2013-04-09 65 views
0

我有data.frames的列表:ř如何作为采样大小采样data.frame的列表,与矢量

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")))) 

如何采样论文data​​.frames并根据向量s调整采样大小:

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

所以对于ldf[[1]],我会得到:

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

ldf[[2]],我禾ULD得到:

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

ECT ......

如何在lapply调用把这个包,我想的东西接近:

lapply(ldf,function(x) x[sample(nrow(x),s),]) # but it's not sampling according to sizes values in s 
+3

听起来像'mapply'的工作。 – Justin 2013-04-09 19:37:51

回答

1

您可以使用mapply(如前所述由@Justin)是这样的:

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)