Conditional replacement of column name in tibble using dplyr

Question

I have the following tibble:

    df <- structure(list(gene_symbol = c("0610005C13Rik", "0610007P14Rik", 
"0610009B22Rik", "0610009L18Rik", "0610009O20Rik", "0610010B08Rik"
), foo.control.cv = c(1.16204038288333, 0.120508045270669, 0.205712615954009, 
0.504508040948641, 0.333956330117591, 0.543693011377001), foo.control.mean = c(2.66407458486012, 
187.137728870855, 142.111269303428, 16.7278587043453, 69.8602872478098, 
4.77769028710622), foo.treated.cv = c(0.905769898934564, 0.186441944401973, 
0.158552512842753, 0.551955061149896, 0.15743983656006, 0.290447431974039
), foo.treated.mean = c(2.40658723367692, 180.846795140269, 139.054032348287, 
11.8584348984435, 76.8141734599118, 2.24088124240385)), .Names = c("gene_symbol", 
"foo.control.cv", "foo.control.mean", "foo.treated.cv", "foo.treated.mean"
), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, 
6L))

Which looks like this:

# A tibble: 6 × 5
    gene_symbol foo.control.cv foo.control.mean foo.treated.cv foo.treated.mean
*         <chr>          <dbl>            <dbl>          <dbl>            <dbl>
1 0610005C13Rik      1.1620404         2.664075      0.9057699         2.406587
2 0610007P14Rik      0.1205080       187.137729      0.1864419       180.846795
3 0610009B22Rik      0.2057126       142.111269      0.1585525       139.054032
4 0610009L18Rik      0.5045080        16.727859      0.5519551        11.858435
5 0610009O20Rik      0.3339563        69.860287      0.1574398        76.814173
6 0610010B08Rik      0.5436930         4.777690      0.2904474         2.240881

What I want to do is to replace all column names with mean in it into mean_expr. Resulting in

    gene_symbol foo.control.cv foo.control.mean_expr foo.treated.cv foo.treated.mean_expr

1 0610005C13Rik      1.1620404         2.664075      0.9057699         2.406587
2 0610007P14Rik      0.1205080       187.137729      0.1864419       180.846795
3 0610009B22Rik      0.2057126       142.111269      0.1585525       139.054032
4 0610009L18Rik      0.5045080        16.727859      0.5519551        11.858435
5 0610009O20Rik      0.3339563        69.860287      0.1574398        76.814173
6 0610010B08Rik      0.5436930         4.777690      0.2904474         2.240881

How can I achieve that?

Answer 1

With current versions of dplyr, you can use rename_at:

library(dplyr)

df %>% rename_at(vars(contains('mean')), funs(sub('mean', 'mean_expr', .)))
#> # A tibble: 6 × 5
#>     gene_symbol foo.control.cv foo.control.mean_expr foo.treated.cv
#> *         <chr>          <dbl>                 <dbl>          <dbl>
#> 1 0610005C13Rik      1.1620404              2.664075      0.9057699
#> 2 0610007P14Rik      0.1205080            187.137729      0.1864419
#> 3 0610009B22Rik      0.2057126            142.111269      0.1585525
#> 4 0610009L18Rik      0.5045080             16.727859      0.5519551
#> 5 0610009O20Rik      0.3339563             69.860287      0.1574398
#> 6 0610010B08Rik      0.5436930              4.777690      0.2904474
#> # ... with 1 more variables: foo.treated.mean_expr <dbl>

Really, you could use rename_all, as well, as names that don't match would be unaffected anyway. Further, you can use a quosure or anything that can be coerced to a function by rlang::as_function for .funs, so you can use purrr-style notation:

df %>% rename_all(~sub('mean', 'mean_expr', .x))

Since a data frame is a list, purrr's set_names can do the same thing:

library(purrr)    # or library(tidyverse)

df %>% set_names(~sub('mean', 'mean_expr', .x))

All return the same thing.

Answer 2

Another option is to paste in rename_at (using the devel version of dplyr)

library(dplyr)
df %>%
    rename_at(vars(matches('mean')), funs(sprintf('%s_expr', .)))
# A tibble: 6 × 5
#    gene_symbol foo.control.cv foo.control.mean_expr foo.treated.cv foo.treated.mean_expr
#*         <chr>          <dbl>                 <dbl>          <dbl>                 <dbl>
#1 0610005C13Rik      1.1620404              2.664075      0.9057699              2.406587    
#2 0610007P14Rik      0.1205080            187.137729      0.1864419            180.846795
#3 0610009B22Rik      0.2057126            142.111269      0.1585525            139.054032
#4 0610009L18Rik      0.5045080             16.727859      0.5519551             11.858435
#5 0610009O20Rik      0.3339563             69.860287      0.1574398             76.814173
#6 0610010B08Rik      0.5436930              4.777690      0.2904474              2.240881

---

Or using rename_if

df %>%
   rename_if(grepl("mean", names(.)), funs(sprintf("%s_expr", .)))

Answer 3

Here is a non-dplyr base R method:

names(df) <- sub("mean$", "mean_expr", names(df))
# or names(df) <- sub("mean", "mean_expr", names(df)) if the mean doesn't have to be at the 
# end of the string

names(df)
#[1] "gene_symbol"           "foo.control.cv"        "foo.control.mean_expr"
#[4] "foo.treated.cv"        "foo.treated.mean_expr"

If you want it to be a part of the pipe, you can make use of setNames function:

df %>% setNames(sub("mean", "mean_expr", names(.))) %>% names(.)
#[1] "gene_symbol"           "foo.control.cv"        "foo.control.mean_expr"
#[4] "foo.treated.cv"        "foo.treated.mean_expr"