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 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]] ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. |
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$fit > [[1]] > function(x) eigen(data.matrix(x)) > <environment: 0x0000017d35f00518> > > [[2]] > function(x) eigen(data.matrix(x)) > <environment: 0x0000017d35ef73d0> > > [[alternative HTML version deleted]] > > ______________________________________________ > [hidden email] mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. |
In reply to this post by Veerappa Chetty
Since purrr:map() is functionally identical to lapply, it is not at all clear to me what you are having trouble with. What output did you want?
On March 31, 2021 5:10:29 PM PDT, Veerappa Chetty <[hidden email]> wrote: >I have to compute characteristic roots for 100s of U.S.counties. I got >help using " lapply". I prefer using MAP and PURR if that is possible. >I am >using them to compute linear regressions. >Here is my test problem and the output. I would appreciate any help. >Thanks >V.K.Chetty >____ >test.dat<- tibble(ID=c(1,2),a=c(1,1),b=c(1,1),c=c(2,2),d=c(4,3)) >test.out<-test.dat %>% nest(-ID) %>% mutate(fit = purrr::map(data,~ >function(x) eigen(data.matrix(x)), data=.)) > >I get this output: >> test.out$fit >[[1]] >function(x) eigen(data.matrix(x)) ><environment: 0x0000017d35f00518> > >[[2]] >function(x) eigen(data.matrix(x)) ><environment: 0x0000017d35ef73d0> > > [[alternative HTML version deleted]] > >______________________________________________ >[hidden email] mailing list -- To UNSUBSCRIBE and more, see >https://stat.ethz.ch/mailman/listinfo/r-help >PLEASE do read the posting guide >http://www.R-project.org/posting-guide.html >and provide commented, minimal, self-contained, reproducible code. -- Sent from my phone. Please excuse my brevity. ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. |
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