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)
Related
I want to visualize the elements of my vectors in a graph. I want to generate a graph with a certain x- and y-axis and then put the values of my vectors as points into the graph. I also want different colors for the values of the different vectors. How do I do that?
For example: I have 10 elements in vector A and want to put those elements into the graph. The first Element of vector A has the y-value A[1] and the x-value 1. The second Element of vector A has the y-value A[2] and the x-value 2. Same with vector B.
vec1 = 1:10
vec2 = 1:10
for(idx in 1:10){
vec1[idx] = runif(1, min=0, max=100)
vec2[idx] = runif(1, min=0, max=100)
}
plot(vec1 and vec2) // How do I do this?
dput output for vec1: c(81.9624423747882, 45.583715592511, 56.2400584807619, 8.25600677635521, 82.0227505406365, 45.6240070518106, 68.7916911672801, 94.491201499477, 22.0095717580989, 4.29550902917981)
dput output for vec2: c(29.5684755546972, 68.0154771078378, 52.2058120695874, 2.48502977192402, 91.9532125117257, 24.7736480785534, 66.5003522532061, 79.014728218317, 47.9641782585531, 20.5593338003382)
Starting with
vec1 = 1:10
vec2 = 1:10
for(idx in 1:10){
vec1[idx] = runif(1, min=0, max=100)
vec2[idx] = runif(1, min=0, max=100)
}
plot(vec1 and vec2) // How do I do this?
Try this:
plot( 1:20, c(vec1,vec2) , col=rep(1:2,10) # just points
lines( 1:20, c(vec1,vec2) ) # add lines
# if you wanted the same x's for both sequences the first argument could be
# rep(1:10, 2) instead of 1:20
Note: Your set up code could have been just two lines (no loop):
vec1 = runif(10, min=0, max=100)
vec2 = runif(10, min=0, max=100)
I think the easiest is to create a data frame, which is usually what most functions expect in R:
library(tidyverse)
vec1 = 1:10
vec2 = 1:10
for(idx in 1:10){
vec1[idx] = runif(1, min=0, max=100)
vec2[idx] = runif(1, min=0, max=100)
}
df <- data.frame(order = 1:10, vec1, vec2) %>%
pivot_longer(!order, names_to = "color", values_to = "value")
plot(df$order, df$value, col = c("red","blue")[df$color %>% as.factor()])
I'm wondering or guessing whether you are aiming for the facility provided by teh base-plotting function arrows? This is the example in the ?arrows page:
x <- stats::runif(12); y <- stats::rnorm(12)
i <- order(x, y); x <- x[i]; y <- y[i]
plot(x,y, main = "arrows(.)" )
## draw arrows from point to point :
s <- seq(length(x)-1) # one shorter than data
arrows(x[s], y[s], x[s+1], y[s+1], col = 1:3)
If you wanted instead to plot with each vector (represented by "arrows") starting from the origin it would be:
x <- stats::runif(12); y <- stats::rnorm(12)
# ordering not needed this time
plot(x,y, main = "arrows(.)", xlim=c(0, max(x)) # to let origin be seen)
## draw arrows from origin to point :
s <- length(x) # one shorter than data
arrows(rep(0,s), rep(0,s), x, y, col = 1:3)
Using the example located here https://www.datacamp.com/community/tutorials/hierarchical-clustering-R and the data located https://archive.ics.uci.edu/ml/datasets/seeds# i am trying to remove the labels at the bottom of the dendrogram when using the color_branches
when plot(hclust_avg, labels=FALSE) it works but not later when using color_branches. is there a way to remove them?
`set.seed(786)
seeds_df <- read.csv("seeds_dataset.txt",sep = '\t',header = FALSE)
feature_name <- c('area','perimeter','compactness','length.of.kernel','width.of.kernal','asymmetry.coefficient','length.of.kernel.groove','type.of.seed')
colnames(seeds_df) <- feature_name
seeds_df<- seeds_df[complete.cases(seeds_df), ]
seeds_label <- seeds_df$type.of.seed
seeds_df$type.of.seed <- NULL
seeds_df_sc <- as.data.frame(scale(seeds_df))
dist_mat <- dist(seeds_df_sc, method = 'euclidean')
hclust_avg <- hclust(dist_mat, method = 'average')
cut_avg <- cutree(hclust_avg, k = 3)
suppressPackageStartupMessages(library(dendextend))
avg_dend_obj <- as.dendrogram(hclust_avg)
avg_col_dend <- color_branches(avg_dend_obj, h = 3)
plot(avg_col_dend)`
Figured this out by colouring the the labels white to the background
avg_dend_obj <- as.dendrogram(hclust_avg)
labels_colors(avg_dend_obj) <- "white"
plot(avg_dend_obj)
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$clrs))
I am looking for a procedure that allows me to generate a sequence of equidistant points (coordinates) along the sides of an arbitrary polygon.
Imaging a polygon defined by the coordinates of its vertexes:
poly.mat <- matrix(c(0,0,
0,1,
0.5,1.5,
0.5,0,
0,0 # last row included to close the polygon
), byrow = T, ncol = 2)
colnames(poly.mat) <- c("x", "y")
plot(poly.mat, type = "l")
If the length of the sequence I want to generate is n (adjustable), how I can produce a sequence, starting at (0,0), of equidistant coordinates.
I got as far as calculating the perimeter of the shape with the geosphere package (which I believe I need)
library(geosphere)
n <- 50 # sequence of length set to be 50
perim <- perimeter(poly.mat)
perim/n # looks like every section needs to be 8210.768 something in length
You will have to write the code yourself. Sorry, there isn't a library function for every last detail of every last assignment. Assuming that each pair of points defines a line segment, you could just generate N points along each segment, as in
begin = [xbegin, ybegin ];
end = [xend, yend ];
xdist = ( xend - xbegin ) / nintervals;
ydist = ( yend - ybegin ) / nintervals;
then your points are given by [ xbegin + i * xdist, ybegin + i * ydist ]
Here is the solution I came up with.
pointDistance <- function(p1, p2){
sqrt((p2[,1]-p1[,1])^2) + sqrt((p2[,2]-p1[,2])^2)
}
getPos <- function(shp.mat, ll){
greaterLL <- shp.mat$cumdis > ll
if(all(greaterLL == FALSE)) return(poly.mat[nrow(poly.mat), c("x", "y")])
smallRow <- min(which(greaterLL)) # the smallest coordinate that has greater length
p.start <- shp.mat[smallRow-1, c("x","y")]
p.end <- shp.mat[smallRow, c("x","y")]
cumVal <- shp.mat$cumdis[smallRow]
prop <- (ll-shp.mat$cumdis[smallRow-1])/(shp.mat$cumdis[smallRow]-shp.mat$cumdis[smallRow-1])
p.start + (prop)* (p.end-p.start)
}
# shp1
poly.mat <- matrix(c(0,0,
0,1,
0.5,1.5,
0.5,0,
0,0
),byrow = T, ncol = 2)
colnames(poly.mat) <- c("x", "y")
poly.mat <- as.data.frame(poly.mat)
# Main fun
pointsOnPath <- function(shp.mat, n){
dist <- vector(mode = "numeric", length = nrow(shp.mat)-1)
for(i in 2:nrow(shp.mat)){
dist[i] <- pointDistance(p1 = shp.mat[i,], p2 = shp.mat[i-1,])
}
shp.mat$dist <- dist
shp.mat$cumdis <- cumsum(shp.mat$dist)
dis <- matrix(seq(from = 0, to = max(shp.mat$cumdis), length.out = n+1), ncol = 1)
out <- lapply(dis, function(x) getPos(shp.mat = shp.mat, ll = x))
out <- do.call("rbind", out)
out$dis <- dis
out[-nrow(out),]
}
df <- pointsOnPath(shp.mat = poly.mat, 5)
# Plot
plot(poly.mat$x, poly.mat$y, type = "l", xlim = c(0,1.5), ylim = c(0,1.5))
points(df$x, df$y, col = "red", lwd = 2)
There is room for improving the code, but it should return the correct result
I'm using RasterVis and levelplot to make a trellis plot of some rasters. I am currently ok for most things but I would like to change the header for each panel from the filename to a chosen string (the filename is convoluted and long, i want to use just a year, for example '2004').
Looking at the levelplot page, it would indicate that levelplot goes looking for some settings as per the argument 'useRaster', either it goes to panel.levelplot or panel.levelplot.raster, but im struggling to use these latter functions.
Any help much appreciated, here's some sample code;
require(rasterVis)
layers <- c(1:4)
s2 <- stack()
for (i in layers) {
r <- raster(nrows=100, ncols=100,ext)
r[] <- sample(seq(from = 1, to = 6, by = 1), size = 10000, replace = TRUE)
rasc <- ratify(r)
rat <- levels(rasc)[[1]]
rat$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)