How to replace values in df that are NOT fulfilling condition

Question

I am trying to clean my data and can't find a way to replace values that are not according to my conditions like in the following example:

df1<-  data.frame(
A=c("a","b","c"),
Ch=c("c1","xyz","c2"),
val=paste0("x",1:3),  stringsAsFactors = FALSE)

all values that are different than c1, c2 I want to change into "other".

I tried:

for( i in 1:length(df)
if (df[i,2]==c1 | c2){
stay same vaue?!?
} else df[i,2] <- "other"

which did not work.

Any suggestions?

Feel free to mark the answer as accepted if it does what you need it to do :) — costebk08
– costebk08, Commented Dec 10, 2019 at 3:54

costebk08 · Accepted Answer · 2019-12-09 20:55:29Z

3

For replicability:

df1<-
tibble(
  A=c("a","b","c"),
  Ch=c("c1","xyz","c2"),
  val=paste0("x",1:3)
)

If you are trying to do it across all columns:

library(tidyverse)

df1%>%mutate_all(list(~ifelse(.=="c1"|.=="c2",.,"other")))

Base R solution:

data.frame(
sapply(
  df1,
  function(x)ifelse(x=="c1"|x=="c2",x,"other")
 )
)

answered Dec 9, 2019 at 20:46

costebk08

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1 Comment

In base R, you can just do df1[df1!="c1" & df1!="c2"] <- "other", assuming that those columns are strings and not factors.

XXavier · Accepted Answer · 2019-12-09 21:08:22Z

0

Here is another base R solution if you want to change only one column

df1<-  data.frame(
A=c("a","b","c"),
Ch=c("c1","xyz","c2"),
val=paste0("x",1:3),  stringsAsFactors = FALSE)

df1$Ch[df1$Ch!="c1" & df1$Ch!="c2"] <- "other"

This will replace the column Ch only

answered Dec 9, 2019 at 21:08

XXavier

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