data.frames列表中元素的平均值 [英] Mean of elements in a list of data.frames
问题描述
假设我有一个data.frames列表(行和列相等)
dat1 = as.data.frame (matrix(rnorm(25),ncol = 5))
/ pre>
dat2 = as.data.frame(matrix(rnorm(25),ncol = 5))
dat3 = as.data.frame (rnorm(25),ncol = 5))
all.dat< -list(dat1 = dat1,dat2 = dat2,dat3 = dat3)
如何返回单个data.frame,它是整个列表中data.frames中每个元素的平均值(或sum等)第一行和第一列的清单1,2,3等等)?我已经在plyr中尝试了lapply和ldply,但是这些返回列表中每个data.frame的统计信息。
编辑:由于某些原因,这被作为家庭作业重新插入。不是这么重要,但这不是一个功课问题,我只是不知道为什么我可以让这个工作。感谢任何见解!
编辑2:进一步澄清
我可以使用循环得到结果,但我希望有一种方法(更简单和更快方式是因为我使用的数据具有12行×100列的data.frame,并且存在这些数据帧中的1000+的列表。)z< -matrix(0,nrow(all.dat $ dat1),ncol(all.dat $ dat1))
for(l in 1:nrow(all.dat $ dat1)){
for(m in 1:ncol(all.dat $ dat1)){
z [l,m]< -mean(unlist(lapply(all.dat,`[ ,i = 1,j = m)))
}
}
其结果是:
> z
[,1] [,2] [,3] [,4] [,5]
[1,] -0.64185488 0.06220447 -0.02153806 0.83567173 0.3978507
[2,] -0.27953054 -0.19567085 0.45718399 -0.02823715 0.4932950
[3,] 0.40506666 0.95157856 1.00017954 0.57434125 -0.5969884
[4,] 0.71972821 -0.29190645 0.16257478 -0.08897047 0.9703909
[5,] -0.05570302 0.62045662 0.93427522 -0.55295824 0.7064439
我想知道是否有更少的笨重和更快的方法来做到这一点。谢谢!
解决方案这是一个带
plyr
的班轮。你可以用你想要的任何其他功能替换mean
。ans1 = aaply(laply(all.dat,as.matrix),c(2,3),mean)
Suppose I had a list of data.frames (of equal rows and columns)
dat1=as.data.frame(matrix(rnorm(25),ncol=5)) dat2=as.data.frame(matrix(rnorm(25),ncol=5)) dat3=as.data.frame(matrix(rnorm(25),ncol=5)) all.dat<-list(dat1=dat1,dat2=dat2,dat3=dat3)
How can I return a single data.frame that is the mean (or sum, etc) for each element in the data.frames across the list (e.g., mean of first row and first column from lists 1, 2, 3 and so on)? I have tried lapply and ldply in plyr but these return the statistic for each data.frame within the list.
Edit: For some reason, this was retagged as homework. Not that it matters either way, but this is not a homework question I just don't know why I can get this to work. Thanks for any insight!
Edit2: For further clarification I can get the results using loops, but I was hoping that there was a way (simpler and faster way because the data I am using has data.frames that are 12 rows by 100 columns and there is a list of 1000+ of these data frames.)
z<-matrix(0,nrow(all.dat$dat1),ncol(all.dat$dat1)) for(l in 1:nrow(all.dat$dat1)){ for(m in 1:ncol(all.dat$dat1)){ z[l,m]<-mean(unlist(lapply(all.dat, `[`, i =l, j = m))) } }
with a result of the means:
> z [,1] [,2] [,3] [,4] [,5] [1,] -0.64185488 0.06220447 -0.02153806 0.83567173 0.3978507 [2,] -0.27953054 -0.19567085 0.45718399 -0.02823715 0.4932950 [3,] 0.40506666 0.95157856 1.00017954 0.57434125 -0.5969884 [4,] 0.71972821 -0.29190645 0.16257478 -0.08897047 0.9703909 [5,] -0.05570302 0.62045662 0.93427522 -0.55295824 0.7064439
I was wondering if there was a less clunky and faster way to do this. Thanks!
解决方案Here is a one liner with
plyr
. You can replacemean
with any other function that you want.ans1 = aaply(laply(all.dat, as.matrix), c(2, 3), mean)
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