Using dplyr functions on variables named "."

Question

Sometimes when generating a data frame from a list, the variable is named "." by default. How can I refer to this variable within dplyr functions, if only to change the variable name to something more appropriate.

# Code that produces my data frame with "." as column name
library(tidyverse)

d <- data.frame(`.` = 1, row.names = "a") 

# Now my code fails because `.` is a poor column name for dplyr functions:
d %>% select(model = rownames(.), outlier = `.`)

If you're open to changing the column names as B Williams suggested above, this post includes various approaches: stackoverflow.com/questions/6081439/… — Z.Lin
– Z.Lin, Commented Aug 31, 2017 at 1:43
. will cause more problems than just with dplyr. Change the variable name. — Hong Ooi
– Hong Ooi, Commented Aug 31, 2017 at 1:49
Hello @HongOoi, I agree, this is just a reprex. The output from another function is naming the column ".". — Joe
– Joe, Commented Sep 1, 2017 at 4:52

Sarah · Accepted Answer · 2017-08-31 01:42:34Z

1

This isn't actually a problem with the column named . its a problem with referencing the rownames in select() see

d <- data.frame(test = 1, row.names = "a")
d %>% select(model = rownames(.), outlier = test)

still returns Error: Strings must match column names. Unknown columns: a

just use

d <- data.frame(`.` = 1, row.names = "a") 
d %>% select(outlier = '.')

will rename the column to outlier

answered Aug 31, 2017 at 1:42

Sarah

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Rich Pauloo · Accepted Answer · 2017-09-01 06:04:41Z

0

Given

d <- data.frame(`.` = 1, row.names = "a")

Base R Solution

colnames(d) <- 'newname'

Dplyr Solution

d %>% rename(newname = '.')

edited Sep 1, 2017 at 6:04

answered Aug 31, 2017 at 1:48

Rich Pauloo

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4 Comments

Rich Pauloo Over a year ago

With the dplyr solution, you can rename and pipe onwards!

alistaire Over a year ago

setNames/purrr::set_names pipes nicely

Joe Over a year ago

When I use your dplyr solution I get "Error in captureDots(strict = __quosured) : the argument has already been evaluated"

Rich Pauloo Over a year ago

@Joe, you're right. I forgot to include quote marks, so I updated the answer. That dplyr solution should work now. Try it out. :)

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