[R] using if else function to complete a column in data frame

[PIKAL Petr]

Hi

Split option seems to me the most effective and ifelse is in this case not necessary.

You can use this function to estimate levels

fff <- function(x) sign(diff(x))
test.s <- split(test, test$ID)
for (i in 1:length(test.s)) test.s[[i]]$movimento[-1] <- fff(test.s[[i]][,2])
test <- do.call(rbind, test.s)
test[,3] <- factor(test[,3], levels=c(-1, 0, 1), labels=c("jusante", "parado", "montante"))
test

To avoid extension of row names you can use
library (plyr)
test<-ldply (test.s, data.frame)

instead of do.call.

I tried assigning factor within for cycle

fff<- function(x) factor(sign(diff(x)), levels=c(-1, 0, 1), labels=c("jusante", "parado", "montante"))
test.s <- split(test, test$ID)
for (i in 1:length(test.s)) test.s[[i]]$movimento[-1] <- fff(test.s[[i]][,2])
test <- do.call(rbind, test.s)
test

but assigned in this case is not factor but numeric vector, which seems to me strange. Maybe somebody could explain this behaviour

Cheers
Petr

> -----Original Message-----
> From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Cacique
> Samurai
> Sent: Monday, January 9, 2017 6:22 AM
> To: R help <r-help at r-project.org>
> Subject: [R] using if else function to complete a column in data frame
>
> Hello all!
>
> I´m trying to complete the "movimento" column in dataframe based in the
> values of "kmr" column in two sequential lines, as below:
>
> data example (dput in the end of email):
>
>          ID    kmr movimento
> 5    10.700 314.20        NA
> 1    10.700 278.74        NA
> 2    10.700 278.74        NA
> 3    10.700 278.74        NA
> 4    10.700 278.74        NA
> 494 100.700 269.94        NA
> 500 100.700 278.74        NA
> 499 100.700 314.20        NA
> 495 100.700 278.74        NA
> 498 100.700 278.74        NA
> 496 100.700 255.40        NA
> 497 100.700 255.10        NA
>
> Once I have different IDs, I wrote this function:
>
> move = function (x){
>
>   for (j in x$ID){
>
>     for (i in 2:length(x$kmr)-1){
>
>       if (x$kmr[i+1]  < x$kmr[i]) {
>         x$movimento[i+1] <- "jusante"
>       } else if (x$kmr[i+1] > x$kmr[i]) {
>         x$movimento[i+1] <- "montante"
>       } else {
>         x$movimento[i+1] <- "parado"
>       }
>
>     }
>
>   }
>
>   return (x)
> }
>
> Worked pretty well with just one ID, but with many IDs the function didn´t
> detach different IDs.
>
>          ID    kmr movimento
> 5    10.700 314.20      <NA>
> 1    10.700 278.74   jusante
> 2    10.700 278.74    parado
> 3    10.700 278.74    parado
> 4    10.700 278.74    parado
> 494 100.700 269.94   jusante <-- this should be <NA>
> 500 100.700 278.74  montante
> 499 100.700 314.20  montante
> 495 100.700 278.74   jusante
> 498 100.700 278.74    parado
> 496 100.700 255.40   jusante
> 497 100.700 255.10   jusante
>
> I also tried remove the first If condition and pass this function using lapply in
> the splitted original data-frame, but didn´t work as well.
>
> Some onde can help?
>
> Thanks in advanced,
>
> Raoni
>
> structure(list(ID = c("10.700", "10.700", "10.700", "10.700", "10.700",
> "100.700", "100.700", "100.700", "100.700", "100.700", "100.700", "100.700"),
> kmr = c(314.2, 278.74, 278.74, 278.74, 278.74, 269.94, 278.74, 314.2, 278.74,
> 278.74, 255.4, 255.1),
>     movimento = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
>     NA)), .Names = c("ID", "kmr", "movimento"), row.names = c(5L, 1L, 2L, 3L,
> 4L, 494L, 500L, 499L, 495L, 498L, 496L, 497L), class = "data.frame")
>
> --
> Raoni Rosa Rodrigues
> Research Associate of Fish Transposition Center CTPeixes Universidade
> Federal de Minas Gerais - UFMG Brasil rodrigues.raoni at gmail.com
>
> ______________________________________________
> R-help at r-project.org 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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