How to Change Factor Labels into String in a Data Frame

Convert data.frame columns from factors to characters

Just following on Matt and Dirk. If you want to recreate your existing data frame without changing the global option, you can recreate it with an apply statement:

bob <- data.frame(lapply(bob, as.character), stringsAsFactors=FALSE)

This will convert all variables to class "character", if you want to only convert factors, see Marek's solution below.

As @hadley points out, the following is more concise.

bob[] <- lapply(bob, as.character)

In both cases, lapply outputs a list; however, owing to the magical properties of R, the use of [] in the second case keeps the data.frame class of the bob object, thereby eliminating the need to convert back to a data.frame using as.data.frame with the argument stringsAsFactors = FALSE.

convert factor and character to numeric in a dataframe

It would help to have some example data to work with, but try:

df$your_factor_variable_now_numeric <- 
                as.numeric(as.character(df$your_old_factor_variable))

And use it only to convert a factor variable, not the complete dataframe. You can also have a look at type.convert. If you want to convert all factors in the dataframe, you can use something along the lines

 df[] <- lapply(df, function(x) as.numeric(as.character(x)))

Note that this converts all factors and might not be what you want if you have factors that do not represent numeric values. If unnecessary conversion is a problem, or if there are non-numeric factors or characters in the data, the following would be appropriate:

 numerify <- function(x) if(is.factor(x)) as.numeric(as.character(x)) else x
 df[] <- lapply(df, numerify)

On a more general point though, the type of your variables should not prevent you from filtering, if, with filtering, you mean subsetting the dataframe. However, the type conversion should be solved with the above code.

Change factor levels and rearrange dataframe

This mistakes is easy to make. You have to supply the column vector to fct_relevel. Like so:

library(dplyr,warn.conflicts = F)
library(forcats)

df <-
  structure(
    list(layer = structure(
      1:5,
      .Label = c(
        'CEOS and managers',
        'Clerks and services',
        'Production',
        'Professionals',
        'Technicians'
      ),
      class = 'factor'
    )),
    row.names = c(NA,-5L),
    class = c('tbl_df', 'tbl', 'data.frame')
  )

df %>%
  mutate(layer = forcats::fct_relevel(
    layer,c(
      'CEOS and managers',
      'Professionals',
      'Technicians',
      'Clerks and services',
      'Production'))) %>% 
  arrange(layer)
#> # A tibble: 5 x 1
#>   layer              
#>   <fct>              
#> 1 CEOS and managers  
#> 2 Professionals      
#> 3 Technicians        
#> 4 Clerks and services
#> 5 Production

^{Created on 2021-01-11 by the reprex package (v0.3.0)}

How to convert data.frame column from Factor to numeric

breast$class <- as.numeric(as.character(breast$class))

If you have many columns to convert to numeric

indx <- sapply(breast, is.factor)
breast[indx] <- lapply(breast[indx], function(x) as.numeric(as.character(x)))

Another option is to use stringsAsFactors=FALSE while reading the file using read.table or read.csv

Just in case, other options to create/change columns

 breast[,'class'] <- as.numeric(as.character(breast[,'class']))

or

 breast <- transform(breast, class=as.numeric(as.character(breast)))

Replace contents of factor column in R dataframe

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"

Pandas: convert categories to numbers

First, change the type of the column:

df.cc = pd.Categorical(df.cc)

Now the data look similar but are stored categorically. To capture the category codes:

df['code'] = df.cc.cat.codes

Now you have:

   cc  temp  code
0  US  37.0     2
1  CA  12.0     1
2  US  35.0     2
3  AU  20.0     0

If you don't want to modify your DataFrame but simply get the codes:

df.cc.astype('category').cat.codes

Or use the categorical column as an index:

df2 = pd.DataFrame(df.temp)
df2.index = pd.CategoricalIndex(df.cc)

Need an efficient way to change factor values from one column of a data frame to another columns

We can use fct_collapse and it returns a factor with new levels

library(dplyr)
library(forcats)
library(magrittr)
df %<>%
    mutate(B = fct_collapse(B, CHANGED = as.character(B)[A== "Kelly"]))

glimpse(df)
#Rows: 7
#Columns: 2
#$ A <fct> Jerry, Kelly, Kelly, Lion, Zebra, Bear, Kelly
#$ B <fct> Eats, CHANGED, CHANGED, Roars, Runs, Sleeps, CHANGED

Convert the factors of a variable into the columns of the dataframe

Use an id variable for rows by group:

dat %>% 
  group_by(Concentration) %>% 
  mutate(id = row_number()) %>% 
  pivot_wider(names_from = Concentration, values_from = Value)

     id   Low Medium  High
  <int> <dbl>  <dbl> <dbl>
1     1  0.21   0.85  2.21
2     2  0.1    0.5   1.85
3     3  0.36  NA    NA   

Convert data.frame column format from character to factor

Hi welcome to the world of R.

mtcars  #look at this built in data set
str(mtcars) #allows you to see the classes of the variables (all numeric)

#one approach it to index with the $ sign and the as.factor function
mtcars$am <- as.factor(mtcars$am)
#another approach
mtcars[, 'cyl'] <- as.factor(mtcars[, 'cyl'])
str(mtcars)  # now look at the classes

This also works for character, dates, integers and other classes

Since you're new to R I'd suggest you have a look at these two websites:

R reference manuals:
http://cran.r-project.org/manuals.html

R Reference card: http://cran.r-project.org/doc/contrib/Short-refcard.pdf

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