935665,
-1.0980231706536, -1.13040590360378, -2.79653431987176, 0.720573498411587,
0.93912102300901, -0.229377746707471, 1.75913134696347, 0.117366786802848
), time = c(-0.353122822287008, 1.40925918161821, 1.69637295955276,
0.128416096258652, 0.376739766712564, 2.30004311672545, 2.20399587729432,
-2.53876460529759, -1.78897494991878, 0.558303494992923)), class = "data.frame", row.names = c(NA,
-10L))
另一种方法（使用您的示例数据rdf）：
mean_of_random_pair(xs)，它从集合xs中随机选择两个项目并计算它们的平均值：mean_of_random_pair <- function(xs){
xs |> sample(size = 2) |> mean(na.rm = TRUE)
}
cumulative_mean 的函数，它计算整体均值 X，其中 X 是现有 X 和新项 x 的平均值:cumulative_mean <- function(xs){
xs |> Reduce(f = \(X, x) mean(c(X, x)),
accumulate = TRUE
)
}
将上述函数链接成管道并在集合 rdf$obs 上运行它 runs 次:
runs = 100
1:runs |>
Map(f = \(i) mean_of_random_pair(rdf$obs)) |>
cumulative_mean()
输出（迭代平均序列）：
[1] 1.1000858 0.8557774 0.3041130 0.4262881 -0.4658256
# ...
检查输出（针对5000次模拟运行）：
runs = 5e3
set.seed(4711)
densities <-
list(obs = 'obs', time = 'time') |>
map(\(var){
1:runs |>
Map(f = \(i) mean_of_random_pair(rdf[[var]])) |>
cumulative_mean() |>
density()
})
densities$time |> plot(col = 'blue', ylim = c(0, 1), xlim = c(-3, 3), main = 'foo')
densities$obs |> lines(col = 'red')
@IRTFM，“Recall”并没有显着改变性能。但一个非常有趣的功能，如果您重命名递归函数，它可以使其继续工作。谢谢! - jay.sf