我有这个清单:
lst
lst <- list(a=c(2.5,9.8,5.0,6.7,6.5,5.2,34.4, 4.2,39.5, 1.3,0.0,0.0,4.1,0.0,0.0,25.5,196.5, 0.0,104.2,0.0,0.0,0.0,0.0,0.0),b=c(147.4,122.9,110.2,142.3))
我想计算列表的每个值和列表的每个元素(a
and b
) z.score 为:(x[i]-mean(x)/sd(x)
,其中 x 是列表中每个元素的所有值(合计),x[i] 是每个列表元素的每个单个组件。
我尝试过lapply
lapply(lst,function (x) as.data.frame(apply(x,2, function(y)- lapply(lst,mean)/lapply(lst,sd))))
但有一个错误...
也许与for
循环为:
lst.new <- vector("list",1)
for (i in 1:length(lst)){
for (j in 1:dim(data.frame(lst[i]))[1]){
res[j] <- (as.numeric(unlist(lst[i]))[j]-mean(as.numeric(unlist(lst[i])))/
sd(as.numeric(unlist(lst[i])))
lst.new[[i]] <- res
}
}
但结果很奇怪(当然我错了lst.new
输出):
[[1]]
[1] -0.3635464 -0.1982809 -0.3069486 -0.2684621 -0.2729899 -0.3024208 0.3586413 -0.3250599 0.4741007 -0.3907133
[11] -0.4201442 -0.4201442 -0.3273238 -0.4201442 -0.4201442 0.1571532 4.0284412 -0.4201442 1.9388512 -0.4201442
[21] -0.4201442 -0.4201442 -0.4201442 -0.4201442
[[2]]
[1] 0.9671130 -0.4517055 -1.1871746 0.6717671 -0.2729899 -0.3024208 0.3586413 -0.3250599 0.4741007 -0.3907133
[11] -0.4201442 -0.4201442 -0.3273238 -0.4201442 -0.4201442 0.1571532 4.0284412 -0.4201442 1.9388512 -0.4201442
[21] -0.4201442 -0.4201442 -0.4201442 -0.4201442
预期结果可以是不同长度的列表或数据帧,如下所示:
a b
-0.36 0.967113
-0.19 -0.45
[...] [...]
等等...
P.S:
0.36 == (2.5- mean(unlist(lst[1])))/sd(unlist(lst[1]))
0.967113 == (147.4 -mean(unlist(lst[2])))/sd(unlist(lst[2]))
我用比较好lapply
(或他的家庭职能)并解决问题