Mapping edgelists to an adjacency matrix (and sum them together)












1















I want to map a number of (undirected) friendship networks (in edgelist format) to an adjacency matrix consisting of all possible nodes (i.e., persons) using R. To begin with, I construct a smaller 4-person circle x <- c(1, 2, 3, 4) which consists of 6 unique edges (1-2, 1-3, 1-4, 2-3, 2-4, 3-4). I then collapsed this set of 6 unique edges into a single list, such that it can be converted into a symmetric matrix using igraph applications (see below).



x = c(1,2,3,4)
x_pairs = combn(x, 2)
List <- split(x_pairs, rep(1:ncol(x_pairs), each = nrow(x_pairs)))
library(purrr)
new_list <- purrr::flatten(List)
g <- make_graph(unlist(new_list), directed = F)
m <- as_adjacency_matrix(g, sparse = F)
m

[,1] [,2] [,3] [,4]
[1,] 0 1 1 1
[2,] 1 0 1 1
[3,] 1 1 0 1
[4,] 1 1 1 0


My dataset has more than one of such smaller friendship circles consist of members out of a total of 50 persons and the memberships of these circles may or may not overlap. So my question is how do I map a series of smaller matrix values like the m above to a 50 by 50 adjacency matrix in two different ways:



(1) without repeating: say, if 3 and 4 are friends in one circle but they are also linked in another circle, the edge between 3 and 4 should remain 1 (but not add up to 2)
(2) cumulatively: if relationship in multiple circles indicates stronger friendship, then it might be more informative to map these circles into a weighted adjacency matrix where each cell in the matrix represents the cumulative counts of row and column id's friendship in different circles. In 3 and 4's situation, their edge value should be 1 + 1 = 2.



I've checked out this and other previous threads but can't seem to figure out how to do this, it will be really appreciated if someone could enlighten me on this.










share|improve this question





























    1















    I want to map a number of (undirected) friendship networks (in edgelist format) to an adjacency matrix consisting of all possible nodes (i.e., persons) using R. To begin with, I construct a smaller 4-person circle x <- c(1, 2, 3, 4) which consists of 6 unique edges (1-2, 1-3, 1-4, 2-3, 2-4, 3-4). I then collapsed this set of 6 unique edges into a single list, such that it can be converted into a symmetric matrix using igraph applications (see below).



    x = c(1,2,3,4)
    x_pairs = combn(x, 2)
    List <- split(x_pairs, rep(1:ncol(x_pairs), each = nrow(x_pairs)))
    library(purrr)
    new_list <- purrr::flatten(List)
    g <- make_graph(unlist(new_list), directed = F)
    m <- as_adjacency_matrix(g, sparse = F)
    m

    [,1] [,2] [,3] [,4]
    [1,] 0 1 1 1
    [2,] 1 0 1 1
    [3,] 1 1 0 1
    [4,] 1 1 1 0


    My dataset has more than one of such smaller friendship circles consist of members out of a total of 50 persons and the memberships of these circles may or may not overlap. So my question is how do I map a series of smaller matrix values like the m above to a 50 by 50 adjacency matrix in two different ways:



    (1) without repeating: say, if 3 and 4 are friends in one circle but they are also linked in another circle, the edge between 3 and 4 should remain 1 (but not add up to 2)
    (2) cumulatively: if relationship in multiple circles indicates stronger friendship, then it might be more informative to map these circles into a weighted adjacency matrix where each cell in the matrix represents the cumulative counts of row and column id's friendship in different circles. In 3 and 4's situation, their edge value should be 1 + 1 = 2.



    I've checked out this and other previous threads but can't seem to figure out how to do this, it will be really appreciated if someone could enlighten me on this.










    share|improve this question



























      1












      1








      1








      I want to map a number of (undirected) friendship networks (in edgelist format) to an adjacency matrix consisting of all possible nodes (i.e., persons) using R. To begin with, I construct a smaller 4-person circle x <- c(1, 2, 3, 4) which consists of 6 unique edges (1-2, 1-3, 1-4, 2-3, 2-4, 3-4). I then collapsed this set of 6 unique edges into a single list, such that it can be converted into a symmetric matrix using igraph applications (see below).



      x = c(1,2,3,4)
      x_pairs = combn(x, 2)
      List <- split(x_pairs, rep(1:ncol(x_pairs), each = nrow(x_pairs)))
      library(purrr)
      new_list <- purrr::flatten(List)
      g <- make_graph(unlist(new_list), directed = F)
      m <- as_adjacency_matrix(g, sparse = F)
      m

      [,1] [,2] [,3] [,4]
      [1,] 0 1 1 1
      [2,] 1 0 1 1
      [3,] 1 1 0 1
      [4,] 1 1 1 0


      My dataset has more than one of such smaller friendship circles consist of members out of a total of 50 persons and the memberships of these circles may or may not overlap. So my question is how do I map a series of smaller matrix values like the m above to a 50 by 50 adjacency matrix in two different ways:



      (1) without repeating: say, if 3 and 4 are friends in one circle but they are also linked in another circle, the edge between 3 and 4 should remain 1 (but not add up to 2)
      (2) cumulatively: if relationship in multiple circles indicates stronger friendship, then it might be more informative to map these circles into a weighted adjacency matrix where each cell in the matrix represents the cumulative counts of row and column id's friendship in different circles. In 3 and 4's situation, their edge value should be 1 + 1 = 2.



      I've checked out this and other previous threads but can't seem to figure out how to do this, it will be really appreciated if someone could enlighten me on this.










      share|improve this question
















      I want to map a number of (undirected) friendship networks (in edgelist format) to an adjacency matrix consisting of all possible nodes (i.e., persons) using R. To begin with, I construct a smaller 4-person circle x <- c(1, 2, 3, 4) which consists of 6 unique edges (1-2, 1-3, 1-4, 2-3, 2-4, 3-4). I then collapsed this set of 6 unique edges into a single list, such that it can be converted into a symmetric matrix using igraph applications (see below).



      x = c(1,2,3,4)
      x_pairs = combn(x, 2)
      List <- split(x_pairs, rep(1:ncol(x_pairs), each = nrow(x_pairs)))
      library(purrr)
      new_list <- purrr::flatten(List)
      g <- make_graph(unlist(new_list), directed = F)
      m <- as_adjacency_matrix(g, sparse = F)
      m

      [,1] [,2] [,3] [,4]
      [1,] 0 1 1 1
      [2,] 1 0 1 1
      [3,] 1 1 0 1
      [4,] 1 1 1 0


      My dataset has more than one of such smaller friendship circles consist of members out of a total of 50 persons and the memberships of these circles may or may not overlap. So my question is how do I map a series of smaller matrix values like the m above to a 50 by 50 adjacency matrix in two different ways:



      (1) without repeating: say, if 3 and 4 are friends in one circle but they are also linked in another circle, the edge between 3 and 4 should remain 1 (but not add up to 2)
      (2) cumulatively: if relationship in multiple circles indicates stronger friendship, then it might be more informative to map these circles into a weighted adjacency matrix where each cell in the matrix represents the cumulative counts of row and column id's friendship in different circles. In 3 and 4's situation, their edge value should be 1 + 1 = 2.



      I've checked out this and other previous threads but can't seem to figure out how to do this, it will be really appreciated if someone could enlighten me on this.







      r igraph adjacency-matrix weighted-graph






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      edited Nov 17 '18 at 11:52









      Henrik

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      42k994110










      asked Nov 15 '18 at 12:34









      Chris T.Chris T.

      365318




      365318
























          1 Answer
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          There are various ways to achieve it. It looks like doing it in graph theoretical terms in igraph is a little more tedious than dealing directly with adjacency matrices. Let



          circles <- list(1:3, 2:4) # Friendship circles with identities 1, ..., n
          n <- max(unlist(circles)) # Total number of people
          nM <- matrix(0, n, n) # n x n matrix of zeroes


          Then



          adjs <- lapply(circles, function(cr) {nM[cr, cr] <- 1; nM[cbind(cr, cr)] <- 0; nM})


          is a list of n x n adjacency matrices for each friendship circle (mostly zeroes in each case).



          Then the two types of aggregate matrices can be obtained by



          (adj1 <- Reduce(`+`, adjs))
          # [,1] [,2] [,3] [,4]
          # [1,] 0 1 1 0
          # [2,] 1 0 2 1
          # [3,] 1 2 0 1
          # [4,] 0 1 1 0
          (adj2 <- 1 * (adj1 > 0))
          # [,1] [,2] [,3] [,4]
          # [1,] 0 1 1 0
          # [2,] 1 0 1 1
          # [3,] 1 1 0 1
          # [4,] 0 1 1 0





          share|improve this answer
























          • @ChrisT., does it answer your question or are you looking for a different approach?

            – Julius Vainora
            Nov 18 '18 at 10:16











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          1 Answer
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          active

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          1














          There are various ways to achieve it. It looks like doing it in graph theoretical terms in igraph is a little more tedious than dealing directly with adjacency matrices. Let



          circles <- list(1:3, 2:4) # Friendship circles with identities 1, ..., n
          n <- max(unlist(circles)) # Total number of people
          nM <- matrix(0, n, n) # n x n matrix of zeroes


          Then



          adjs <- lapply(circles, function(cr) {nM[cr, cr] <- 1; nM[cbind(cr, cr)] <- 0; nM})


          is a list of n x n adjacency matrices for each friendship circle (mostly zeroes in each case).



          Then the two types of aggregate matrices can be obtained by



          (adj1 <- Reduce(`+`, adjs))
          # [,1] [,2] [,3] [,4]
          # [1,] 0 1 1 0
          # [2,] 1 0 2 1
          # [3,] 1 2 0 1
          # [4,] 0 1 1 0
          (adj2 <- 1 * (adj1 > 0))
          # [,1] [,2] [,3] [,4]
          # [1,] 0 1 1 0
          # [2,] 1 0 1 1
          # [3,] 1 1 0 1
          # [4,] 0 1 1 0





          share|improve this answer
























          • @ChrisT., does it answer your question or are you looking for a different approach?

            – Julius Vainora
            Nov 18 '18 at 10:16
















          1














          There are various ways to achieve it. It looks like doing it in graph theoretical terms in igraph is a little more tedious than dealing directly with adjacency matrices. Let



          circles <- list(1:3, 2:4) # Friendship circles with identities 1, ..., n
          n <- max(unlist(circles)) # Total number of people
          nM <- matrix(0, n, n) # n x n matrix of zeroes


          Then



          adjs <- lapply(circles, function(cr) {nM[cr, cr] <- 1; nM[cbind(cr, cr)] <- 0; nM})


          is a list of n x n adjacency matrices for each friendship circle (mostly zeroes in each case).



          Then the two types of aggregate matrices can be obtained by



          (adj1 <- Reduce(`+`, adjs))
          # [,1] [,2] [,3] [,4]
          # [1,] 0 1 1 0
          # [2,] 1 0 2 1
          # [3,] 1 2 0 1
          # [4,] 0 1 1 0
          (adj2 <- 1 * (adj1 > 0))
          # [,1] [,2] [,3] [,4]
          # [1,] 0 1 1 0
          # [2,] 1 0 1 1
          # [3,] 1 1 0 1
          # [4,] 0 1 1 0





          share|improve this answer
























          • @ChrisT., does it answer your question or are you looking for a different approach?

            – Julius Vainora
            Nov 18 '18 at 10:16














          1












          1








          1







          There are various ways to achieve it. It looks like doing it in graph theoretical terms in igraph is a little more tedious than dealing directly with adjacency matrices. Let



          circles <- list(1:3, 2:4) # Friendship circles with identities 1, ..., n
          n <- max(unlist(circles)) # Total number of people
          nM <- matrix(0, n, n) # n x n matrix of zeroes


          Then



          adjs <- lapply(circles, function(cr) {nM[cr, cr] <- 1; nM[cbind(cr, cr)] <- 0; nM})


          is a list of n x n adjacency matrices for each friendship circle (mostly zeroes in each case).



          Then the two types of aggregate matrices can be obtained by



          (adj1 <- Reduce(`+`, adjs))
          # [,1] [,2] [,3] [,4]
          # [1,] 0 1 1 0
          # [2,] 1 0 2 1
          # [3,] 1 2 0 1
          # [4,] 0 1 1 0
          (adj2 <- 1 * (adj1 > 0))
          # [,1] [,2] [,3] [,4]
          # [1,] 0 1 1 0
          # [2,] 1 0 1 1
          # [3,] 1 1 0 1
          # [4,] 0 1 1 0





          share|improve this answer













          There are various ways to achieve it. It looks like doing it in graph theoretical terms in igraph is a little more tedious than dealing directly with adjacency matrices. Let



          circles <- list(1:3, 2:4) # Friendship circles with identities 1, ..., n
          n <- max(unlist(circles)) # Total number of people
          nM <- matrix(0, n, n) # n x n matrix of zeroes


          Then



          adjs <- lapply(circles, function(cr) {nM[cr, cr] <- 1; nM[cbind(cr, cr)] <- 0; nM})


          is a list of n x n adjacency matrices for each friendship circle (mostly zeroes in each case).



          Then the two types of aggregate matrices can be obtained by



          (adj1 <- Reduce(`+`, adjs))
          # [,1] [,2] [,3] [,4]
          # [1,] 0 1 1 0
          # [2,] 1 0 2 1
          # [3,] 1 2 0 1
          # [4,] 0 1 1 0
          (adj2 <- 1 * (adj1 > 0))
          # [,1] [,2] [,3] [,4]
          # [1,] 0 1 1 0
          # [2,] 1 0 1 1
          # [3,] 1 1 0 1
          # [4,] 0 1 1 0






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 17 '18 at 12:50









          Julius VainoraJulius Vainora

          38.1k76585




          38.1k76585













          • @ChrisT., does it answer your question or are you looking for a different approach?

            – Julius Vainora
            Nov 18 '18 at 10:16



















          • @ChrisT., does it answer your question or are you looking for a different approach?

            – Julius Vainora
            Nov 18 '18 at 10:16

















          @ChrisT., does it answer your question or are you looking for a different approach?

          – Julius Vainora
          Nov 18 '18 at 10:16





          @ChrisT., does it answer your question or are you looking for a different approach?

          – Julius Vainora
          Nov 18 '18 at 10:16




















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