应用数学函数时，如何从具有不同属性的元素的R列表中提取完整数据？

Question

I performed a kmeansvar() on mtcars dataset and a quick peek at the contents of the output is shown below.

>require(ClustOfVar)
>fit <- kmeansvar(X.quanti = mtcars,init = 3)

>var_fit <- fit$var
> var_fit
$cluster1
     squared loading
mpg        0.8823383
cyl        0.8933205
disp       0.9184043
hp         0.7650133
wt         0.8245741

$cluster2
     squared loading
drat       0.7793067
am         0.8509076
gear       0.8415168

$cluster3
     squared loading
qsec       0.8388092
vs         0.7755302
carb       0.7018491

My next step requires me to extract the variable name with the maximum value of squared loading .

I have obviously used 'lapply' but my resultant output is this as shown below.

$cluster1
[1] 0.9184043

$cluster2
[1] 0.8509076

$cluster3
[1] 0.8388092

---

My expected output is :

$cluster1
     squared loading
disp       0.9184043

$cluster2
     squared loading

am         0.8509076


$cluster3
     squared loading
qsec       0.8388092

or, to be even more precise, my required values are :

cluster1 disp
cluster2 am
cluster3 qsec

How do I extract the attributes of an element in a list?

Answer 1

We can loop over the list with sapply , use which.max to get the index of the max values, extract the row names corresponding to the index.

sapply(var_fit, function(x) row.names(x)[which.max(x)])
#cluster1 cluster2 cluster3 
# "disp"     "am"   "qsec" 

Answer 2

I would also do it with lapply , with a little twist:

res1 <- lapply(var_fit, function(x) x[which.max(x),,drop=FALSE])

To get the names of the rows, I would have pretty much used the same method as the one proposed by @akrun. But the same result can be obtained by working on res1 :

res2 <- sapply(res1, rownames)