Joining factor levels of two columns

Question

I have 2 columns of data with the same type of data (Strings).

I want to join the levels of the columns. ie. we have:

col1   col2
Bob    John
Tom             


        

Answer 1

You want the factors to include all the unique names from both columns.

col1 <- factor(c("Bob", "Tom", "Frank", "Jim", "Tom"))
col2 <- factor(c("John", "Bob", "Jane", "Bob", "Bob"))
mynames <- unique(c(levels(col1), levels(col2)))
fcol1 <- factor(col1, levels = mynames)
fcol2 <- factor(col2, levels = mynames)

EDIT: a little nicer if you replace the third line with this:

mynames <- union(levels(col1), levels(col2))

Answer 2

Could have sworn this didn't work when I was writing the abomination below, but it does now:

## self contained example:
txt <- "col1   col2
Bob    John
Tom    Bob
Frank  Jane
Jim    Bob
Tom    Bob"
dat <- read.table(textConnection(txt), header = TRUE)

Just compute unique set of levels and coerce each colX to a factor:

> dat3 <- dat
> lev <- as.character(unique(unlist(sapply(dat, levels))))
> dat3 <- within(dat3, col1 <- factor(col1, levels = lev))
> dat3 <- within(dat3, col2 <- factor(col2, levels = lev))
> str(dat3)
'data.frame':   5 obs. of  2 variables:
 $ col1: Factor w/ 6 levels "Bob","Tom","Frank",..: 1 2 3 4 2
 $ col2: Factor w/ 6 levels "Bob","Tom","Frank",..: 5 1 6 1 1
> data.matrix(dat3)
     col1 col2
[1,]    1    5
[2,]    2    1
[3,]    3    6
[4,]    4    1
[5,]    2    1

[Original: to show how stupidly complex and obfuscated one can write R code it one tries really hard!] Not sure this is particularly elegant (and it isn't), but...

We first unlist the data:

tmp <- unlist(dat)

then compute the unique levels

lev <- as.character(unique(tmp))

and then restructure tmp (from above) back into the same dimensions as the original data, convert to data.frame (preserving the strings), lapply over this data frame, creating a factor with levels lev computed above, and finally coerce to a data frame.

dat2 <- data.frame(lapply(data.frame(matrix(tmp, ncol = ncol(dat)), 
                                     stringsAsFactors = FALSE), 
                          FUN = factor, levels = lev))

Which gives:

> dat2
     X1   X2
1   Bob John
2   Tom  Bob
3 Frank Jane
4   Jim  Bob
5   Tom  Bob
> sapply(dat2, levels)
     X1      X2     
[1,] "Bob"   "Bob"  
[2,] "Tom"   "Tom"  
[3,] "Frank" "Frank"
[4,] "Jim"   "Jim"  
[5,] "John"  "John" 
[6,] "Jane"  "Jane" 
> data.matrix(dat2)
     X1 X2
[1,]  1  5
[2,]  2  1
[3,]  3  6
[4,]  4  1
[5,]  2  1

Answer 3

x <- structure(list(col1 = structure(c(1L, 4L, 2L, 3L, 4L), .Label = c("Bob", "Frank", "Jim", "Tom"), class = "factor"), col2 = structure(c(3L, 1L, 2L, 1L, 1L), .Label = c("Bob", "Jane", "John"), class = "factor")), .Names = c("col1", "col2"), class = "data.frame", row.names = c(NA, -5L))

Make a simple union of factor names:

both <- union(levels(x$col1), levels(x$col2))

And relevel the two factors:

x$col1 <- factor(x$col1, levels=both)
x$col2 <- factor(x$col2, levels=both)

---

After editing: added example to make numeric values from factors

You could simply transform the factor levels to numeric values, e.g.:

as.numeric(x$col1)

Or a more simpler, nicer solution based on @Gavin Simpson's hint below in one step:

data.matrix(x)