Kappa applied to all pairwise combinations of raters, and store the result












0















For a visualisation, I need to compute the weighted kappa of all combinations / pairs of seven raters. So, if I use some sample data with seven columns:



ex <- structure(list(`1` = c(1, 1, 2, 1, 2, 1, 1, 1, 2, 3, 1, 1, 1, 

1, 1, 1, 2, 1, 2, 2), `2` = c(1, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 

2, 1, 1, 2, 2, 1, 2, 2, 2), `3` = c(1, 1, 2, 1, 2, 1, 1, 1, 2, 

3, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2), `4` = c(1, 1, 2, 2, 2, 1, 2, 

2, 2, 3, 2, 1, 1, 2, 2, 2, 1, 1, 2, 2), pa = c(1, 2, 1, 2, 3, 

1, 2, 2, 2, 2, 2, 2, 1, 2, 1, 2, 2, 1, 3, 2), ta = c(2, 2, 2, 

1, 3, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 1, 1, 3), ka = c(1, 

1, 2, 1, 2, 1, 1, 1, 2, 3, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2)), row.names = c(NA, -20L), class = c("tbl_df", "tbl", "data.frame"))


I would like to get a structure that captures the output of irr::kappa2 on all combinations of columns. A structure like:



out <- data.frame("1_2"=irr::kappa2(ex[c(1,2)]),

"1_3"=irr::kappa2(ex[c(1,3)]),"1_4"=irr::kappa2(ex[c(1,3)]),....


(for all unique combinations of columns).



Any ideas?






r functional-programming tidyverse







share|improve this question























  • Just to be clear, do you need the whole output of irr::kappa2() or just the value of Kappa?

    – Ric S
    Jan 19 at 16:05











  • The kappa value is the what I want primarily.

    – Fredrik Karlsson
    Jan 19 at 16:16











  • Ok then, I'll edit my answer below in the case you want only the kappa value

    – Ric S
    Jan 19 at 16:18











  • Great thank you very much!

    – Fredrik Karlsson
    Jan 19 at 16:24











  • You're welcome! :)

    – Ric S
    Jan 19 at 16:27
















0















For a visualisation, I need to compute the weighted kappa of all combinations / pairs of seven raters. So, if I use some sample data with seven columns:



ex <- structure(list(`1` = c(1, 1, 2, 1, 2, 1, 1, 1, 2, 3, 1, 1, 1, 

1, 1, 1, 2, 1, 2, 2), `2` = c(1, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 

2, 1, 1, 2, 2, 1, 2, 2, 2), `3` = c(1, 1, 2, 1, 2, 1, 1, 1, 2, 

3, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2), `4` = c(1, 1, 2, 2, 2, 1, 2, 

2, 2, 3, 2, 1, 1, 2, 2, 2, 1, 1, 2, 2), pa = c(1, 2, 1, 2, 3, 

1, 2, 2, 2, 2, 2, 2, 1, 2, 1, 2, 2, 1, 3, 2), ta = c(2, 2, 2, 

1, 3, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 1, 1, 3), ka = c(1, 

1, 2, 1, 2, 1, 1, 1, 2, 3, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2)), row.names = c(NA, -20L), class = c("tbl_df", "tbl", "data.frame"))


I would like to get a structure that captures the output of irr::kappa2 on all combinations of columns. A structure like:



out <- data.frame("1_2"=irr::kappa2(ex[c(1,2)]),

"1_3"=irr::kappa2(ex[c(1,3)]),"1_4"=irr::kappa2(ex[c(1,3)]),....


(for all unique combinations of columns).



Any ideas?






r functional-programming tidyverse







share|improve this question























  • Just to be clear, do you need the whole output of irr::kappa2() or just the value of Kappa?

    – Ric S
    Jan 19 at 16:05











  • The kappa value is the what I want primarily.

    – Fredrik Karlsson
    Jan 19 at 16:16











  • Ok then, I'll edit my answer below in the case you want only the kappa value

    – Ric S
    Jan 19 at 16:18











  • Great thank you very much!

    – Fredrik Karlsson
    Jan 19 at 16:24











  • You're welcome! :)

    – Ric S
    Jan 19 at 16:27














0












0








0








For a visualisation, I need to compute the weighted kappa of all combinations / pairs of seven raters. So, if I use some sample data with seven columns:



ex <- structure(list(`1` = c(1, 1, 2, 1, 2, 1, 1, 1, 2, 3, 1, 1, 1, 

1, 1, 1, 2, 1, 2, 2), `2` = c(1, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 

2, 1, 1, 2, 2, 1, 2, 2, 2), `3` = c(1, 1, 2, 1, 2, 1, 1, 1, 2, 

3, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2), `4` = c(1, 1, 2, 2, 2, 1, 2, 

2, 2, 3, 2, 1, 1, 2, 2, 2, 1, 1, 2, 2), pa = c(1, 2, 1, 2, 3, 

1, 2, 2, 2, 2, 2, 2, 1, 2, 1, 2, 2, 1, 3, 2), ta = c(2, 2, 2, 

1, 3, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 1, 1, 3), ka = c(1, 

1, 2, 1, 2, 1, 1, 1, 2, 3, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2)), row.names = c(NA, -20L), class = c("tbl_df", "tbl", "data.frame"))


I would like to get a structure that captures the output of irr::kappa2 on all combinations of columns. A structure like:



out <- data.frame("1_2"=irr::kappa2(ex[c(1,2)]),

"1_3"=irr::kappa2(ex[c(1,3)]),"1_4"=irr::kappa2(ex[c(1,3)]),....


(for all unique combinations of columns).



Any ideas?






r functional-programming tidyverse







share|improve this question














For a visualisation, I need to compute the weighted kappa of all combinations / pairs of seven raters. So, if I use some sample data with seven columns:



ex <- structure(list(`1` = c(1, 1, 2, 1, 2, 1, 1, 1, 2, 3, 1, 1, 1, 

1, 1, 1, 2, 1, 2, 2), `2` = c(1, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 

2, 1, 1, 2, 2, 1, 2, 2, 2), `3` = c(1, 1, 2, 1, 2, 1, 1, 1, 2, 

3, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2), `4` = c(1, 1, 2, 2, 2, 1, 2, 

2, 2, 3, 2, 1, 1, 2, 2, 2, 1, 1, 2, 2), pa = c(1, 2, 1, 2, 3, 

1, 2, 2, 2, 2, 2, 2, 1, 2, 1, 2, 2, 1, 3, 2), ta = c(2, 2, 2, 

1, 3, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 1, 1, 3), ka = c(1, 

1, 2, 1, 2, 1, 1, 1, 2, 3, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2)), row.names = c(NA, -20L), class = c("tbl_df", "tbl", "data.frame"))


I would like to get a structure that captures the output of irr::kappa2 on all combinations of columns. A structure like:



out <- data.frame("1_2"=irr::kappa2(ex[c(1,2)]),

"1_3"=irr::kappa2(ex[c(1,3)]),"1_4"=irr::kappa2(ex[c(1,3)]),....


(for all unique combinations of columns).



Any ideas?






r functional-programming tidyverse




r functional-programming tidyverse






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Jan 19 at 15:50









Fredrik KarlssonFredrik Karlsson

345314




345314













  • Just to be clear, do you need the whole output of irr::kappa2() or just the value of Kappa?

    – Ric S
    Jan 19 at 16:05











  • The kappa value is the what I want primarily.

    – Fredrik Karlsson
    Jan 19 at 16:16











  • Ok then, I'll edit my answer below in the case you want only the kappa value

    – Ric S
    Jan 19 at 16:18











  • Great thank you very much!

    – Fredrik Karlsson
    Jan 19 at 16:24











  • You're welcome! :)

    – Ric S
    Jan 19 at 16:27



















  • Just to be clear, do you need the whole output of irr::kappa2() or just the value of Kappa?

    – Ric S
    Jan 19 at 16:05











  • The kappa value is the what I want primarily.

    – Fredrik Karlsson
    Jan 19 at 16:16











  • Ok then, I'll edit my answer below in the case you want only the kappa value

    – Ric S
    Jan 19 at 16:18











  • Great thank you very much!

    – Fredrik Karlsson
    Jan 19 at 16:24











  • You're welcome! :)

    – Ric S
    Jan 19 at 16:27

















Just to be clear, do you need the whole output of irr::kappa2() or just the value of Kappa?

– Ric S
Jan 19 at 16:05





Just to be clear, do you need the whole output of irr::kappa2() or just the value of Kappa?

– Ric S
Jan 19 at 16:05













The kappa value is the what I want primarily.

– Fredrik Karlsson
Jan 19 at 16:16





The kappa value is the what I want primarily.

– Fredrik Karlsson
Jan 19 at 16:16













Ok then, I'll edit my answer below in the case you want only the kappa value

– Ric S
Jan 19 at 16:18





Ok then, I'll edit my answer below in the case you want only the kappa value

– Ric S
Jan 19 at 16:18













Great thank you very much!

– Fredrik Karlsson
Jan 19 at 16:24





Great thank you very much!

– Fredrik Karlsson
Jan 19 at 16:24













You're welcome! :)

– Ric S
Jan 19 at 16:27





You're welcome! :)

– Ric S
Jan 19 at 16:27












1 Answer
1






active

oldest

votes


















2














A solution would be to store the whole output structure of function kappa2 into an element of a list, and to have an element for each possible combination of columns:



# initialization

out_list <- list()

column <- 1



# cycle for storing kappa2's output structure

for (i in 1:(ncol(ex)-1)){

    for (j in (i+1):ncol(ex)){

        out_list[[column]] <- irr::kappa2(ex[,c(i,j)])

        # renaming the elements

        names(out_list)[column] <- paste0(i, "_", j)

        column <- column + 1

    }

}



In case you just want the Kappa value for each pair of columns, like you said in the comments, you can use the following (very similar to before) code:



# initialization

# the number of columns of "out" is from mathematics

out <- as.data.frame(matrix(0, nrow = 1, ncol = ncol(ex) * (ncol(ex)-1) / 2))

column <- 1



# cycle for calculation kappa

for (i in 1:(ncol(ex)-1)){

    for (j in (i+1):ncol(ex)){

        out[1,column] <- irr::kappa2(ex[,c(i,j)])$value

        colnames(out)[column] <- paste0(i, "_", j)

        column <- column + 1

    }

}





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    1 Answer
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    active

    oldest

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    active

    oldest

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    2














    A solution would be to store the whole output structure of function kappa2 into an element of a list, and to have an element for each possible combination of columns:



    # initialization
    
out_list <- list()
    
column <- 1
    

    
# cycle for storing kappa2's output structure
    
for (i in 1:(ncol(ex)-1)){
    
    for (j in (i+1):ncol(ex)){
    
        out_list[[column]] <- irr::kappa2(ex[,c(i,j)])
    
        # renaming the elements
    
        names(out_list)[column] <- paste0(i, "_", j)
    
        column <- column + 1
    
    }
    
}



    In case you just want the Kappa value for each pair of columns, like you said in the comments, you can use the following (very similar to before) code:



    # initialization
    
# the number of columns of "out" is from mathematics
    
out <- as.data.frame(matrix(0, nrow = 1, ncol = ncol(ex) * (ncol(ex)-1) / 2))
    
column <- 1
    

    
# cycle for calculation kappa
    
for (i in 1:(ncol(ex)-1)){
    
    for (j in (i+1):ncol(ex)){
    
        out[1,column] <- irr::kappa2(ex[,c(i,j)])$value
    
        colnames(out)[column] <- paste0(i, "_", j)
    
        column <- column + 1
    
    }
    
}





    share|improve this answer






























      2














      A solution would be to store the whole output structure of function kappa2 into an element of a list, and to have an element for each possible combination of columns:



      # initialization
      
out_list <- list()
      
column <- 1
      

      
# cycle for storing kappa2's output structure
      
for (i in 1:(ncol(ex)-1)){
      
    for (j in (i+1):ncol(ex)){
      
        out_list[[column]] <- irr::kappa2(ex[,c(i,j)])
      
        # renaming the elements
      
        names(out_list)[column] <- paste0(i, "_", j)
      
        column <- column + 1
      
    }
      
}



      In case you just want the Kappa value for each pair of columns, like you said in the comments, you can use the following (very similar to before) code:



      # initialization
      
# the number of columns of "out" is from mathematics
      
out <- as.data.frame(matrix(0, nrow = 1, ncol = ncol(ex) * (ncol(ex)-1) / 2))
      
column <- 1
      

      
# cycle for calculation kappa
      
for (i in 1:(ncol(ex)-1)){
      
    for (j in (i+1):ncol(ex)){
      
        out[1,column] <- irr::kappa2(ex[,c(i,j)])$value
      
        colnames(out)[column] <- paste0(i, "_", j)
      
        column <- column + 1
      
    }
      
}





      share|improve this answer




























        2












        2








        2







        A solution would be to store the whole output structure of function kappa2 into an element of a list, and to have an element for each possible combination of columns:



        # initialization
        
out_list <- list()
        
column <- 1
        

        
# cycle for storing kappa2's output structure
        
for (i in 1:(ncol(ex)-1)){
        
    for (j in (i+1):ncol(ex)){
        
        out_list[[column]] <- irr::kappa2(ex[,c(i,j)])
        
        # renaming the elements
        
        names(out_list)[column] <- paste0(i, "_", j)
        
        column <- column + 1
        
    }
        
}



        In case you just want the Kappa value for each pair of columns, like you said in the comments, you can use the following (very similar to before) code:



        # initialization
        
# the number of columns of "out" is from mathematics
        
out <- as.data.frame(matrix(0, nrow = 1, ncol = ncol(ex) * (ncol(ex)-1) / 2))
        
column <- 1
        

        
# cycle for calculation kappa
        
for (i in 1:(ncol(ex)-1)){
        
    for (j in (i+1):ncol(ex)){
        
        out[1,column] <- irr::kappa2(ex[,c(i,j)])$value
        
        colnames(out)[column] <- paste0(i, "_", j)
        
        column <- column + 1
        
    }
        
}





        share|improve this answer















        A solution would be to store the whole output structure of function kappa2 into an element of a list, and to have an element for each possible combination of columns:



        # initialization
        
out_list <- list()
        
column <- 1
        

        
# cycle for storing kappa2's output structure
        
for (i in 1:(ncol(ex)-1)){
        
    for (j in (i+1):ncol(ex)){
        
        out_list[[column]] <- irr::kappa2(ex[,c(i,j)])
        
        # renaming the elements
        
        names(out_list)[column] <- paste0(i, "_", j)
        
        column <- column + 1
        
    }
        
}



        In case you just want the Kappa value for each pair of columns, like you said in the comments, you can use the following (very similar to before) code:



        # initialization
        
# the number of columns of "out" is from mathematics
        
out <- as.data.frame(matrix(0, nrow = 1, ncol = ncol(ex) * (ncol(ex)-1) / 2))
        
column <- 1
        

        
# cycle for calculation kappa
        
for (i in 1:(ncol(ex)-1)){
        
    for (j in (i+1):ncol(ex)){
        
        out[1,column] <- irr::kappa2(ex[,c(i,j)])$value
        
        colnames(out)[column] <- paste0(i, "_", j)
        
        column <- column + 1
        
    }
        
}






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Jan 19 at 16:27

























        answered Jan 19 at 16:13









        Ric SRic S

        669418




        669418






























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