This question is an unexpected follow-up from Draw vertical ending of error bar line in dotplot. While the quoted question was succesfully resolved - there is a caveat. When I introduce more then three conditions to dotplot it doesn't want to draw the vertical ticks |--o--| in the endings of error bars.
As #Josh suggested in the comments, I injected browser() into first line of function that draws updated panel.Dotplot to see what goes wrong, but it didn't come out with anything that helps me to solve it. Here is an example code for four-conditions Dotplot() with updated panel.Dotplot function that doesn't work. It will work, if you decrease number of conditions (check answer for the question quoted above):
require(Hmisc)
#Fake conditions
mean = c(1:18)
lo = mean-0.2
up = mean+0.2
name = c("a","b","c")
cond1 = c("A","B","C")
cond2 = c(rep("E1",9),rep("E2",9))
d = data.frame (name = rep(name,6), mean, lo, up,
cond1=rep(cond1,each=3,times=2), cond2)
# Create the customized panel function
mypanel.Dotplot <- function(x, y, ...) {
panel.Dotplot(x,y,...)
tips <- attr(x, "other")
panel.arrows(x0 = tips[,1], y0 = y,x1 = tips[,2],
y1 = y,length = 0.1, unit = "native",
angle = 90, code = 3)
}
#Draw Dotplot - `panel.Dotplot` doesn't change anything
setTrellis()
Dotplot(name ~ Cbind(mean,lo,up) | cond1 * cond2, data=d, ylab="", xlab="",col=1,
panel = mypanel.Dotplot)
The error bars are in fact being rendered, but are not visible due to their very short length (± 0.2 units). Increasing the error to ± 1 results in the following (I've also increased the length specified in panel.arrows - i.e. the error bar cap length - to 0.5):
If your true data is so precise relative to the range of x-values then you might want to consider smaller points (so they aren't as prone to obscuring the error bars) or a layout that exaggerates the x axis. For example, the following uses your original error of ± 0.2 units, and your original arrow cap length of 0.1:
Dotplot(name ~ Cbind(mean,lo,up) | cond1 * cond2, data=d, ylab="", xlab="",
col=1, panel = mypanel.Dotplot, pch=20, cex=0.4, layout=c(1, 6), strip=FALSE,
strip.left=strip.custom(par.strip.text=list(cex=0.75), bg=0, fg=0))
Related
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.
How to create multiple boxplot with value shown in R ?
Now I'm using this code
boxplot(Data_frame[ ,2] ~ Data_frame[ ,3], )
I tried to use this
boxplot(Data_frame[ ,2] ~ Data_frame[ ,3], )
text(y=fivenum(Data_frame$x), labels =fivenum(Data_frame$x), x=1.25)
But only first boxplot have value. How to show value in all boxplot in one graph.
Thank you so much!
As far as I understand your question (it is not clear how the fivenum summary should be displayed) here is one solution. It presents the summary using the top axis.
x <- data.frame(
Time = c(1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3),
Value = c(5,10,15,20,30,50,70,80,100,5,7,9,11,15,17,19,17,19,100,200,300,400,500,700,1000,200))
boxplot(x$Value ~ x$Time)
fivenums <- aggregate(x$Value, by=list(Time=x$Time), FUN=fivenum)
labels <- apply(fivenums[,-1], 1, function(x) paste(x[-1], collapse = ", "))
axis(3, at=fivenums[,1],labels=labels, las=1, col.axis="red")
Of course you can additionally play with the font size or rotation for this summary. Moreover you can break the line in one place, so the label will have smaller width.
Edit
In order to get what have you posted in the comment below you can add
text(x = 3 + 0.5, y = fivenums[3,-1], labels=fivenums[3,-1])
and you will get
however it won't be readable for other boxplots.
I am quite new to programming/R and I'm having a very unusual problem. I've made a scatterplot and I would like to simply put the x y axis at 0 on the plot. However, when I use abline they are slightly off. I managed to get them to 0 using trial and error, but trying to plot other lines becomes impossible.
library('car')
scatterplot(cost~qaly, reg.line=FALSE, smooth=FALSE, spread=FALSE,
boxplots='xy', span=0.5, xlab="QALY", ylab="COST", main="Bootstrap",
cex=0.5, data=scat2, xlim=c(-.05,.05), grid=FALSE)
abline(v = 0, h = 0)
This gives lines which are slightly to the left and below 0.
here is an image of what this returns:
(I can't post an image since I'm new apparently)
I found that these values put the lines on 0:
abline(v=0.003)
abline(h=3000)
Thanks in advance for the help!
Using #Laterow's example, reproduce the issue
require(car)
set.seed(10)
x <- rnorm(1000); y <- rnorm(1000)
scatterplot(y ~ x)
abline(v=0, h=0)
scatterplot seems to be resetting the par settings on exit. You can sort of check this with locator(1) around some point, eg, for {-3,-3} I get
# $x
# [1] -2.469414
#
# $y
# [1] -2.223922
Option 1
As #joran points out, reset.par = FALSE is the easiest way
scatterplot(y ~ x, reset.par = FALSE)
abline(v=0, h=0)
Option 2
In ?scatterplot, it says that ... is passed to plot meaning you can use plot's very useful panel.first and panel.last arguments (among others).
scatterplot(y ~ x, panel.first = {grid(); abline(v = 0)}, grid = FALSE)
Note that if you were to do the basic
scatterplot(y ~ x, panel.first = abline(v = 0))
you would be unable to see the line because the default scatterplot grid covers it up, so you can turn that off, plot a grid first then do the abline.
You could also do the abline in panel.last, but this would be on top of your points, so maybe not as desirable.
I can not figure out how the lattice levelplot works. I have played with this now for some time, but could not find reasonable solution.
Sample data:
Data <- data.frame(x=seq(0,20,1),y=runif(21,0,1))
Data.mat <- data.matrix(Data)
Plot with levelplot:
rgb.palette <- colorRampPalette(c("darkgreen","yellow", "red"), space = "rgb")
levelplot(Data.mat, main="", xlab="Time", ylab="", col.regions=rgb.palette(100),
cuts=100, at=seq(0,1,0.1), ylim=c(0,2), scales=list(y=list(at=NULL)))
This is the outcome:
Since, I do not understand how this levelplot really works, I can not make it work. What I would like to have is the colour strips to fill the whole window of the corresponding x (Time).
Alternative solution with other method.
Basically, I'm trying here to plot the increasing risk over time, where the red is the highest risk = 1. I would like to visualize the sequence of possible increase or clustering risk over time.
From ?levelplot we're told that if the first argument is a matrix then "'x' provides the
'z' vector described above, while its rows and columns are
interpreted as the 'x' and 'y' vectors respectively.", so
> m = Data.mat[, 2, drop=FALSE]
> dim(m)
[1] 21 1
> levelplot(m)
plots a levelplot with 21 columns and 1 row, where the levels are determined by the values in m. The formula interface might look like
> df <- data.frame(x=1, y=1:21, z=runif(21))
> levelplot(z ~ y + x, df)
(these approaches do not quite result in the same image).
Unfortunately I don't know much about lattice, but I noted your "Alternative solution with other method", so may I suggest another possibility:
library(plotrix)
color2D.matplot(t(Data[ , 2]), show.legend = TRUE, extremes = c("yellow", "red"))
Heaps of things to do to make it prettier. Still, a start. Of course it is important to consider the breaks in your time variable. In this very simple attempt, regular intervals are implicitly assumed, which happens to be the case in your example.
Update
Following the advice in the 'Details' section in ?color2D.matplot: "The user will have to adjust the plot device dimensions to get regular squares or hexagons, especially when the matrix is not square". Well, well, quite ugly solution.
par(mar = c(5.1, 4.1, 0, 2.1))
windows(width = 10, height = 2.5)
color2D.matplot(t(Data[ , 2]),
show.legend = TRUE,
axes = TRUE,
xlab = "",
ylab = "",
extremes = c("yellow", "red"))
I am drawing dotplot() using lattice or Dotplot() using Hmisc. When I use default parameters, I can plot error bars without small vertical endings
--o--
but I would like to get
|--o--|
I know I can get
|--o--|
when I use centipede.plot() from plotrix or segplot() from latticeExtra, but those solutions don't give me such nice conditioning options as Dotplot(). I was trying to play with par.settings of plot.line, which works well for changing error bar line color, width, etc., but so far I've been unsuccessful in adding the vertical endings:
require(Hmisc)
mean = c(1:5)
lo = mean-0.2
up = mean+0.2
d = data.frame (name = c("a","b","c","d","e"), mean, lo, up)
Dotplot(name ~ Cbind(mean,lo,up),data=d,ylab="",xlab="",col=1,cex=1,
par.settings = list(plot.line=list(col=1),
layout.heights=list(bottom.padding=20,top.padding=20)))
Please, don't give me solutions that use ggplot2...
I've had this same need in the past, with barchart() instead of with Dotplot().
My solution then was to create a customized panel function that: (1) first executes the original panel function ; and (2) then uses panel.arrows() to add the error bar (using a two-headed arrow, in which the edges of the head form a 90 degree angle with the shaft).
Here's what that might look like with Dotplot():
# Create the customized panel function
mypanel.Dotplot <- function(x, y, ...) {
panel.Dotplot(x,y,...)
tips <- attr(x, "other")
panel.arrows(x0 = tips[,1], y0 = y,
x1 = tips[,2], y1 = y,
length = 0.15, unit = "native",
angle = 90, code = 3)
}
# Use almost the same call as before, replacing the default panel function
# with your customized function.
Dotplot(name ~ Cbind(mean,lo,up),data=d,ylab="",xlab="",col=1,cex=1,
panel = mypanel.Dotplot,
par.settings = list(plot.line=list(col=1),
layout.heights=list(bottom.padding=20,top.padding=20)))