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我有一个 data.frames 列表：

ldf <- list(structure(c(2, 0.15, 0.52, 5, 6, 2, 1), .Dim = c(1L, 7L), .Dimnames = list(
    "M01", c("fitness", "cMRc", "cMRcg", "pMR", "cMEnr", "gMR", 
    "sex"))), structure(c(4, 4, 9, 0.22, 0.19, 0.05, 0.555, 0.495, 
0.605, 8, 7, 4, 15, 10, 4, 3, 3, 3, 1, 1, 1), .Dim = c(3L, 7L
), .Dimnames = list(c("M03", "M06", "M12"), c("fitness", "cMRc", 
"cMRcg", "pMR", "cMEnr", "gMR", "sex"))), structure(c(4, 4, 7, 
7, 0.145, 0.09, 0.2, 0.195, 0.46, 0.68, 0.45, 0.48, 6, 3, 7, 
5, 8, 3, 10, 9, 4, 4, 4, 4, 1, 1, 1, 1), .Dim = c(4L, 7L), .Dimnames = list(
    c("M05", "M08", "M09", "M10"), c("fitness", "cMRc", "cMRcg", 
    "pMR", "cMEnr", "gMR", "sex"))), structure(c(10, 10, 0.145, 
0.18, 0.725, 0.4, 6, 5, 9, 5, 5, 5, 1, 1), .Dim = c(2L, 7L), .Dimnames = list(
    c("M02", "M04"), c("fitness", "cMRc", "cMRcg", "pMR", "cMEnr", 
    "gMR", "sex"))))

如何对这些 data.frames 进行采样并根据向量 s 调整采样大小：

s <- c(1,2,3,1)

所以ldf[[1]]，我会得到：

ldf[[1]][sample(nrow(ldf[[1]]),size=1),]

因为ldf[[2]]，我会得到：

ldf[[2]][sample(nrow(ldf[[2]]),size=2),]

等等……

如何将其包装在 lapply 电话中，我猜接近：

lapply(ldf,function(x) x[sample(nrow(x),s),]) # but it's not sampling according to sizes values in s
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1 回答 1

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您可以mapply像这样使用（如@Justin 所述）：

f <- function(x,y){
  rep <- y > nrow(x)    ## repeat if y > nrow(x)
  x[sample(nrow(x),size=y,rep=rep),]
}
s <- c(1,2,3,1)
mapply(FUN=f,ldf,s )
于 2013-04-09T20:13:50.547 回答