替换 R 数据框中因子列的内容

Question

I need to replace the levels of a factor column in a dataframe. Using the iris dataset as an example, how would I replace any cells which contain virginica with setosa in the Species column?

I expected the following to work, but it generates a warning message and simply inserts NAs:

iris$Species[iris$Species == 'virginica'] <- 'setosa'

Answer 1

I bet the problem is when you are trying to replace values with a new one, one that is not currently part of the existing factor's levels:

levels(iris$Species)
# [1] "setosa"     "versicolor" "virginica" 

Your example was bad, this works:

iris$Species[iris$Species == 'virginica'] <- 'setosa'

This is what more likely creates the problem you were seeing with your own data:

iris$Species[iris$Species == 'virginica'] <- 'new.species'
# Warning message:
# In `[<-.factor`(`*tmp*`, iris$Species == "virginica", value = c(1L,  :
#   invalid factor level, NAs generated

It will work if you first increase your factor levels:

levels(iris$Species) <- c(levels(iris$Species), "new.species")
iris$Species[iris$Species == 'virginica'] <- 'new.species'

---

If you want to replace "species A" with "species B" you'd be better off with

levels(iris$Species)[match("oldspecies",levels(iris$Species))] <- "newspecies"

Answer 2

For the things that you are suggesting you can just change the levels using the levels :

levels(iris$Species)[3] <- 'new'

Answer 3

You can use the function revalue from the package plyr to replace values in a factor vector.

In your example to replace the factor virginica by setosa :

 data(iris)
 library(plyr)
 revalue(iris$Species, c("virginica" = "setosa")) -> iris$Species

Answer 4

I had the same problem. This worked better:

Identify which level you want to modify: levels(iris$Species)

    "setosa" "versicolor" "virginica" 

So, setosa is the first.

Then, write this:

     levels(iris$Species)[1] <-"new name"

Answer 5

Using dlpyr::mutate and forcats::fct_recode :

library(dplyr)
library(forcats)

iris <- iris %>%  
  mutate(Species = fct_recode(Species,
    "Virginica" = "virginica",
    "Versicolor" = "versicolor"
  )) 

iris %>% 
  count(Species)

# A tibble: 3 x 2
     Species     n
      <fctr> <int>
1     setosa    50
2 Versicolor    50
3  Virginica    50   

Answer 6

A more general solution that works with all the data frame at once and where you don't have to add new factors levels is:

data.mtx <- as.matrix(data.df)
data.mtx[which(data.mtx == "old.value.to.replace")] <- "new.value"
data.df <- as.data.frame(data.mtx)

A nice feature of this code is that you can assign as many values as you have in your original data frame at once, not only one "new.value" , and the new values can be random values. Thus you can create a complete new random data frame with the same size as the original.

Answer 7

You want to replace the values in a dataset column, but you're getting an error like this:

invalid factor level, NA generated

Try this instead:

levels(dataframe$column)[levels(dataframe$column)=='old_value'] <- 'new_value'

Answer 8

In case you have to replace multiple values and if you don't mind "refactoring" your variable with as.factor(as.character(...)) you could try the following:

replace.values <- function(search, replace, x){
  stopifnot(length(search) == length(replace))
  xnew <- replace[ match(x, search) ]
  takeOld <- is.na(xnew) & !is.na(x)
  xnew[takeOld] <- x[takeOld]
  return(xnew)
}

iris$Species <- as.factor(search=c("oldValue1","oldValue2"),
                          replace=c("newValue1","newValue2"),
                          x=as.character(iris$Species))