I am trying to plot small networks using igraph in R. Each vertex in the network has a name, which is equivalent to its label. I would like to make each vertex have a rectangular symbol that is just large enough to fit its label.
This is my main inspiration.
What is the best way to do this with igraph?
Edit: more information
The code is here
jsonToNM <- function(jfile, directed=TRUE) {
# Requires "rjson" and "igraph"
nm.json <- fromJSON(file=jfile)
nm.graph <- c()
# Initialize the graph with the given nodes
g <- graph.empty(n=length(nm.json), directed=directed)
# Add their names
V(g)$name <- names(nm.json)
V(g)$label <- V(g)$name
# Now, add the edges
for(i in 1:length(nm.json)) {
# If the node has a "connected" field,
# then we note the connections by looking
# the names up.
if(length(nm.json[[i]]$connected > 0)) {
for(j in 1:length(nm.json[[i]]$connected)) {
# Add the entry
g <- g + edge(names(nm.json)[i],
nm.json[[i]]$connected[j])
}
}
}
plot(g, vertex.label.dist=1.5)
}
And the current output is below.
My goal is to place the labels inside of the vertex graphic, and expand the width of the vertex to accommodate the label.
Here is an example. Among some dirty tricks (i.e. multiplying the vertex size by 200), the key is to use two plot commands, so that we can measure the width (and height) of the labels with strwidth(), after the plot size is set with the first (empty) plot.
library(igraph)
camp <- graph.formula(Harry:Steve:Don:Bert - Harry:Steve:Don:Bert,
Pam:Brazey:Carol:Pat - Pam:Brazey:Carol:Pat,
Holly - Carol:Pat:Pam:Jennie:Bill,
Bill - Pauline:Michael:Lee:Holly,
Pauline - Bill:Jennie:Ann,
Jennie - Holly:Michael:Lee:Ann:Pauline,
Michael - Bill:Jennie:Ann:Lee:John,
Ann - Michael:Jennie:Pauline,
Lee - Michael:Bill:Jennie,
Gery - Pat:Steve:Russ:John,
Russ - Steve:Bert:Gery:John,
John - Gery:Russ:Michael)
V(camp)$label <- V(camp)$name
set.seed(42) ## to make this reproducable
co <- layout.auto(camp)
plot(0, type="n", ann=FALSE, axes=FALSE, xlim=extendrange(co[,1]),
ylim=extendrange(co[,2]))
plot(camp, layout=co, rescale=FALSE, add=TRUE,
vertex.shape="rectangle",
vertex.size=(strwidth(V(camp)$label) + strwidth("oo")) * 100,
vertex.size2=strheight("I") * 2 * 100)
Btw. this does not really work well with SVG graphics, because there is no way to measure the width of the text from R, the SVG device only makes guesses.
I know that this is not a direct answer to your question but I would suggest to use a different tool for visualization. yEd is very good at adjusting the nodes' width to the label's size. You can also manipulate the visualization easily, and export it to SVG for a final polish. It can be obtained for free from www.yworks.com (Disclaimer: I am not working there).
To export the graph in a well-readable format (yEd does not understand igraph's gml-format), use graphml:
write.graph(graph, "test.graphml", format="graphml")
Open it in yEd. Go to edit-> properties mapper and click on "new configuration (Node)" (the green "plus" symbol, upper left). In the middle of the fram, under "data source", search for the name of your labels (should be 'name'). In the middle tab called "map to" choose "label text" and in the right column leave the "conversion" be set to "Automatic".
Now choose Tools -> fit node to label (the default parameters are fine for a first try) and then choose your favourite layout. You can export to various image-formats but to my knowledge all are implemented using a bitmap-intermediate. Thus, I normally export to SVG and do the polishing in inkscape. If anyone knows a more efficient procedure to get good-looking layouts of medium-sized graphs produced in igraph, let me know.
Related
I'm currently struggling with some image analysis. I have images of zebrafish embryo vasculature, and I want to measure the distance between certain features (the highest point to the lowest etc).
I have processed the images to be more visible (higher contrast) using EBImage
.
I would appreciate any guidance.
Since you are using R and EBImage, I would presume that there is more analysis intended than just extracting measurements from an image. If that is all you intend, other software such as Fiji or the more streamline precursor, ImageJ, may be more user-friendly.
To answer the question, don't use display() for the image as you show here. Rather, use the plot() method that uses the option method = raster as the default. With the image plotted in a graphic window, you can use all the tools of R to interact with the plot. The resolution you have is determined by the size of your image and display. All values are returned in pixels and obviously need to be scaled appropriately.
This example uses locator() in a small helper function to measure diagonal distances between vascular junctions (?) in the image.
This simple helper function marks two points and measures the distance between the points. End the call to locator() with a right-click control-click or the escape key. In RStudio, you may have to explicitly press another button in the window and the points/lines may not be drawn until all calls to locator() are terminated.
p2p <- function(n = 512) # end with ctrl-click or Esc
{
ans <- numeric()
while (n > 0) {
# this call to locator places 2 points as crosses
# and connects them with a line
p <- locator(2, type = "o", pch = 3, col = "magenta")
if (is.null(p)) break
ans <- c(ans, sqrt(sum(sapply(p, diff)^2)))
n <- n - 1
}
return(ans) # return the vector of point-to-point distances
}
Now replot the image in the question (without the elements from the browser display) and then interact with the image.
plot(img) # not 'display(img)'
d <- p2p() # interact with the image, collecting distances
Here's the image after selecting six pairs of points with the distances measured between each pair of points.
round(d, 1)
> [1] 113.4 99.2 109.4 110.8 120.6 122.7
mean(d)
> 112.6736
Have fun!
Not dumb at all. Yes, it is in pixels—EBImage and R gives you fractional pixels.
I am automatically generating graphs whose nodes need to be in fixed positions. For example:
There is actually an arc from node V4 to node V16, but we annot see it because there are also arcs from V4 to V10 and from V10 to V16.
Note that both the nodes and the arcs are generated automatically, and that the positions may vary, so I would need an automated way to curve arcs that are hidden behind other arcs.
Also note that none of these solutions are valid: igraph: Resolving tight overlapping nodes ; Using igraph, how to force curvature when arrows point in opposite directions. The first one simply places de nodes in a certain way, but my nodes need to be fixed. The second one simply deals with pairs of nodes that have two arcs connecting them going in the opposite direction.
UPDATE: The construction of the graph is the result of the learning process of the graph that forms a Bayesian Network using bnlearn library, so I am not very sure how could I produce a reproducible example. The positions of the nodes are fixed because they represent positions. I actually need some magic, some kind of detection of overlapping arcs: If two arcs overlap, curve one of them slightly so that it can be seen. I know from the linked questions that curving an arc is an option, so I thought maybe this kind of magic could be achieved
One solution would be to use the qgraph package. In the example below it automatically curves the bidirectional edges:
library(igraph)
library(qgraph)
# the raster layout
layout <- cbind(1:3, rep(1:3, each = 3))
# fully connected network
adj <- matrix(1, 9, 9)
# plot directed and undirected network
layout(matrix(1:2, 1, 2))
qgraph(adj, layout = layout, directed = FALSE, title = "undirected")
qgraph(adj, layout = layout, directed = TRUE, title = "directed") # automatically curves the bidirectional arrows
To convert an igraph object to something qgraph can use, all you need is an edgelist or adjacency matrix:
g <- make_ring(9)
edgeList <- do.call(rbind, igraph::get.adjedgelist(g))
qgraph(edgeList)
If you also want to include the axes, you can do so using axis() since qgraph uses base graphics. However, you probably have to tinker with par() as well to make it look nice.
I will try 3D printing data to make some nice visual illustration for a binary classification example.
Here is my 3D plot:
require(rgl)
#Get example data from mtcars and normalize to range 0:1
fun_norm <- function(k){(k-min(k))/(max(k)-min(k))}
x_norm <- fun_norm(mtcars$drat)
y_norm <- fun_norm(mtcars$mpg)
z_norm <- fun_norm(mtcars$qsec)
#Plot nice big spheres with rgl that I hope will look good after 3D printing
plot3d(x_norm, y_norm, z_norm, type="s", radius = 0.02, aspect = T)
#The sticks are meant to suspend the spheres in the air
plot3d(x_norm, y_norm, z_norm, type="h", lwd = 5, aspect = T, add = T)
#Nice thick gridline that will also be printed
grid3d(c("x","y","z"), lwd = 5)
Next, I wanted to add a z=0 plane, inspired by this blog here describing the r2stl written by Ian Walker. It is supposed to be the foundation of the printed structure that holds everything together.
planes3d(a=0, b=0, c=1, d=0)
However, it has no volume, it is a thin slab with height=0. I want it to form a solid base for the printed structure, which is meant to keep everything together (check out the aforementioned blog for more details, his examples are great). How do I increase the thickness of my z=0 plane to achieve the same effect?
Here is the final step to exporting as STL:
writeSTL("test.stl")
One can view the final product really nicely using the open source Meshlab as recommended by Ian in the blog.
Additional remark: I noticed that the thin plane is also separate from the grids that I added on the -z face of the cube and is floating. This might also cause a problem when printing. How can I merge the grids with the z=0 plane? (I will be sending the STL file to a friend who will print for me, I want to make things as easy for him as possible)
You can't make a plane thicker. You can make a solid shape (extrude3d() is the function to use). It won't adapt itself to the bounding box the way a plane does, so you would need to draw it last.
For example,
example(plot3d)
bbox <- par3d("bbox")
slab <- translate3d(extrude3d(bbox[c(1,2,2,1)], bbox[c(3,3,4,4)], 0.5),
0,0, bbox[5])
shade3d(slab, col = "gray")
produces this output:
This still isn't printable (the points have no support), but it should get you started.
In the matlib package, there's a function regvec3d() that draws a vector space representation of a 2-predictor multiple regression model. The plot method for the result of the function has an argument show.base that draws the base x1-x2 plane, and draws it thicker if show.base >0.
It is a simple hack that just draws a second version of the plane at a small offset. Maybe this will be enough for your application.
if (show.base > 0) planes3d(0, 0, 1, 0, color=col.plane, alpha=0.2)
if (show.base > 1) planes3d(0, 0, 1, -.01, color=col.plane, alpha=0.1)
from a data.frame (or any other R object type), with 3 Columns: "Node, Parent and text", I'd like to plot a tree with rows from "Node" to "Parent" and "text" as label.
Can anyone suggest a good library to use and example code, if possible.
I've been looking at the igraph library, but all examples I could find plot trees with sequential numbers or letters as nodes and its not simple to set the tree layout.
Any help would be greatly appreciated
Thanks
EDIT:
Thanks guys for all your help, I really appreciate it.
Some extra comments, if you can help further
#md1630, I tried your suggestion but that's not what I'm looking for. The fist code plots the tree with the root on top and the arrows from root to leaf and the second corrects the arrows but inverts the tree. What I'd like is root on top and arrow from leafs to root (I understand that may not be a tree per say - but that's the requirement
#user20650 your solution looks correct but the image starts to get crowded as the number of nodes increase. Any idea on how to add more space between them?
#math Am I using the function you provided correctly? I called plot(layout.binary(g)) and got the result on the left. The one on the right is the output of plot(g)
upgrade comment
library(igraph)
# some example data
dat <- data.frame(parent=rep(letters[1:3], each=2),
node=letters[2:7],
text=paste0("lab", 1:6))
# create graph
g <- graph.data.frame(dat)
# plot
# layout.reingold.tilford gives a tree structure
# edge and vertx labels can be defined in the plot command or alternatively
# you can add them to the graph via V(g)$name and E(g($label assignments
plot(g, layout = layout.reingold.tilford,
edge.label=E(g)$text, vertex.label=paste0("v_lab",1:7))
EDIT re comment
If you want the direction to go from the leaves towards the root; you can first, get the tree layout coordinates from the more standard tree structure, and then reverse the edges.
# get tree layout coords
g <- graph.data.frame(dat)
lay = layout.reingold.tilford(g)
# redraw graph with edges reversed
g2 <- graph.data.frame(dat[2:1], vertices = get.data.frame(g, what="vertices"))
par(mar=rep(0,4), mfrow=c(1,2))
plot(g, layout=lay)
plot(g2, layout=lay)
You can use rgraphviz. Here's the code to plot the tree from a dataframe df with columns "Node, Parent and text". I didn't run this on my computer so there may be bugs. But roughly this is the idea:
source("http://bioconductor.org/biocLite.R")
biocLite("Rgraphviz")
library("Rgraphviz")
#first set up the graph with just the nodes
nodes<- unique(df['Node'])
gR <- new("graphNEL", nodes = nodes, edgemode = "directed")
#add edges for each row in df
for (j in (1:nrow(df))) {
gR <- addEdge(df[j,2], df[j,1], gR, 1)
}
#add text labels
nAttrs <- list()
z <- df['text']
nAttrs$label <- z
#plot
plot(gR, nodeAttrs = nAttrs) #you can specify more attributes here
You can use igraph to get a network with your data (supposing your dataframe is dd):
g = graph(t(dd[,2:1]))
V(g)$label = as.character(dd$text)
plot(g, layout=layout.binary)
I supposed your root (with no parents) is not in the dataframe, otherwise use dd[-1,2:1] instead.
If you want to have a tree, you can easily produce a layout, it is simply a function that takes a graph and return a matrix. For a binary tree :
layout.binary = function(graph) {
layout = c()
r_vertex = length(V(graph))
depth = ceiling(log2(r_vertex+1))
for (ii in 0:(depth-1)) {
for (jj in 1:min(2^ii, r_vertex)) {
layout = rbind(layout, c(ii, (2*(jj-1)+1)/(2^(ii+1))))
}
r_vertex = r_vertex - 2^ii
}
return(layout)
}
It will plot an horizontal tree, use c((2*(jj-1)+1)/(2^(ii+1)), ii) if you want it to be vertical.
I need to draw a network with 5 nodes and 20 directed edges (an edge connecting each 2 nodes) using R, but I need two features to exist:
To be able to control the thickness of each edge.
The edges not to be overlapping (i.e.,the edge form A to B is not drawn over the edge from B to A)
I've spent hours looking for a solution, and tried many packages, but there's always a problem.
Can anybody suggest a solution please and provide a complete example as possible?
Many Thanks in advance.
If it is ok for the lines to be curved then I know two ways. First I create an edgelist:
Edges <- data.frame(
from = rep(1:5,each=5),
to = rep(1:5,times=5),
thickness = abs(rnorm(25)))
Edges <- subset(Edges,from!=to)
This contains the node of origin at the first column, node of destination at the second and weight at the third. You can use my pacake qgraph to plot a weighted graph using this. By default the edges are curved if there are multiple edges between two nodes:
library("qgraph")
qgraph(Edges,esize=5,gray=TRUE)
However this package is not really intended for this purpose and you can't change the edge colors (yet, working on it:) ). You can only make all edges black with a small trick:
qgraph(Edges,esize=5,gray=TRUE,minimum=0,cut=.Machine$double.xmin)
For more control you can use the igraph package. First we make the graph:
library("igraph")
g <- graph.edgelist(as.matrix(Edges[,-3]))
Note the conversion to matrix and subtracting one because the first node is 0. Next we define the layout:
l <- layout.fruchterman.reingold(g)
Now we can change some of the edge parameters with the E()function:
# Define edge widths:
E(g)$width <- Edges$thickness * 5
# Define arrow widths:
E(g)$arrow.width <- Edges$thickness * 5
# Make edges curved:
E(g)$curved <- 0.2
And finally plot the graph:
plot(g,layout=l)
While not an R answer specifically, I would recommend using Cytoscape to generate the network.
You can automate it using a RCytoscape.
http://bioconductor.org/packages/release/bioc/html/RCytoscape.html
The package informatively named 'network' can draw directed networks fairly well, and handle your issues.
ex.net <- rbind(c(0, 1, 1, 1), c(1, 0, 0, 1), c(0, 0, 0, 1), c(1, 0, 1, 0))
plot(network(ex.net), usecurve = T, edge.curve = 0.00001,
edge.lwd = c(4, rep(1, 7)))
The edge.curve argument, if set very low and combined with usecurve=T, separates the edges, although there might be a more direct way of doing this, and edge.lwd can take a vector as its argument for different sizes.
It's not always the prettiest result, I admit. But it's fairly easy to get decent looking network plots that can be customized in a number of different ways (see ?network.plot).
The 'non overlapping' constraint on edges is the big problem here. First, your network has to be 'planar' otherwise it's impossible in 2-dimensions (you cant connect three houses to gas, electric, phone company buildings without crossovers).
I think an algorithm for planar graph layout essentially solves the 4-colour problem. Have fun with that. Heuristics exist, search for planar graph layout, and force-directed, and read Planar Graph Layouts