I’ve been playing with the igraph package a bit lately (see previous post HERE) and wanted to approach a problem I once visited in the past. The basic gist of the problem is this:
Students in a class are asked their top three favorite students to work with (rank order). After a social intervention this same question is posed again to students. The intended outcome of the intervention is that the distribution of students receiving many or very few choices will diminish. In other words the dorks will become less dorky and the popular students will become less popular. The idea is to visual this relationship.
Here is a script of one such visualization. It’s a bit light on annotations but merely experimenting with the code should give a good sense of what is occurring.
library(igraph) set.seed(101) #create a data set X <-lapply(1:10, function(i) sample(LETTERS[c(1:10)[-i]], 3)) Y <- data.frame(person = LETTERS[1:10], sex = rbinom(10, 1, .5), do.call(rbind, X)) names(Y)[3:5] <- paste0("choice.", 1:3) #reshape the data to long format Z <- reshape(Y, direction="long", varying=3:5) colnames(Z)[3:4] <- c("choice.no", "choice") rownames(Z) <- NULL Z <- Z[, c(1, 4, 3, 2)] #turn the data into a graph structure edges <- as.matrix(Z[, 1:2]) g <- graph.data.frame(edges, directed=TRUE) V(g)$label <- V(g)$name #change label size based on number of votes SUMS <- data.frame(table(Z$choice)) SUMS$Var1 <- as.character(SUMS$Var1) SUMS <- SUMS[order(as.character(SUMS$Var1)), ] SUMS$Freq <- as.integer(SUMS$Freq) label.size <- 2 V(g)$label.cex <- log(scale(SUMS$Freq) + max(abs(scale(SUMS$Freq)))+ label.size) #Color edges that are reciprocal red x <- t(apply(edges, 1, sort)) x <- paste0(x[, 1], x[, 2]) y <- x[duplicated(x)] COLS <- ifelse(x %in% y, "red", "gray40") E(g)$color <- COLS #reverse score the choices.no and weight E(g)$width <- (4 - Z$choice.no)*2 #color vertex based on sex V(g)$gender <- Y$sex V(g)$color <- ifelse(V(g)$gender==0, "pink", "lightblue") #plot it opar <- par()$mar; par(mar=rep(0, 4)) #Give the graph lots of room plot.igraph(g, layout=layout.auto(g)) par(mar=opar)
For an additional script of this analysis with 20 students click here.
For helpful igraph documentation click here