[R] using MAP and PURR to compute characteristic roots
David Winsemius
dw|n@em|u@ @end|ng |rom comc@@t@net
Thu Apr 1 02:42:11 CEST 2021
On 3/31/21 5:10 PM, Veerappa Chetty wrote:
> I have to compute characteristic roots for 100s of U.S.counties. I got
> help using " lapply". I prefer using MAP and PURR if that is possible. I am
> using them to compute linear regressions.
> Here is my test problem and the output. I would appreciate any help.
> Thanks
> V.K.Chetty
> ____
> test.dat<- tibble(ID=c(1,2),a=c(1,1),b=c(1,1),c=c(2,2),d=c(4,3))
> test.out<-test.dat %>% nest(-ID) %>% mutate(fit = purrr::map(data,~
> function(x) eigen(data.matrix(x)), data=.))
I think you will find that there is a bias on R-help toward using "base"
R functions where it seems "natural". The venue where "tidyverse"
solutions are typically returned is StackOverflow.)
It wasn't stated clearly what you wanted from this. The `eigen`
function returns a list object so the apply function neatly handles row
based operations:
> apply( test.dat[-1], 1, function(x){eigen(matrix(x,2))})
[[1]]
eigen() decomposition
$values
[1] 4.5615528 0.4384472
$vectors
[,1] [,2]
[1,] -0.4896337 -0.9627697
[2,] -0.8719282 0.2703230
[[2]]
eigen() decomposition
$values
[1] 3.7320508 0.2679492
$vectors
[,1] [,2]
[1,] -0.5906905 -0.9390708
[2,] -0.8068982 0.3437238
If you instead wanted only the eigenvalues you might have chosen to
extract them at the time of the `eigen` call:
apply( test.dat[-1], 1, function(x){eigen(matrix(x,2))$values})
[,1] [,2]
[1,] 4.5615528 3.7320508
[2,] 0.4384472 0.2679492
Do note that the values are returned column-wise.
(I see that you noted in an earlier (duplicate) rhelp posting that you
were expecting "younger" respondents. Since I'm in my eighth decade, I
probably don't qualify.)
--
David Winsemius, MD
>
> I get this output:
>> test.out$fit
> [[1]]
> function(x) eigen(data.matrix(x))
> <environment: 0x0000017d35f00518>
>
> [[2]]
> function(x) eigen(data.matrix(x))
> <environment: 0x0000017d35ef73d0>
>
> [[alternative HTML version deleted]]
>
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