R library(BradleyTerry2). Error in Diff(player1, player2... must be factors with the same levels

Question

I have a problem using BradleyTerry2 package in R. My data looks like as below. I created the data from another code, and I think the format is same to the "citations.sf" in BradleyTerry exmple (https://cran.r-project.org/web/packages/BradleyTerry2/vignettes/BradleyTerry.pdf)

   player1 player2 win1 win2
1        1       2   10    0
2        1       3   10    0
3        1       4    5    5
4        1       5   10    0
5        1       6    9    1
6        2       3    6    4
7        2       4    4    6
8        2       5    5    5
9        2       6    8    2
10       3       4    2    8
11       3       5    7    3
12       3       6    6    4
13       4       5   10    0
14       4       6    9    1
15       5       6    4    6

However, when I run speedModel <- BTm(cbind(win1, win2), player1, player2, data = dat), It shows an error message as below.

Error in Diff(player1, player2, formula, id, data, separate.ability, refcat, : 'player1$..' and 'player2$..' must be factors with the same levels

I looked at another page on StackOverflow (Updated with data: Error in Diff...must be factors with the same levels), and I tried the below code. (I don't understand what it does, though.)

levels(dat[,1]) <- dat(c(dat[,1], dat[,2]))
levels(dat[,2]) <- dat(c(dat[,1], dat[,2]))

However, the BTm() function throws the same message. Can anyone tell me what I can do?

Here is the result of dput(dat)

structure(list(player1 = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 
2L, 2L, 2L, 3L, 3L, 3L, 4L, 4L, 5L), .Label = 1:6), player2 = structure(c(2L, 
3L, 4L, 5L, 6L, 3L, 4L, 5L, 6L, 4L, 5L, 6L, 5L, 6L, 6L), .Label = 1:6), 
win1 = c(10L, 10L, 5L, 10L, 9L, 6L, 4L, 5L, 8L, 2L, 7L, 6L, 
10L, 9L, 4L), win2 = c(0L, 0L, 5L, 0L, 1L, 4L, 6L, 5L, 2L, 
8L, 3L, 4L, 0L, 1L, 6L)), row.names = c(NA, -15L), class = "data.frame")

Answer 1

Your problem is with the definition of the player factors in your data frame. They need to have the exact same levels but because player1 only contains the values 1, ..., 5, while player2 contains the values 2, ..., 6 then they will have different levels.

You need to force the levels to be identical for the two factors by providing the exact levels. Here's one way to do it by creating two new variables with the right factor levels.

dat$p1 <- factor(dat$player1, levels=unique(c(dat$player1, dat$player2)))  
dat$p2 <- factor(dat$player2, levels=unique(c(dat$player1, dat$player2)))  

Then we can run

> speedModel <- BTm(cbind(win1, win2), p1, p2, data = dat)
> speedModel
Bradley Terry model fit by glm.fit 

Call:  BTm(outcome = cbind(win1, win2), player1 = p1, player2 = p2, 
    data = indata)

Coefficients:
    ..2      ..3      ..4      ..5      ..6  
-2.1433  -2.4885  -0.7286  -3.1201  -2.9323  

Degrees of Freedom: 15 Total (i.e. Null);  10 Residual
Null Deviance:      81.14 
Residual Deviance: 13.71    AIC: 51.6