r: how to replace same specific part in many column names simultaneously

Question

I have my dataset t comprising 1850 columns. Most of these contain "fu1" in their name. I want to replace all fu1 to "v1"

A sample of t looks like:

> t
  symp_pre_rh_fu1___1 symp_pre_rh_fu1___2 symp_pre_rh_fu1___3 symp_pre_rh_fu1___4
1                   0                   0                   0                   0
2                   1                   0                   0                   0
3                   1                   1                   0                   1
4                   1                   1                   0                   0
5                   0                   0                   0                   0

But I want it to print:

> t
  symp_pre_rh_v1___1 symp_pre_rh_v1___2 symp_pre_rh_v1___3 symp_pre_rh_v1___4
1                   0                   0                   0                   0
2                   1                   0                   0                   0
3                   1                   1                   0                   1
4                   1                   1                   0                   0
5                   0                   0                   0                   0

I would prefer a solution in dplyr if possible.

t <- structure(list(symp_pre_rh_fu1___1 = c(0L, 1L, 1L, 1L, 0L), symp_pre_rh_fu1___2 = c(0L, 
0L, 1L, 1L, 0L), symp_pre_rh_fu1___3 = c(0L, 0L, 0L, 0L, 0L), 
    symp_pre_rh_fu1___4 = c(0L, 0L, 1L, 0L, 0L), symp_pre_rh_fu1___5 = c(0L, 
    0L, 1L, 0L, 0L)), row.names = c(NA, -5L), class = "data.frame")

names(t) <- sub("fu1", "v1", names(t)) or even names(t) <- gsub("fu1", "v1", names(t)) but use sub in this case — Onyambu
– Onyambu, Commented Dec 16, 2020 at 19:38

Ian Campbell · Accepted Answer · 2020-12-16 19:51:20Z

1

@Onyambu provided the easy way. A tidyverse approach might be to use dplyr::rename_all.

The second argument is .funs which will accept a function to apply to the names. In this case, we can use stringr::str_replace to replace the string.

library(dplyr)
library(stringr)
t %>% rename_all(~str_replace(.,"fu1","v1"))
  symp_pre_rh_v1___1 symp_pre_rh_v1___2 symp_pre_rh_v1___3 symp_pre_rh_v1___4 symp_pre_rh_v1___5
1                  0                  0                  0                  0                  0
2                  1                  0                  0                  0                  0
3                  1                  1                  0                  1                  1
4                  1                  1                  0                  0                  0
5                  0                  0                  0                  0                  0

answered Dec 16, 2020 at 19:51

Ian Campbell

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1 Comment

cmirian Over a year ago

Thank you, exactly what I requested - a dplyr solution.

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