Replace given value in vector

Question

I'm looking for a function which will replace all occurrences of one value with another value. For example I'd like to replace all zeros with ones. I don't want to have to store the result in a variable, but want to be able to use the vector anonymously as part of a larger expression.

I know how to write a suitable function myself:

> vrepl <- function(haystack, needle, replacement) {
+   haystack[haystack == needle] <- replacement
+   return(haystack)
+ }
> 
> vrepl(c(3, 2, 1, 0, 4, 0), 0, 1)
[1] 3 2 1 1 4 1

But I'm wondering whether there is some standard function to do this job, preferrably from the base package, as an alternative from some other commonly used package. I believe that using such a standard will likely make my code more readable, and I won't have to redefine that function wherever I need it.

Answer 1

Perhaps replace is what you are looking for:

> x = c(3, 2, 1, 0, 4, 0)
> replace(x, x==0, 1)
[1] 3 2 1 1 4 1

Or, if you don't have x (any specific reason why not?):

replace(c(3, 2, 1, 0, 4, 0), c(3, 2, 1, 0, 4, 0)==0, 1)

Many people are familiar with gsub, so you can also try either of the following:

as.numeric(gsub(0, 1, x))
as.numeric(gsub(0, 1, c(3, 2, 1, 0, 4, 0)))

Update

After reading the comments, perhaps with is an option:

with(data.frame(x = c(3, 2, 1, 0, 4, 0)), replace(x, x == 0, 1))

Answer 2

Another simpler option is to do:

 > x = c(1, 1, 2, 4, 5, 2, 1, 3, 2)
 > x[x==1] <- 0
 > x
 [1] 0 0 2 4 5 2 0 3 2

Answer 3

A simple way to do this is using base ifelse, which is vectorized (there's also if_else from dplyr that handles missing values). If the condition is satisfied, we use a replacement value, otherwise we use the original value.

v <- c(3, 2, 1, 0, 4, 0)
ifelse(v == 0, 1, v)

We can avoid a named variable by using a pipe.

c(3, 2, 1, 0, 4, 0) %>% ifelse(. == 0, 1, .)

A common task is to do multiple replacements. Instead of nested ifelse statements, we can use case_when from dplyr:

case_when(v == 0 ~ 1,
          v == 1 ~ 2,
          TRUE ~ v)

---

Old answer:

For factor or character vectors, we can use revalue from plyr:

> revalue(c("a", "b", "c"), c("b" = "B"))
[1] "a" "B" "c"

This has the advantage of only specifying the input vector once, so we can use a pipe like

x %>% revalue(c("b" = "B"))

Answer 4

To replace more than one number:

vec <- 1:10
replace(vec, vec== c(2,6), c(0,9)) #2 and 6 will be replaced by 0 and 9.

Edit:

for a continous series, you can do this vec <- c(1:10); replace(vec, vec %in% c(2,6), c(0,9)) but for vec <- c(1:10,2,2,2); replace(vec, vec %in% c(2,6), 0) we can replace multiple values with one value.

Answer 5

Why the fuss?

replace(haystack, haystack %in% needles, replacements)

Demo:

haystack <- c("q", "w", "e", "r", "t", "y")
needles <- c("q", "w")
replacements <- c("a", "z")

replace(haystack, haystack %in% needles, replacements)
#> [1] "a" "z" "e" "r" "t" "y"

Answer 6

If someone is looking for a dplyr alternative to replace, case_match is a simple variant to case_when that works on vectors. It is available since 1.1.0.

x = c(3, 2, 1, 0, 4, 0)
case_match(x, 0 ~ 1, .default = x)
#[1] 3 2 1 1 4 1

It can also work with multiple replacements:

x = c(3, 2, 1, 0, 4, 0)
case_match(x, 
           c(0, 2, 4) ~ "even",
           c(1, 3) ~ "odd")
#[1] "odd"  "even" "odd"  "even" "even" "even"

Answer 7

The ifelse function would be a quick and easy way to do this.

Answer 8

If you want to replace lot of values in single go, you can use 'library(car)'.

Example

library(car)

x <- rep(1:5,3)

xr <- recode(x, '3=1; 4=2')

x
## [1] 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5
xr
## [1] 1 2 1 2 5 1 2 1 2 5 1 2 1 2 5