Question

我有一个概率矢量，比如

prob=c(0.1,0.8,0.1)

和一个数据框：df=cbind(c("A","B","A"),c(1,2,3),c("q","v","z"))

我想用替换从df采样n对象，第一列的概率为0.1，第二列的概率为0.8，第三列的概率为0.1

Answer 1

我们将取消列出data.frame，并即时修改prob向量，使其具有适当的长度。

df <- data.frame(c("A","B","A"), c(1,2,3), c("q","v","z"), stringsAsFactors = F)

n <- 5
set.seed(1)
unname(sample(unlist(df), n, replace = TRUE, prob= rep(prob, each = nrow(df))))
# [1] "3" "1" "A" "z" "2"

如果您真的是从矩阵开始的，而不是一个短一点的data.frame：

df=cbind(c("A","B","A"),c(1,2,3),c("q","v","z"))
set.seed(1)
sample(df, n, replace = TRUE, prob= rep(prob, each = nrow(df)))
# [1] "3" "1" "A" "z" "2"

从列表中（回答评论）

l =list(c("A","B"),c(1,2,3),c("q","v","z","w"))
set.seed(1)
sample(unlist(l), n, replace = TRUE, prob= rep(prob/lengths(l), lengths(l)))
# [1] "3" "2" "1" "v" "3" "B" "q"

Answer 2

这是基于以下假设：一列内的样本概率是统一的：

我们首先使用向量n中的概率对prob列位置进行采样；

df=cbind(c("A","B","A"),c(1,2,3),c("q","v","z"))
prob=c(0.1,0.8,0.1)
n = 10

set.seed(1)
colselect <- sample(1:ncol(df), size = n, replace = TRUE, prob = prob)

[1] 2 2 2 1 2 3 1 2 2 2

然后，我们遍历列位置，并从相应列中分别采样一个元素：

sapply(colselect, function(x) sample(df[,x], 1))

[1] "1" "1" "3" "B" "3" "v" "A" "3" "2" "3"

来自R中不同列的样本

2 个答案: