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











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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")









share|improve this question


























    up vote
    0
    down vote

    favorite












    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")









    share|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      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")









      share|improve this 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")






      r






      share|improve this question













      share|improve this question











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      share|improve this question










      asked Nov 9 at 16:06









      pico123

      31




      31
























          1 Answer
          1






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          accepted










          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





          share|improve this answer





















          • Thank you ekstroem. It works well for me! I still don't understand what is happening, because p1 and p2 look same to player1 and player2 on the matrix. ...R is fascinating!
            – pico123
            Nov 14 at 12:03












          • You should read up on R factors and their factor levels - that's where the problem was
            – ekstroem
            Nov 14 at 19:06










          • If it works then please accept the answer.
            – ekstroem
            Nov 14 at 19:06










          • Sorry, I didn't know the function. I just did and thank you again!
            – pico123
            Nov 16 at 9:36











          Your Answer






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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          0
          down vote



          accepted










          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





          share|improve this answer





















          • Thank you ekstroem. It works well for me! I still don't understand what is happening, because p1 and p2 look same to player1 and player2 on the matrix. ...R is fascinating!
            – pico123
            Nov 14 at 12:03












          • You should read up on R factors and their factor levels - that's where the problem was
            – ekstroem
            Nov 14 at 19:06










          • If it works then please accept the answer.
            – ekstroem
            Nov 14 at 19:06










          • Sorry, I didn't know the function. I just did and thank you again!
            – pico123
            Nov 16 at 9:36















          up vote
          0
          down vote



          accepted










          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





          share|improve this answer





















          • Thank you ekstroem. It works well for me! I still don't understand what is happening, because p1 and p2 look same to player1 and player2 on the matrix. ...R is fascinating!
            – pico123
            Nov 14 at 12:03












          • You should read up on R factors and their factor levels - that's where the problem was
            – ekstroem
            Nov 14 at 19:06










          • If it works then please accept the answer.
            – ekstroem
            Nov 14 at 19:06










          • Sorry, I didn't know the function. I just did and thank you again!
            – pico123
            Nov 16 at 9:36













          up vote
          0
          down vote



          accepted







          up vote
          0
          down vote



          accepted






          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





          share|improve this answer












          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






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 9 at 19:49









          ekstroem

          3,52021230




          3,52021230












          • Thank you ekstroem. It works well for me! I still don't understand what is happening, because p1 and p2 look same to player1 and player2 on the matrix. ...R is fascinating!
            – pico123
            Nov 14 at 12:03












          • You should read up on R factors and their factor levels - that's where the problem was
            – ekstroem
            Nov 14 at 19:06










          • If it works then please accept the answer.
            – ekstroem
            Nov 14 at 19:06










          • Sorry, I didn't know the function. I just did and thank you again!
            – pico123
            Nov 16 at 9:36


















          • Thank you ekstroem. It works well for me! I still don't understand what is happening, because p1 and p2 look same to player1 and player2 on the matrix. ...R is fascinating!
            – pico123
            Nov 14 at 12:03












          • You should read up on R factors and their factor levels - that's where the problem was
            – ekstroem
            Nov 14 at 19:06










          • If it works then please accept the answer.
            – ekstroem
            Nov 14 at 19:06










          • Sorry, I didn't know the function. I just did and thank you again!
            – pico123
            Nov 16 at 9:36
















          Thank you ekstroem. It works well for me! I still don't understand what is happening, because p1 and p2 look same to player1 and player2 on the matrix. ...R is fascinating!
          – pico123
          Nov 14 at 12:03






          Thank you ekstroem. It works well for me! I still don't understand what is happening, because p1 and p2 look same to player1 and player2 on the matrix. ...R is fascinating!
          – pico123
          Nov 14 at 12:03














          You should read up on R factors and their factor levels - that's where the problem was
          – ekstroem
          Nov 14 at 19:06




          You should read up on R factors and their factor levels - that's where the problem was
          – ekstroem
          Nov 14 at 19:06












          If it works then please accept the answer.
          – ekstroem
          Nov 14 at 19:06




          If it works then please accept the answer.
          – ekstroem
          Nov 14 at 19:06












          Sorry, I didn't know the function. I just did and thank you again!
          – pico123
          Nov 16 at 9:36




          Sorry, I didn't know the function. I just did and thank you again!
          – pico123
          Nov 16 at 9:36


















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