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Is there a restriction on how many raster I can plot using using the plot function? For e.g
library(raster)
mystack <- stack()
for(i in 1:25){
df <- data.frame( x = rep( 0:1, each=2 ),
y = rep( 0:1, 2),
l = rnorm( 4 ))
dfr <- rasterFromXYZ(df)
mystack <- stack(mystack, dfr)
}
plot(mystack)
It plots only 16 plots no matter how many iterations I do.
You were on the right track, from ?`plot,Raster,ANY-method`:
## S4 method for signature 'Raster,ANY' plot(x, y, maxpixels=500000, col, alpha=NULL, colNA=NA, add=FALSE, ext=NULL, useRaster=TRUE,
interpolate=FALSE, addfun=NULL, nc, nr, maxnl=16, main, npretty=0,
...)
Notice maxnl = 16. All you need is to change that to the desired number of layers:
plot(mystack, maxnl=25)
I wish to visualize how well a clustering algorithm is doing (with certain distance metric). I have samples and their corresponding classes.
To visualize, I cluster and I wish to color the branches of a dendrogram by the items in the cluster. The color will be the color most items in the hierarchical cluster correspond to (given by the data\classes).
Example: If my clustering algorithm chose indexes 1,21,24 to be a certain cluster (at a certain level) and I have a csv file containing a class number in each row corresponding to lets say 1,2,1. I want this edge to be coloured 1.
Example Code:
require(cluster)
suppressPackageStartupMessages(library(dendextend))
dir <- 'distance_metrics/'
filename <- 'aligned.csv'
my.data <- read.csv(paste(dir, filename, sep=""), header = T, row.names = 1)
my.dist <- as.dist(my.data)
real.clusters <-read.csv("clusters", header = T, row.names = 1)
clustered <- diana(my.dist)
# dend <- colour_branches(???dend, max(real.clusters)???)
plot(dend)
EDIT:
another example partial code
dir <- 'distance_metrics/' # csv in here contains a symmetric matrix
clust.dir <- "clusters/" #csv in here contains a column vector with classes
my.data <- read.csv(paste(dir, filename, sep=""), header = T, row.names = 1)
filename <- 'table.csv'
my.dist <- as.dist(my.data)
real.clusters <-read.csv(paste(clust.dir, filename, sep=""), header = T, row.names = 1)
clustered <- diana(my.dist)
dnd <- as.dendrogram(clustered)
Both node and edge color attributes can be set recursively on "dendrogram" objects (which are just deeply nested lists) using dendrapply. The cluster package also features an as.dendrogram method for "diana" class objects, so conversion between the object types is seamless. Using your diana clustering and borrowing some code from #Edvardoss iris example, you can create the colored dendrogram as follows:
library(cluster)
set.seed(999)
iris2 <- iris[sample(x = 1:150,size = 50,replace = F),]
clust <- diana(iris2)
dnd <- as.dendrogram(clust)
## Duplicate rownames aren't allowed, so we need to set the "labels"
## attributes recursively. We also label inner nodes here.
rectify_labels <- function(node, df){
newlab <- df$Species[unlist(node, use.names = FALSE)]
attr(node, "label") <- (newlab)
return(node)
}
dnd <- dendrapply(dnd, rectify_labels, df = iris2)
## Create a color palette as a data.frame with one row for each spp
uniqspp <- as.character(unique(iris$Species))
colormap <- data.frame(Species = uniqspp, color = rainbow(n = length(uniqspp)))
colormap[, 2] <- c("red", "blue", "green")
colormap
## Now color the inner dendrogram edges
color_dendro <- function(node, colormap){
if(is.leaf(node)){
nodecol <- colormap$color[match(attr(node, "label"), colormap$Species)]
attr(node, "nodePar") <- list(pch = NA, lab.col = nodecol)
attr(node, "edgePar") <- list(col = nodecol)
}else{
spp <- attr(node, "label")
dominantspp <- levels(spp)[which.max(tabulate(spp))]
edgecol <- colormap$color[match(dominantspp, colormap$Species)]
attr(node, "edgePar") <- list(col = edgecol)
}
return(node)
}
dnd <- dendrapply(dnd, color_dendro, colormap = colormap)
## Plot the dendrogram
plot(dnd)
The function you are looking for is color_brances from the dendextend R package, using the arguments clusters and col. Here is an example (based on Shaun Wilkinson's example):
library(cluster)
set.seed(999)
iris2 <- iris[sample(x = 1:150,size = 50,replace = F),]
clust <- diana(iris2)
dend <- as.dendrogram(clust)
temp_col <- c("red", "blue", "green")[as.numeric(iris2$Species)]
temp_col <- temp_col[order.dendrogram(dend)]
temp_col <- factor(temp_col, unique(temp_col))
library(dendextend)
dend %>% color_branches(clusters = as.numeric(temp_col), col = levels(temp_col)) %>%
set("labels_colors", as.character(temp_col)) %>%
plot
there are suspicions that misunderstood the question however I'll try to answer:
from my previous objectives were rewritten by the example of iris
clrs <- rainbow(n = 3) # create palette
clrs <- clrs[iris$Species] # assign colors
plot(x = iris$Sepal.Length,y = iris$Sepal.Width,col=clrs) # simple test colors
# cluster
dt <- cbind(iris,clrs)
dt <- dt[sample(x = 1:150,size = 50,replace = F),] # create short dataset for visualization convenience
empty.labl <- gsub("."," ",dt$Species) # create a space vector with length of names intended for reserve place to future text labels
dst <- dist(x = scale(dt[,1:4]),method = "manhattan")
hcl <- hclust(d = dst,method = "complete")
plot(hcl,hang=-1,cex=1,labels = empty.labl, xlab = NA,sub=NA)
dt <- dt[hcl$order,] # sort rows for order objects in dendrogramm
text(x = seq(nrow(dt)), y=-.5,labels = dt$Species,srt=90,cex=.8,xpd=NA,adj=c(1,0.7),col=as.character(dt$clrs))
I'd like to be able to facet an R plotly heatmap.
Here's what I mean:
I have a hierarchically-clustered gene expression dataset:
require(permute)
set.seed(1)
mat <- rbind(cbind(matrix(rnorm(2500,2,1),nrow=25,ncol=500),matrix(rnorm(2500,-2,1),nrow=25,ncol=500)),
cbind(matrix(rnorm(2500,-2,1),nrow=25,ncol=500),matrix(rnorm(2500,2,1),nrow=25,ncol=500)))
rownames(mat) <- paste("g",1:50,sep=".")
colnames(mat) <- paste("s",1:1000,sep=".")
hc.col <- hclust(dist(t(mat)))
dd.col <- as.dendrogram(hc.col)
col.order <- order.dendrogram(dd.col)
hc.row <- hclust(dist(mat))
dd.row <- as.dendrogram(hc.row)
row.order <- order.dendrogram(dd.row)
mat <- mat[row.order,col.order]
I then discretize it to specific expression ranges because that happens to help the resolution of colors for my case. I'm also creating other structures to help me plot the colorbar the way I want it to:
require(RColorBrewer)
mat.intervals <- cut(mat,breaks=6)
interval.mat <- matrix(mat.intervals,nrow=50,ncol=1000,dimnames=list(rownames(mat),colnames(mat)))
interval.cols <- brewer.pal(6,"Set2")
names(interval.cols) <- levels(mat.intervals)
require(reshape2)
interval.df <- reshape2::melt(interval.mat,varnames=c("gene","sample"),value.name="expr")
interval.cols2 <- rep(interval.cols, each=ncol(mat))
color.df <- data.frame(range=c(0:(2*length(interval.cols)-1)),colors=c(0:(2*length(interval.cols)-1)))
color.df <- setNames(data.frame(color.df$range,color.df$colors),NULL)
for (i in 1:(2*length(interval.cols))) {
color.df[[2]][[i]] <- interval.cols[[(i + 1) / 2]]
color.df[[1]][[i]] <- i/(2*length(interval.cols))-(i %% 2)/(2*length(interval.cols))
}
They way I generated the data I know that samples 1-500 are one cluster and samples 501:1000 are the other, so I label them:
interval.df$cluster <- NA
interval.df$cluster[which(interval.df$sample %in% paste("s",1:500,sep="."))] <- "A"
interval.df$cluster[which(interval.df$sample %in% paste("s",501:1000,sep="."))] <- "B"
I thought that adding a sample with not color and interval will create a white column in the heatmap plot that will look like a facet border:
divider.df <- data.frame(gene=unique(interval.df$gene),sample=NA,expr=NA,cluster=NA)
interval.df <- rbind(dplyr::filter(interval.df,cluster == "A"),divider.df,dplyr::filter(interval.df,cluster == "B"))
And now I try plotting:
#add ticks for each cluster
tick.vals <- c("s.158","s.617")
tick.text <- c("A","B")
require(plotly)
plot_ly(z=c(interval.df$expr),x=interval.df$sample,y=interval.df$gene,colors=interval.cols2,type="heatmap",colorscale=color.df,
colorbar=list(title="score",tickmode="array",tickvals=c(1:6),ticktext=names(interval.cols),len=0.2,outlinecolor="white",bordercolor="white",borderwidth=5,bgcolor="white")) %>%
layout(xaxis = list(title = 'Cluster',tickmode = 'array',tickvals = tick.vals,ticktext = tick.text))
But I don't see any separation between the clusters:
Any idea how to achieve such a facet border between the two clusters?
Your example is quite involved so I have reduced it down to a minimal example to focus on the gap you are looking for in the quadrants of your heatmap.
Modified from the examples on the plotly site, here.
library(plotly)
m <- matrix(rnorm(9), nrow = 3, ncol = 3)
p <- plot_ly(
x = c("a", "b", "c"), y = c("d", "e", "f"),
z = m, type = "heatmap"
)
subplot(p, p, p, p, shareX = TRUE, shareY = TRUE, nrows = 2)
If you create a plotly object for each of the quadrants and then use subplot, you will get a result looking similar to this:
N.B. I have cropped out the legend because it was duplicated for the facets, you could merge these into one.
I'm using RasterVis and levelplot to make a trellis plot of some rasters. I am currently ok for most things but I would like to change the header for each panel from the filename to a chosen string (the filename is convoluted and long, i want to use just a year, for example '2004').
Looking at the levelplot page, it would indicate that levelplot goes looking for some settings as per the argument 'useRaster', either it goes to panel.levelplot or panel.levelplot.raster, but im struggling to use these latter functions.
Any help much appreciated, here's some sample code;
require(rasterVis)
layers <- c(1:4)
s2 <- stack()
for (i in layers) {
r <- raster(nrows=100, ncols=100,ext)
r[] <- sample(seq(from = 1, to = 6, by = 1), size = 10000, replace = TRUE)
rasc <- ratify(r)
rat <- levels(rasc)[[1]]
rat$legend <- c("A","B","C","D","E","F")
levels(rasc) <- rat
s2 <- stack(s2, rasc)
}
levelplot(s2, col.regions=rev(terrain.colors(6)),main = "example")
In the above e.g., I would like "layer.1.1" to be "2004", and so on through to 2007
require(rasterVis)
layers <- c(1:4)
s2 <- stack()
for (i in layers) {
r <- raster(nrows=100, ncols=100)
r[] <- sample(seq(from = 1, to = 6, by = 1), size = 10000, replace = TRUE)
rasc <- ratify(r)
rat <- levels(rasc)[[1]]
rat$legend <- c("A","B","C","D","E","F")
levels(rasc) <- rat
s2 <- stack(s2, rasc)
}
levelplot(s2, col.regions=rev(terrain.colors(6)),main = "example", names.attr=2004:2007)
p.strip <- list(cex=1.5, lines=1, col="blue", fontfamily='Serif')
levelplot(s2, col.regions=rev(terrain.colors(6)), main = "example",
names.attr=2004:2007, par.strip.text=p.strip)
I want to modify the properties of the leaves in a dendrogram produced from plot of an hclust object. Minimally, I want to change the colors, but any help you can provide will be appreciated.
I did try to google the answer, but but every solution that I saw seemed alot harder than what I would have guessed.
A while ago, Joris Meys kindly provided me with this snippet of code that changes the color of leaves. Modify it to reflect your attributes.
clusDendro <- as.dendrogram(Clustering)
labelColors <- c("red", "blue", "darkgreen", "darkgrey", "purple")
## function to get colorlabels
colLab <- function(n) {
if(is.leaf(n)) {
a <- attributes(n)
# clusMember - a vector designating leaf grouping
# labelColors - a vector of colors for the above grouping
labCol <- labelColors[clusMember[which(names(clusMember) == a$label)]]
attr(n, "nodePar") <- c(a$nodePar, lab.col = labCol)
}
n
}
## Graph
clusDendro <- dendrapply(clusDendro, colLab)
op <- par(mar = par("mar") + c(0,0,0,2))
plot(clusDendro,
main = "Major title",
horiz = T, type = "triangle", center = T)
par(op)
Here is a solution for this question using a new package called "dendextend", built exactly for this sort of thing.
You can see many examples in the presentations and vignettes of the package, in the "usage" section in the following URL: https://github.com/talgalili/dendextend
Here is the solution for this question:
# define dendrogram object to play with:
dend <- as.dendrogram(hclust(dist(USArrests[1:3,]), "ave"))
# loading the package
install.packages('dendextend') # it is now on CRAN
library(dendextend)# let's add some color:
labels_colors(dend) <- 2:4
labels_colors(dend)
plot(dend)
It is not clear what you want to use it for, but I often need to identify a branch in a dendrogram. I've hacked the rect.hclust method to add a density and label input.
You would call it like this:
k <- 3 # number of branches to identify
labels.to.identify <- c('1','2','3')
required.density <- 10 # the density of shading lines, in lines per inch
rect.hclust.nice(tree, k, labels=labels.to.identify, density=density.required)
Here is the function
rect.hclust.nice = function (tree, k = NULL, which = NULL, x = NULL, h = NULL, border = 2,
cluster = NULL, density = NULL,labels = NULL, ...)
{
if (length(h) > 1 | length(k) > 1)
stop("'k' and 'h' must be a scalar")
if (!is.null(h)) {
if (!is.null(k))
stop("specify exactly one of 'k' and 'h'")
k <- min(which(rev(tree$height) < h))
k <- max(k, 2)
}
else if (is.null(k))
stop("specify exactly one of 'k' and 'h'")
if (k < 2 | k > length(tree$height))
stop(gettextf("k must be between 2 and %d", length(tree$height)),
domain = NA)
if (is.null(cluster))
cluster <- cutree(tree, k = k)
clustab <- table(cluster)[unique(cluster[tree$order])]
m <- c(0, cumsum(clustab))
if (!is.null(x)) {
if (!is.null(which))
stop("specify exactly one of 'which' and 'x'")
which <- x
for (n in 1L:length(x)) which[n] <- max(which(m < x[n]))
}
else if (is.null(which))
which <- 1L:k
if (any(which > k))
stop(gettextf("all elements of 'which' must be between 1 and %d",
k), domain = NA)
border <- rep(border, length.out = length(which))
labels <- rep(labels, length.out = length(which))
retval <- list()
for (n in 1L:length(which)) {
rect(m[which[n]] + 0.66, par("usr")[3L], m[which[n] +
1] + 0.33, mean(rev(tree$height)[(k - 1):k]), border = border[n], col = border[n], density = density, ...)
text((m[which[n]] + m[which[n] + 1]+1)/2, grconvertY(grconvertY(par("usr")[3L],"user","ndc")+0.02,"ndc","user"),labels[n])
retval[[n]] <- which(cluster == as.integer(names(clustab)[which[n]]))
}
invisible(retval)
}