[R] using factor variable in the "DO" function

[Jeff Newmiller]

You have competing non-standard evaluation tools in play. I would decouple them with a function (untested):

my fit <- function( DF ) { 
    data.frame( HOSP_NRD = .$HOSP_NRD[1]
        , fitHosp = lm(log(y)~ log(x)+I(log(x)^2)+NCHRONIC+AGE+sex
      , data=DF)
      )
}
dfhosp <- ( dat.2.wide.sub
  %>% group_by(HOSP_NRD)
  %>% do( my_fit( . ) )
-- 
Sent from my phone. Please excuse my brevity.

On March 7, 2016 10:48:28 AM PST, Veerappa Chetty <chettyvk at gmail.com> wrote:
>Hi,
>
>The following call does not work:
>
>dfhosp=dat.2.wide.sub %>% group_by(HOSP_NRD)%>%
>  do(fitHosp=lm(log(y)~ log(x)+I(log(x)^2)+NCHRONIC+AGE+sex ,data=.))
>
>Error in `$<-.data.frame`(`*tmp*`, "sex", value = integer(0)) :
>  replacement has 0 rows, data has 25174
>
>When I use "sex" as a binary variable with values "0" and "1", it
>works.
>
>Is there a way to use factor in this call?
>Thanks.
>
>V.K.Chetty
>
>-- 
>Professor of Family Medicine
>Boston University
>Tel: 617-414-6221, Fax:617-414-3345
>emails: chettyvk at gmail.com,vchetty at bu.edu
>
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