avoiding over-crowding of labels in r graphs - r

I am working on avoid over crowding of the labels in the following plot:
set.seed(123)
position <- c(rep (0,5), rnorm (5,1,0.1), rnorm (10, 3,0.1), rnorm (3, 4, 0.2), 5, rep(7,5), rnorm (3, 8,2), rnorm (10,9,0.5),
rep (0,5), rnorm (5,1,0.1), rnorm (10, 3,0.1), rnorm (3, 4, 0.2), 5, rep(7,5), rnorm (3, 8,2), rnorm (10,9,0.5))
group <- c(rep (1, length (position)/2),rep (2, length (position)/2) )
mylab <- paste ("MR", 1:length (group), sep = "")
barheight <- 0.5
y.start <- c(group-barheight/2)
y.end <- c(group+barheight/2)
mydf <- data.frame (position, group, barheight, y.start, y.end, mylab)
plot(0,type="n",ylim=c(0,3),xlim=c(0,10),axes=F,ylab="",xlab="")
#Create two horizontal lines
require(fields)
yline(1,lwd=4)
yline(2,lwd=4)
#Create text for the lines
text(10,1.1,"Group 1",cex=0.7)
text(10,2.1,"Group 2",cex=0.7)
#Draw vertical bars
lng = length(position)/2
lg1 = lng+1
lg2 = lng*2
segments(mydf$position[1:lng],mydf$y.start[1:lng],y1=mydf$y.end[1:lng])
segments(mydf$position[lg1:lg2],mydf$y.start[lg1:lg2],y1=mydf$y.end[lg1:lg2])
text(mydf$position[1:lng],mydf$y.start[1:lng]+0.65, mydf$mylab[1:lng], srt = 90)
text(mydf$position[lg1:lg2],mydf$y.start[lg1:lg2]+0.65, mydf$mylab[lg1:lg2], srt = 90)
You can see some areas are crowed with the labels - when x value is same or similar. I want just to display only one label (when there is multiple label at same point). For example,
mydf$position[1:5] are all 0,
but corresponding labels mydf$mylab[1:5] -
MR1 MR2 MR3 MR4 MR5
I just want to display the first one "MR1".
Similarly the following points are too close (say the difference of 0.35), they should be considered a single cluster and first label will be displayed. In this way I would be able to get rid of overcrowding of labels. How can I achieve it ?

If you space the labels out and add some extra lines you can label every marker.
clpl <- function(xdata, names, y=1, dy=0.25, add=FALSE){
o = order(xdata)
xdata=xdata[o]
names=names[o]
if(!add)plot(0,type="n",ylim=c(y-1,y+2),xlim=range(xdata),axes=F,ylab="",xlab="")
abline(h=1,lwd=4)
dy=0.25
segments(xdata,y-dy,xdata,y+dy)
tpos = seq(min(xdata),max(xdata),len=length(xdata))
text(tpos,y+2*dy,names,srt=90,adj=0)
segments(xdata,y+dy,tpos,y+2*dy)
}
Then using your data:
clpl(mydf$position[lg1:lg2],mydf$mylab[lg1:lg2])
gives:
You could then think about labelling clusters underneath the main line.
I've not given much thought to doing multiple lines in a plot, but I think with a bit of mucking with my code and the add parameter it should be possible. You could also use colour to show clusters. I'm fairly sure these techniques are present in some of the clustering packages for R...
Obviously with a lot of markers even this is going to get smushed, but with a lot of clusters the same thing is going to happen. Maybe you end up labelling clusters with a this technique?

In general, I agree with #Joran that cluster labelling can't be automated but you've said that labelling a group of lines with the first label in the cluster would be OK, so it is possible to automate some of the process.
Putting the following code after the line lg2 = lng*2 gives the result shown in the image below:
clust <- cutree(hclust(dist(mydf$position[1:lng])),h=0.75)
u <- rep(T,length(unique(clust)))
clust.labels <- sapply(c(1:lng),function (i)
{
if (u[clust[i]])
{
u[clust[i]] <<- F
as.character(mydf$mylab)[i]
}
else
{
""
}
})
segments(mydf$position[1:lng],mydf$y.start[1:lng],y1=mydf$y.end[1:lng])
segments(mydf$position[lg1:lg2],mydf$y.start[lg1:lg2],y1=mydf$y.end[lg1:lg2])
text(mydf$position[1:lng],mydf$y.start[1:lng]+0.65, clust.labels, srt = 90)
text(mydf$position[lg1:lg2],mydf$y.start[lg1:lg2]+0.65, mydf$mylab[lg1:lg2], srt = 90)
(I've only labelled the clusters on the lower line -- the same principle could be applied to the upper line too). The parameter h of cutree() might have to be adjusted case-by-case to give the resolution of labels that you want, but this approach is at least easier than labelling every cluster by hand.

Related

non-linear 2d object transformation by horizontal axis

How can such a non-linear transformation be done?
here is the code to draw it
my.sin <- function(ve,a,f,p) a*sin(f*ve+p)
s1 <- my.sin(1:100, 15, 0.1, 0.5)
s2 <- my.sin(1:100, 21, 0.2, 1)
s <- s1+s2+10+1:100
par(mfrow=c(1,2),mar=rep(2,4))
plot(s,t="l",main = "input") ; abline(h=seq(10,120,by = 5),col=8)
plot(s*7,t="l",main = "output")
abline(h=cumsum(s)/10*2,col=8)
don't look at the vector, don't look at the values, only look at the horizontal grid, only the grid matters
####UPDATE####
I see that my question is not clear to many people, I apologize for that...
Here are examples of transformations only along the vertical axis, maybe now it will be more clear to you what I want
link Source
#### UPDATE 2 ####
Thanks for your answer, this looks like what I need, but I have a few more questions if I may.
To clarify, I want to explain why I need this, I want to compare vectors with each other that are non-linearly distorted along the horizontal axis .. Maybe there are already ready-made tools for this?
You mentioned that there are many ways to do such non-linear transformations, can you name a few of the best ones in my case?
how to make the function f() more non-linear, so that it consists, for example, not of one sinusoid, but of 10 or more. Тhe figure shows that the distortion is quite simple, it corresponds to one sinusoid
and how to make the function f can be changed with different combinations of sinusoids.
set.seed(126)
par(mar = rep(2, 4),mfrow=c(1,3))
s <- cumsum(rnorm(100))
r <- range(s)
gridlines <- seq(r[1]*2, r[2]*2, by = 0.2)
plot(s, t = "l", main = "input")
abline(h = gridlines, col = 8)
f <- function(x) 2 * sin(x)/2 + x
plot(s, t = "l", main = "input+new greed")
abline(h = f(gridlines), col = 8)
plot(f(s), t = "l", main = "output")
abline(h = f(gridlines), col = 8)
If I understand you correctly, you wish to map the vector s from the regular spacing defined in the first image to the irregular spacing implied by the second plot.
Unfortunately, your mapping is not well-defined, since there is no clear correspondence between the horizontal lines in the first image and the second image. There are in fact an infinite number of ways to map the first space to the second.
We can alter your example a bit to make it a bit more rigorous.
If we start with your function and your data:
my.sin <- function(ve, a, f, p) a * sin(f * ve + p)
s1 <- my.sin(1:100, 15, 0.1, 0.5)
s2 <- my.sin(1:100, 21, 0.2, 1)
s <- s1 + s2 + 10 + 1:100
Let us also create a vector of gridlines that we will draw on the first plot:
gridlines <- seq(10, 120, by = 2.5)
Now we can recreate your first plot:
par(mar = rep(2, 4))
plot(s, t = "l", main = "input")
abline(h = gridlines, col = 8)
Now, suppose we have a function that maps our y axis values to a different value:
f <- function(x) 2 * sin(x/5) + x
If we apply this to our gridlines, we have something similar to your second image:
plot(s, t = "l", main = "input")
abline(h = f(gridlines), col = 8)
Now, what we want to do here is effectively transform our curve so that it is stretched or compressed in such a way that it crosses the gridlines at the same points as the gridlines in the original image. To do this, we simply apply our mapping function to s. We can check the correspondence to the original gridlines by plotting our new curves with a transformed axis :
plot(f(s), t = "l", main = "output", yaxt = "n")
axis(2, at = f(20 * 1:6), labels = 20 * 1:6)
abline(h = f(gridlines), col = 8)
It may be possible to create a mapping function using the cumsum(s)/10 * 2 that you have in your original example, but it is not clear how you want this to correspond to the original y axis values.
Response to edits
It's not clear what you mean by comparing two vectors. If one is a non-linear deformation of the other, then presumably you want to find the underlying function that produces the deformation. It is possible to create a function that applies the deformation empirically simply by doing f <- approxfun(untransformed_vector, transformed_vector).
I didn't say there were many ways of doing non-linear transformations. What I meant is that in your original example, there is no correspondence between the grid lines in the original picture and the second picture, so there is an infinite choice for which gridines in the first picture correspond to which gridlines in the second picture. There is therefore an infinite choice of mapping functions that could be specified.
The function f can be as complicated as you like, but in this scenario it should at least be everywhere non-decreasing, such that any value of the function's output can be mapped back to a single value of its input. For example, function(x) x + sin(x)/4 + cos(3*(x + 2))/5 would be a complex but ever-increasing sinusoidal function.

Clusters labels in dendrogram

I'm wondering - is there any way how to add labels of clusters into dendrogram. See simple example:
hc = hclust(dist(mtcars))
plot(hc, hang = -1)
rect.hclust(hc, k = 3, border = "red")
Desired output should look like this:
Thanks for any suggestions!
You need to get the coordinates of the place to put your clusters' labels:
First axis:
As you are calling rect.hclust, you might as well assign the result so you can use it to find the beginning of clusters (the first one begins at 1 the 2nd at 1 + the length of the first, etc.
rh <- rect.hclust(hc, k = 3, border = "red")
beg_clus <- head(cumsum(c(1, lengths(rh))), -1)
Second axis:
You just want to be above the red rectangle, which is at the middle of the height where you have k-1 clusters and the height where you have k clusters. Let's say you're aiming at 4/5 of the distance instead of 1/2:
y_clus <- weighted.mean(rev(hc$height)[2:3], c(4, 1))
Putting the labels:
text(x=beg_clus, y=y_clus, col="red", labels=LETTERS[1:3], font=2)
An alternative to adding text labels is in the mjcgraphics package that deals with cluster labels. See https://github.com/drmjc/mjcgraphics and https://rdrr.io/github/drmjc/mjcgraphics/man/rect.hclust.labels.html
rect.hclust.labels(hc, k=3, border = 1 ) # adds labels to clusters

R - Histogram Doesn't show density due to magnitude of the Data

I have a vector called data with length 444000 approximately, and most of the numeric values are between 1 and 100 (almost all of them). I want to draw the histogram and draw the the appropriate density on it. However, when I draw the histogram I get this:
hist(data,freq=FALSE)
What can I do to actually see a more detailed histogram? I tried to use the breaks code, it helped, but it's really hard do see the histogram, because it's so small. For example I used breaks = 2000 and got this:
Is there something that I can do? Thanks!
Since you don't show data, I'll generate some random data:
d <- c(rexp(1e4, 100), runif(100, max=5e4))
hist(d)
Dealing with outliers like this, you can display the histogram of the logs, but that may difficult to interpret:
If you are okay with showing a subset of the data, then you can filter the outliers out either dynamically (perhaps using quantile) or manually. The important thing when showing this visualization in your analysis is that if you must remove data for the plot, then be up-front when the removal. (This is terse ... it would also be informative to include the range and/or other properties of the omitted data, but that's subjective and will differ based on the actual data.)
quantile(d, seq(0, 1, len=11))
d2 <- d[ d < quantile(d, 0.90) ]
hist(d2)
txt <- sprintf("(%d points shown, %d excluded)", length(d2), length(d) - length(d2))
mtext(txt, side = 1, line = 3, adj = 1)
d3 <- d[ d < 10 ]
hist(d3)
txt <- sprintf("(%d points shown, %d excluded)", length(d3), length(d) - length(d3))
mtext(txt, side = 1, line = 3, adj = 1)

Plotting a chessboard with no external libraries

I'd like if someone could help me with this problem I've been hours trying to solve.
I have to plot a chessboard with no external libraries (using only the default graphical functions in R).
My attempt is working with black squares till I have to filter and paint the white squares:
plot(c(1:9),c(1:9),type="n")
for (i in 1:8){
rect(i,1:9,i+1,9,col="black",border="white")
}
I could do it manually in this way, but I know there's a simpler way:
plot(c(1:9),c(1:9),type="n")
rect(1, 2, 2, 1,col="black",border="white")
rect(4, 1, 3, 2,col="black",border="white")
rect(6, 1, 5, 2,col="black",border="white")
rect(7, 1, 8, 2,col="black",border="white")
(...)
I've tried adding a function to filter even numbers inside the loop but doesn't seems to works for me.
I would appreciate any suggestion!
Use image and just repeat 0:1 over and over. Then you can mess with the limits a bit to make it fit nice.
image(matrix(1:0, 9, 9), col=0:1, xlim=c(-.05,.93), ylim=c(-.05,.93))
Just change the col= argument in your solution as shown. Also note that c(1:9) can be written as just 1:9 :
plot(1:9, 1:9, type = "n")
for (i in 1:8) {
col <- if (i %% 2) c("white", "black") else c("black", "white")
rect(i, 1:9, i+1, 9, col = col, border = "white")
}
remembering Jeremy Kun's post
https://jeremykun.com/2018/03/25/a-parlor-trick-for-set/ on Set helped
me figure the hard part (for me) of this question. i realized that
diagonals on the board (what bishops move on) have a constant color.
and, so, their Y-intercept (where they hit the Y-axis) will uniquely
determine their color, and adjacent Y values will have different
colors. for a square at (x,y), the y intercept (since the slope is 1)
will be at Y == (y-x). since the parity is the same for addition as
for subtraction, and i'm never sure which mod functions (in which
languages) may give a negative result, i use "(x+y) %% 2".
b <- matrix(nrow=8,ncol=8) # basic board
colorindex <- (col(b)+row(b))%%2 # parity of the Y-intercept
# for each square
colors <- c("red", "white")[colorindex+1] # choose colors
side <- 1/8 # side of one square
ux <- col(b)*side # upper x values
lx <- ux-side # lower x values
uy <- row(b)*side # upper y
ly <- uy-side # upper y
plot.new() # initialize R graphics
rect(lx, ly, ux, uy, col=colors, asp=1) # draw the board

spplot() - make color.key look nice

I'm afraid I have a spplot() question again.
I want the colors in my spplot() to represent absolute values, not automatic values as spplot does it by default.
I achieve this by making a factor out of the variable I want to draw (using the command cut()). This works very fine, but the color-key doesn't look good at all.
See it yourself:
library(sp)
data(meuse.grid)
gridded(meuse.grid) = ~x+y
meuse.grid$random <- rnorm(nrow(meuse.grid), 7, 2)
meuse.grid$random[meuse.grid$random < 0] <- 0
meuse.grid$random[meuse.grid$random > 10] <- 10
# making a factor out of meuse.grid$ random to have absolute values plotted
meuse.grid$random <- cut(meuse.grid$random, seq(0, 10, 0.1))
spplot(meuse.grid, c("random"), col.regions = rainbow(100, start = 4/6, end = 1))
How can I have the color.key on the right look good - I'd like to have fewer ticks and fewer labels (maybe just one label on each extreme of the color.key)
Thank you in advance!
[edit]
To make clear what I mean with absolute values: Imagine a map where I want to display the sea height. Seaheight = 0 (which is the min-value) should always be displayed blue. Seaheight = 10 (which, just for the sake of the example, is the max-value) should always be displayed red. Even if there is no sea on the regions displayed on the map, this shouldn't change.
I achieve this with the cut() command in my example. So this part works fine.
THIS IS WHAT MY QUESTION IS ABOUT
What I don't like is the color description on the right side. There are 100 ticks and each tick has a label. I want fewer ticks and fewer labels.
The way to go is using the attribute colorkey. For example:
## labels
labelat = c(1, 2, 3, 4, 5)
labeltext = c("one", "two", "three", "four", "five")
## plot
spplot(meuse.grid,
c("random"),
col.regions = rainbow(100, start = 4/6, end = 1),
colorkey = list(
labels=list(
at = labelat,
labels = labeltext
)
)
)
First, it's not at all clear what you are wanting here. There are many ways to make the color.key look "nice" and that is to understand first the data being passed to spplot and what is being asked of it. cut() is providing fully formatted intervals like (2.3, 5.34] which will need to be handled a different way, increasing the margins in the plot, specific formatting and spacing for the labels, etc. etc. This just may not be what you ultimately want.
Perhaps you just want integer values, rounded from the input values?
library(sp)
data(meuse.grid)
gridded(meuse.grid) = ~x+y
meuse.grid$random <- rnorm(nrow(meuse.grid), 7, 2)
Round the values (or trunc(), ceil(), floor() them . . .)
meuse.grid$rclass <- round(meuse.grid$random)
spplot(meuse.grid, c("rclass"), col.regions = rainbow(100, start = 4/6, end = 1))

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