Executing a function with paste to create a new variable in a dataframe in R

Question

Data:

structure(list(`p value` = c(0.00151124736422317, 0.804709799937324, 
0.0192537412780042, 0.000467854188597731, 4.80216666553605e-06, 
0.0231434946595433), significance = c(TRUE, FALSE, TRUE, TRUE, 
TRUE, TRUE)), .Names = c("p value", "significance"), row.names = c("Q5.i", 
"Q5.ii", "Q5.iii", "Q5.iv", "Q5.v", "Q5.vi"), class = "data.frame")

Objective: To create a function that would take input of dataframe name and a (new) variabe name. The function would:

create a new variable that is based on the row name of the dataframe
delete the row name
reorder the variable so that the newly created column is first column

Challenges: I am stuck at the first step.

I've searched the internet and stackoverflow for snippets of code that would help and I've managed to hammer something although it couldn't work. What have I tried:

row2col<-function(df, varname){
 eval(parse(text=paste(df, "[[", "'", varname, "'", "]]", "<-row.names(", df, ")", sep="")))
}


row2col<-function(df, varname){
  assign(parse(text=paste(df, varname, sep="$")), row.names(df))
}

Results:

nothing happened (not even an error message)
a character vector of row names (rather than a variable within the dataframe) was created

Thanks for your help and attention to this post.

R rule of thumb - if you are ever using eval assign parse.... don't. :-) — thelatemail
– thelatemail, Commented Jun 28, 2016 at 6:34
This function exists already, as dplyr::add_rownames() and tibble::rownames_to_column() — alistaire
– alistaire, Commented Jun 28, 2016 at 6:53
@alistaire Thanks for your sharing. I use the dplyr package extensively and didn't even notice its already in there. — Raphael Lee
– Raphael Lee, Commented Jun 28, 2016 at 6:55

talat · Accepted Answer · 2016-06-28 06:21:12Z

4

You don't need to use eval, parse, assign - that's in many cases not the right approach. Here's a simple alternative:

row2col <- function(dat, varname) {
  dat[[varname]] <- row.names(dat)  
  row.names(dat) <- NULL
  dat[, c(varname, setdiff(names(dat), varname))]
}

And then you can test it:

> row2col(df, "testcol")
#  testcol      p value significance
#1    Q5.i 1.511247e-03         TRUE
#2   Q5.ii 8.047098e-01        FALSE
#3  Q5.iii 1.925374e-02         TRUE
#4   Q5.iv 4.678542e-04         TRUE
#5    Q5.v 4.802167e-06         TRUE
#6   Q5.vi 2.314349e-02         TRUE

answered Jun 28, 2016 at 6:21

talat

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1 Comment

Raphael Lee Over a year ago

And I struggled for half a day with the first objective only...thanks very very much!!!

mkearney · Accepted Answer · 2016-06-28 06:23:02Z

0

Create new var using row names.

data$new_var <- row.names(data)

Reset row names

row.names(data) <- NULL

Reorder data frame with new var first

data <- data[, c(ncol(data):(ncol(data) - 1))]

answered Jun 28, 2016 at 6:23

mkearney

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