Convert a DataFrame into Adjacency/Weights Matrix in R












7















I have a DataFrame, df.



n is a column denoting the number of groups in the x column.
x is a column containing the comma-separated groups. 



df <- data.frame(n = c(2, 3, 2, 2), 

                 x = c("a, b", "a, c, d", "c, d", "d, b"))



> df

n        x

2     a, b

3  a, c, d

2     c, d

2     d, b


I would like to convert this DataFrame into a weights matrix where the row and column names are the unique values of the groups in df$c, and the elements represent the number of times each of the groups appear together in df$c.



The output should look like this: 



m <- matrix(c(0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 2, 1, 1, 2, 0), nrow = 4, ncol = 4)

rownames(m) <- letters[1:4]; colnames(m) <- letters[1:4]



> m

  a b c d

a 0 1 1 1

b 1 0 0 1

c 1 0 0 2

d 1 1 2 0





r matrix adjacency-matrix







share|improve this question


















  • 1





    your question is unclear. I can't see c in df. it only has n and x

    – YOLO
    4 hours ago













  • c is one of the x values. Its a frequency table of how often different letters appear in the same line in x

    – RAB
    4 hours ago











  • Do you mean df$x instead of df$c in the bolded part of the question?

    – mikoontz
    2 hours ago
















7















I have a DataFrame, df.



n is a column denoting the number of groups in the x column.
x is a column containing the comma-separated groups. 



df <- data.frame(n = c(2, 3, 2, 2), 

                 x = c("a, b", "a, c, d", "c, d", "d, b"))



> df

n        x

2     a, b

3  a, c, d

2     c, d

2     d, b


I would like to convert this DataFrame into a weights matrix where the row and column names are the unique values of the groups in df$c, and the elements represent the number of times each of the groups appear together in df$c.



The output should look like this: 



m <- matrix(c(0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 2, 1, 1, 2, 0), nrow = 4, ncol = 4)

rownames(m) <- letters[1:4]; colnames(m) <- letters[1:4]



> m

  a b c d

a 0 1 1 1

b 1 0 0 1

c 1 0 0 2

d 1 1 2 0





r matrix adjacency-matrix







share|improve this question


















  • 1





    your question is unclear. I can't see c in df. it only has n and x

    – YOLO
    4 hours ago













  • c is one of the x values. Its a frequency table of how often different letters appear in the same line in x

    – RAB
    4 hours ago











  • Do you mean df$x instead of df$c in the bolded part of the question?

    – mikoontz
    2 hours ago














7












7








7








I have a DataFrame, df.



n is a column denoting the number of groups in the x column.
x is a column containing the comma-separated groups. 



df <- data.frame(n = c(2, 3, 2, 2), 

                 x = c("a, b", "a, c, d", "c, d", "d, b"))



> df

n        x

2     a, b

3  a, c, d

2     c, d

2     d, b


I would like to convert this DataFrame into a weights matrix where the row and column names are the unique values of the groups in df$c, and the elements represent the number of times each of the groups appear together in df$c.



The output should look like this: 



m <- matrix(c(0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 2, 1, 1, 2, 0), nrow = 4, ncol = 4)

rownames(m) <- letters[1:4]; colnames(m) <- letters[1:4]



> m

  a b c d

a 0 1 1 1

b 1 0 0 1

c 1 0 0 2

d 1 1 2 0





r matrix adjacency-matrix







share|improve this question














I have a DataFrame, df.



n is a column denoting the number of groups in the x column.
x is a column containing the comma-separated groups. 



df <- data.frame(n = c(2, 3, 2, 2), 

                 x = c("a, b", "a, c, d", "c, d", "d, b"))



> df

n        x

2     a, b

3  a, c, d

2     c, d

2     d, b


I would like to convert this DataFrame into a weights matrix where the row and column names are the unique values of the groups in df$c, and the elements represent the number of times each of the groups appear together in df$c.



The output should look like this: 



m <- matrix(c(0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 2, 1, 1, 2, 0), nrow = 4, ncol = 4)

rownames(m) <- letters[1:4]; colnames(m) <- letters[1:4]



> m

  a b c d

a 0 1 1 1

b 1 0 0 1

c 1 0 0 2

d 1 1 2 0





r matrix adjacency-matrix




r matrix adjacency-matrix






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 4 hours ago









Rich PaulooRich Pauloo

2,188930




2,188930








  • 1





    your question is unclear. I can't see c in df. it only has n and x

    – YOLO
    4 hours ago













  • c is one of the x values. Its a frequency table of how often different letters appear in the same line in x

    – RAB
    4 hours ago











  • Do you mean df$x instead of df$c in the bolded part of the question?

    – mikoontz
    2 hours ago














  • 1





    your question is unclear. I can't see c in df. it only has n and x

    – YOLO
    4 hours ago













  • c is one of the x values. Its a frequency table of how often different letters appear in the same line in x

    – RAB
    4 hours ago











  • Do you mean df$x instead of df$c in the bolded part of the question?

    – mikoontz
    2 hours ago








1




1





your question is unclear. I can't see c in df. it only has n and x

– YOLO
4 hours ago







your question is unclear. I can't see c in df. it only has n and x

– YOLO
4 hours ago















c is one of the x values. Its a frequency table of how often different letters appear in the same line in x

– RAB
4 hours ago





c is one of the x values. Its a frequency table of how often different letters appear in the same line in x

– RAB
4 hours ago













Do you mean df$x instead of df$c in the bolded part of the question?

– mikoontz
2 hours ago





Do you mean df$x instead of df$c in the bolded part of the question?

– mikoontz
2 hours ago












2 Answers
2






active

oldest

votes


















4














Here's a very rough and probably pretty inefficient solution using tidyverse for wrangling and combinat to generate permutations. 



library(tidyverse)

library(combinat)



df <- data.frame(n = c(2, 3, 2, 2), 

                 x = c("a, b", "a, c, d", "c, d", "d, b"))



df %>% 

    ## Parse entries in x into distinct elements

    mutate(split = map(x, str_split, pattern = ', '), 

           flat = flatten(split)) %>% 

    ## Construct 2-element subsets of each set of elements

    mutate(combn = map(flat, combn, 2, simplify = FALSE)) %>% 

    unnest(combn) %>% 

    ## Construct permutations of the 2-element subsets

    mutate(perm = map(combn, permn)) %>% 

    unnest(perm) %>% 

    ## Parse the permutations into row and column indices

    mutate(row = map_chr(perm, 1), 

           col = map_chr(perm, 2)) %>% 

    count(row, col) %>% 

    ## Long to wide representation

    spread(key = col, value = nn, fill = 0) %>% 

    ## Coerce to matrix

    column_to_rownames(var = 'row') %>% 

    as.matrix()





share|improve this answer































    2














    Using Base R, you could do something like below



    a = strsplit(as.character(df$x),', ')
    
b = unique(unlist(a))
    
d = unlist(sapply(a,combn,2,toString))
    
e = data.frame(table(factor(d,c(paste(b,b,sep=','),combn(b,2,toString)))))
    
f = read.table(text = do.call(paste,c(sep =',', e)),sep=',',strip.white = T)
    
g = xtabs(V3~V1+V2,f)
    
g[lower.tri(g)] = t(g)[lower.tri(g)]
    
g
    
   V2
    
V1  a b c d
    
  a 0 1 1 1
    
  b 1 0 0 0
    
  c 1 0 0 2
    
  d 1 0 2 0





    share|improve this answer























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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4














      Here's a very rough and probably pretty inefficient solution using tidyverse for wrangling and combinat to generate permutations. 



      library(tidyverse)
      
library(combinat)
      

      
df <- data.frame(n = c(2, 3, 2, 2), 
      
                 x = c("a, b", "a, c, d", "c, d", "d, b"))
      

      
df %>% 
      
    ## Parse entries in x into distinct elements
      
    mutate(split = map(x, str_split, pattern = ', '), 
      
           flat = flatten(split)) %>% 
      
    ## Construct 2-element subsets of each set of elements
      
    mutate(combn = map(flat, combn, 2, simplify = FALSE)) %>% 
      
    unnest(combn) %>% 
      
    ## Construct permutations of the 2-element subsets
      
    mutate(perm = map(combn, permn)) %>% 
      
    unnest(perm) %>% 
      
    ## Parse the permutations into row and column indices
      
    mutate(row = map_chr(perm, 1), 
      
           col = map_chr(perm, 2)) %>% 
      
    count(row, col) %>% 
      
    ## Long to wide representation
      
    spread(key = col, value = nn, fill = 0) %>% 
      
    ## Coerce to matrix
      
    column_to_rownames(var = 'row') %>% 
      
    as.matrix()





      share|improve this answer




























        4














        Here's a very rough and probably pretty inefficient solution using tidyverse for wrangling and combinat to generate permutations. 



        library(tidyverse)
        
library(combinat)
        

        
df <- data.frame(n = c(2, 3, 2, 2), 
        
                 x = c("a, b", "a, c, d", "c, d", "d, b"))
        

        
df %>% 
        
    ## Parse entries in x into distinct elements
        
    mutate(split = map(x, str_split, pattern = ', '), 
        
           flat = flatten(split)) %>% 
        
    ## Construct 2-element subsets of each set of elements
        
    mutate(combn = map(flat, combn, 2, simplify = FALSE)) %>% 
        
    unnest(combn) %>% 
        
    ## Construct permutations of the 2-element subsets
        
    mutate(perm = map(combn, permn)) %>% 
        
    unnest(perm) %>% 
        
    ## Parse the permutations into row and column indices
        
    mutate(row = map_chr(perm, 1), 
        
           col = map_chr(perm, 2)) %>% 
        
    count(row, col) %>% 
        
    ## Long to wide representation
        
    spread(key = col, value = nn, fill = 0) %>% 
        
    ## Coerce to matrix
        
    column_to_rownames(var = 'row') %>% 
        
    as.matrix()





        share|improve this answer


























          4












          4








          4







          Here's a very rough and probably pretty inefficient solution using tidyverse for wrangling and combinat to generate permutations. 



          library(tidyverse)
          
library(combinat)
          

          
df <- data.frame(n = c(2, 3, 2, 2), 
          
                 x = c("a, b", "a, c, d", "c, d", "d, b"))
          

          
df %>% 
          
    ## Parse entries in x into distinct elements
          
    mutate(split = map(x, str_split, pattern = ', '), 
          
           flat = flatten(split)) %>% 
          
    ## Construct 2-element subsets of each set of elements
          
    mutate(combn = map(flat, combn, 2, simplify = FALSE)) %>% 
          
    unnest(combn) %>% 
          
    ## Construct permutations of the 2-element subsets
          
    mutate(perm = map(combn, permn)) %>% 
          
    unnest(perm) %>% 
          
    ## Parse the permutations into row and column indices
          
    mutate(row = map_chr(perm, 1), 
          
           col = map_chr(perm, 2)) %>% 
          
    count(row, col) %>% 
          
    ## Long to wide representation
          
    spread(key = col, value = nn, fill = 0) %>% 
          
    ## Coerce to matrix
          
    column_to_rownames(var = 'row') %>% 
          
    as.matrix()





          share|improve this answer













          Here's a very rough and probably pretty inefficient solution using tidyverse for wrangling and combinat to generate permutations. 



          library(tidyverse)
          
library(combinat)
          

          
df <- data.frame(n = c(2, 3, 2, 2), 
          
                 x = c("a, b", "a, c, d", "c, d", "d, b"))
          

          
df %>% 
          
    ## Parse entries in x into distinct elements
          
    mutate(split = map(x, str_split, pattern = ', '), 
          
           flat = flatten(split)) %>% 
          
    ## Construct 2-element subsets of each set of elements
          
    mutate(combn = map(flat, combn, 2, simplify = FALSE)) %>% 
          
    unnest(combn) %>% 
          
    ## Construct permutations of the 2-element subsets
          
    mutate(perm = map(combn, permn)) %>% 
          
    unnest(perm) %>% 
          
    ## Parse the permutations into row and column indices
          
    mutate(row = map_chr(perm, 1), 
          
           col = map_chr(perm, 2)) %>% 
          
    count(row, col) %>% 
          
    ## Long to wide representation
          
    spread(key = col, value = nn, fill = 0) %>% 
          
    ## Coerce to matrix
          
    column_to_rownames(var = 'row') %>% 
          
    as.matrix()






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 3 hours ago









          Dan HicksDan Hicks

          1876




          1876

























              2














              Using Base R, you could do something like below



              a = strsplit(as.character(df$x),', ')
              
b = unique(unlist(a))
              
d = unlist(sapply(a,combn,2,toString))
              
e = data.frame(table(factor(d,c(paste(b,b,sep=','),combn(b,2,toString)))))
              
f = read.table(text = do.call(paste,c(sep =',', e)),sep=',',strip.white = T)
              
g = xtabs(V3~V1+V2,f)
              
g[lower.tri(g)] = t(g)[lower.tri(g)]
              
g
              
   V2
              
V1  a b c d
              
  a 0 1 1 1
              
  b 1 0 0 0
              
  c 1 0 0 2
              
  d 1 0 2 0





              share|improve this answer




























                2














                Using Base R, you could do something like below



                a = strsplit(as.character(df$x),', ')
                
b = unique(unlist(a))
                
d = unlist(sapply(a,combn,2,toString))
                
e = data.frame(table(factor(d,c(paste(b,b,sep=','),combn(b,2,toString)))))
                
f = read.table(text = do.call(paste,c(sep =',', e)),sep=',',strip.white = T)
                
g = xtabs(V3~V1+V2,f)
                
g[lower.tri(g)] = t(g)[lower.tri(g)]
                
g
                
   V2
                
V1  a b c d
                
  a 0 1 1 1
                
  b 1 0 0 0
                
  c 1 0 0 2
                
  d 1 0 2 0





                share|improve this answer


























                  2












                  2








                  2







                  Using Base R, you could do something like below



                  a = strsplit(as.character(df$x),', ')
                  
b = unique(unlist(a))
                  
d = unlist(sapply(a,combn,2,toString))
                  
e = data.frame(table(factor(d,c(paste(b,b,sep=','),combn(b,2,toString)))))
                  
f = read.table(text = do.call(paste,c(sep =',', e)),sep=',',strip.white = T)
                  
g = xtabs(V3~V1+V2,f)
                  
g[lower.tri(g)] = t(g)[lower.tri(g)]
                  
g
                  
   V2
                  
V1  a b c d
                  
  a 0 1 1 1
                  
  b 1 0 0 0
                  
  c 1 0 0 2
                  
  d 1 0 2 0





                  share|improve this answer













                  Using Base R, you could do something like below



                  a = strsplit(as.character(df$x),', ')
                  
b = unique(unlist(a))
                  
d = unlist(sapply(a,combn,2,toString))
                  
e = data.frame(table(factor(d,c(paste(b,b,sep=','),combn(b,2,toString)))))
                  
f = read.table(text = do.call(paste,c(sep =',', e)),sep=',',strip.white = T)
                  
g = xtabs(V3~V1+V2,f)
                  
g[lower.tri(g)] = t(g)[lower.tri(g)]
                  
g
                  
   V2
                  
V1  a b c d
                  
  a 0 1 1 1
                  
  b 1 0 0 0
                  
  c 1 0 0 2
                  
  d 1 0 2 0






                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 2 hours ago









                  OnyambuOnyambu

                  15.5k1520




                  15.5k1520






























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