I am trying to visualize some data and in order to do it I am using R's hist.
Bellow are my data
jancoefabs <- as.numeric(as.vector(abs(Janmodelnorm$coef)))
jancoefabs
[1] 1.165610e+00 1.277929e-01 4.349831e-01 3.602961e-01 7.189458e+00
[6] 1.856908e-04 1.352052e-05 4.811291e-05 1.055744e-02 2.756525e-04
[11] 2.202706e-01 4.199914e-02 4.684091e-02 8.634340e-01 2.479175e-02
[16] 2.409628e-01 5.459076e-03 9.892580e-03 5.378456e-02
Now as the more cunning of you might have guessed these are the absolute values of some model's coefficients.
What I need is an histogram that will have for axes:
x will be the number (count or length) of coefficients which is 19 in total, along with their names.
y will show values of each column (as breaks?) having a ylim="" set, according to min and max of those values (or something similar).
Note that Janmodelnorm$coef simply produces the following
(Intercept) LON LAT ME RAT
1.165610e+00 -1.277929e-01 -4.349831e-01 -3.602961e-01 -7.189458e+00
DS DSA DSI DRNS DREW
-1.856908e-04 1.352052e-05 4.811291e-05 -1.055744e-02 -2.756525e-04
ASPNS ASPEW SI CUR W_180_270
-2.202706e-01 -4.199914e-02 4.684091e-02 -8.634340e-01 -2.479175e-02
W_0_360 W_90_180 W_0_180 NDVI
2.409628e-01 5.459076e-03 -9.892580e-03 -5.378456e-02
So far and consulting ?hist, I am trying to play with the code bellow without success. Therefore I am taking it from scratch.
# hist(jancoefabs, col="lightblue", border="pink",
# breaks=8,
# xlim=c(0,10), ylim=c(20,-20), plot=TRUE)
When plot=FALSE is set, I get a bunch of somewhat useful info about the set. I also find hard to use breaks argument efficiently.
Any suggestion will be appreciated. Thanks.
Rather than using hist, why not use a barplot or a standard plot. For example,
## Generate some data
set.seed(1)
y = rnorm(19, sd=5)
names(y) = c("Inter", LETTERS[1:18])
Then plot the cofficients
barplot(y)
Alternatively, you could use a scatter plot
plot(1:19, y, axes=FALSE, ylim=c(-10, 10))
axis(2)
axis(1, 1:19, names(y))
and add error bars to indicate the standard errors (see for example Add error bars to show standard deviation on a plot in R)
Are you sure you want a histogram for this? A lattice barchart might be pretty nice. An example with the mtcars built-in data set.
> coef <- lm(mpg ~ ., data = mtcars)$coef
> library(lattice)
> barchart(coef, col = 'lightblue', horizontal = FALSE,
ylim = range(coef), xlab = '',
scales = list(y = list(labels = coef),
x = list(labels = names(coef))))
A base R dotchart might be good too,
> dotchart(coef, pch = 19, xlab = 'value')
> text(coef, seq(coef), labels = round(coef, 3), pos = 2)
Related
I am working on a script where I need to calculate the coordinates for a beeswarm plot without immediately plotting. When I use beeswarm, I get x-coordinates that aren't swarmed, and more or less the same value:
But if I generate the same plot again it swarms correctly:
And if I use dev.off() I again get no swarming:
The code I used:
n <- 250
df = data.frame(x = floor(runif(n, 0, 5)),
y = rnorm(n = n, mean = 500, sd = 100))
#Plot 1:
A = with(df, beeswarm(y ~ x, do.plot = F))
plot(x = A$x, y=A$y)
#Plot 2:
A = with(df, beeswarm(y ~ x, do.plot = F))
plot(x = A$x, y=A$y)
dev.off()
#Plot 3:
A = with(df, beeswarm(y ~ x, do.plot = F))
plot(x = A$x, y=A$y)
It seems to me like beeswarm uses something like the current plot parameters (or however it is called) to do the swarming and therefore chokes when a plot isn't showing. I have tried to play around with beeswarm parameters such as spacing, breaks, corral, corralWidth, priority, and xlim, but it does not make a difference. FYI: If do.plot is set to TRUE the x-coordinates are calculated correctly, but this is not helpful as I don't want to plot immediately.
Any tips or comments are greatly appreciated!
You're right; beeswarm uses the current plot parameters to calculate the amount of space to leave between points. It seems that setting "do.plot=FALSE" does not do what one would expect, and I'm not sure why I included this parameter.
If you want to control the parameters manually, you could use the functions swarmx or swarmy instead. These functions must be applied to each group separately, e.g.
dfsplitswarmed <- by(df, df$x, function(aa) swarmx(aa$x, aa$y, xsize = 0.075, ysize = 7.5, cex = 1, log = ""))
dfswarmed <- do.call(rbind, dfsplitswarmed)
plot(dfswarmed)
In this case, I set the xsize and ysize values based on what the function would default to for this particular data set. If you can find a set of xsize/ysize values that work for your data, this approach might work for you.
Otherwise, perhaps a simpler approach would be to leave do.plot=TRUE, and then discard the plots.
I am running quantile regressions for several independent variables separately (same dependent). I want to plot only the slope estimates over several quantiles of each variable in a single plot.
Here's a toy data:
set.seed(1988)
y <- rnorm(50, 5, 3)
x1 <- rnorm(50, 3, 1)
x2 <- rnorm(50, 1, 0.5)
# Running Quantile Regression
require(quantreg)
fit1 <- summary(rq(y~x1, tau=1:9/10), se="boot")
fit2 <- summary(rq(y~x2, tau=1:9/10), se="boot")
I want to plot only the slope estimates over quantiles. Hence, I am giving parm=2 in plot.
plot(fit1, parm=2)
plot(fit2, parm=2)
Now, I want to combine both these plots in a single page.
What I have tried so far;
I tried setting par(mfrow=c(2,2)) and plotting them. But it's producing a blank page.
I have tried using gridExtra and gridGraphics without success. Tried to convert base graphs into Grob objects as stated here
Tried using function layout function as in this document
I am trying to look into the source code of plot.rqs. But I am unable to understand how it's plotting confidence bands (I'm able to plot only the coefficients over quantiles) or to change mfrow parameter there.
Can anybody point out where am I going wrong? Should I look into the source code of plot.rqs and change any parameters there?
While quantreg::plot.summary.rqs has an mfrow parameter, it uses it to override par('mfrow') so as to facet over parm values, which is not what you want to do.
One alternative is to parse the objects and plot manually. You can pull the tau values and coefficient matrix out of fit1 and fit2, which are just lists of values for each tau, so in tidyverse grammar,
library(tidyverse)
c(fit1, fit2) %>% # concatenate lists, flattening to one level
# iterate over list and rbind to data.frame
map_dfr(~cbind(tau = .x[['tau']], # from each list element, cbind the tau...
coef(.x) %>% # ...and the coefficient matrix,
data.frame(check.names = TRUE) %>% # cleaned a little
rownames_to_column('term'))) %>%
filter(term != '(Intercept)') %>% # drop intercept rows
# initialize plot and map variables to aesthetics (positions)
ggplot(aes(x = tau, y = Value,
ymin = Value - Std..Error,
ymax = Value + Std..Error)) +
geom_ribbon(alpha = 0.5) +
geom_line(color = 'blue') +
facet_wrap(~term, nrow = 2) # make a plot for each value of `term`
Pull more out of the objects if you like, add the horizontal lines of the original, and otherwise go wild.
Another option is to use magick to capture the original images (or save them with any device and reread them) and manually combine them:
library(magick)
plots <- image_graph(height = 300) # graphics device to capture plots in image stack
plot(fit1, parm = 2)
plot(fit2, parm = 2)
dev.off()
im1 <- image_append(plots, stack = TRUE) # attach images in stack top to bottom
image_write(im1, 'rq.png')
The function plot used by quantreg package has it's own mfrow parameter. If you do not specify it, it enforces some option which it chooses on it's own (and thus overrides your par(mfrow = c(2,2)).
Using the mfrow parameter within plot.rqs:
# make one plot, change the layout
plot(fit1, parm = 2, mfrow = c(2,1))
# add a new plot
par(new = TRUE)
# create a second plot
plot(fit2, parm = 2, mfrow = c(2,1))
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"))
Refer to the above plot. I have drawn the equations in excel and then shaded by hand. You can see it is not very neat. You can see there are six zones, each bounded by two or more equations. What is the easiest way to draw inequalities and shade the regions using hatched patterns ?
To build up on #agstudy's answer, here's a quick-and-dirty way to represent inequalities in R:
plot(NA,xlim=c(0,1),ylim=c(0,1), xaxs="i",yaxs="i") # Empty plot
a <- curve(x^2, add = TRUE) # First curve
b <- curve(2*x^2-0.2, add = TRUE) # Second curve
names(a) <- c('xA','yA')
names(b) <- c('xB','yB')
with(as.list(c(b,a)),{
id <- yB<=yA
# b<a area
polygon(x = c(xB[id], rev(xA[id])),
y = c(yB[id], rev(yA[id])),
density=10, angle=0, border=NULL)
# a>b area
polygon(x = c(xB[!id], rev(xA[!id])),
y = c(yB[!id], rev(yA[!id])),
density=10, angle=90, border=NULL)
})
If the area in question is surrounded by more than 2 equations, just add more conditions:
plot(NA,xlim=c(0,1),ylim=c(0,1), xaxs="i",yaxs="i") # Empty plot
a <- curve(x^2, add = TRUE) # First curve
b <- curve(2*x^2-0.2, add = TRUE) # Second curve
d <- curve(0.5*x^2+0.2, add = TRUE) # Third curve
names(a) <- c('xA','yA')
names(b) <- c('xB','yB')
names(d) <- c('xD','yD')
with(as.list(c(a,b,d)),{
# Basically you have three conditions:
# curve a is below curve b, curve b is below curve d and curve d is above curve a
# assign to each curve coordinates the two conditions that concerns it.
idA <- yA<=yD & yA<=yB
idB <- yB>=yA & yB<=yD
idD <- yD<=yB & yD>=yA
polygon(x = c(xB[idB], xD[idD], rev(xA[idA])),
y = c(yB[idB], yD[idD], rev(yA[idA])),
density=10, angle=0, border=NULL)
})
In R, there is only limited support for fill patterns and they can only be
applied to rectangles and polygons.This is and only within the traditional graphics, no ggplot2 or lattice.
It is possible to fill a rectangle or polygon with a set of lines drawn
at a certain angle, with a specific separation between the lines. A density
argument controls the separation between the lines (in terms of lines per inch)
and an angle argument controls the angle of the lines.
here an example from the help:
plot(c(1, 9), 1:2, type = "n")
polygon(1:9, c(2,1,2,1,NA,2,1,2,1),
density = c(10, 20), angle = c(-45, 45))
EDIT
Another option is to use alpha blending to differentiate between regions. Here using #plannapus example and gridBase package to superpose polygons, you can do something like this :
library(gridBase)
vps <- baseViewports()
pushViewport(vps$figure,vps$plot)
with(as.list(c(a,b,d)),{
grid.polygon(x = xA, y = yA,gp =gpar(fill='red',lty=1,alpha=0.2))
grid.polygon(x = xB, y = yB,gp =gpar(fill='green',lty=2,alpha=0.2))
grid.polygon(x = xD, y = yD,gp =gpar(fill='blue',lty=3,alpha=0.2))
}
)
upViewport(2)
There are several submissions on the MATLAB Central File Exchange that will produce hatched plots in various ways for you.
I think a tool that will come handy for you here is gnuplot.
Take a look at the following demos:
feelbetween
statistics
some tricks
I'd like to superpose a histogram and an xyplot representing the cumulative distribution function using r's lattice package.
I've tried to accomplish this with custom panel functions, but can't seem to get it right--I'm getting hung up on one plot being univariate and one being bivariate I think.
Here's an example with the two plots I want stacked vertically:
set.seed(1)
x <- rnorm(100, 0, 1)
discrete.cdf <- function(x, decreasing=FALSE){
x <- x[order(x,decreasing=FALSE)]
result <- data.frame(rank=1:length(x),x=x)
result$cdf <- result$rank/nrow(result)
return(result)
}
my.df <- discrete.cdf(x)
chart.hist <- histogram(~x, data=my.df, xlab="")
chart.cdf <- xyplot(100*cdf~x, data=my.df, type="s",
ylab="Cumulative Percent of Total")
graphics.off()
trellis.device(width = 6, height = 8)
print(chart.hist, split = c(1,1,1,2), more = TRUE)
print(chart.cdf, split = c(1,2,1,2))
I'd like these superposed in the same frame, rather than stacked.
The following code doesn't work, nor do any of the simple variations of it that I have tried:
xyplot(cdf~x,data=cdf,
panel=function(...){
panel.xyplot(...)
panel.histogram(~x)
})
You were on the right track with your custom panel function. The trick is passing the correct arguments to the panel.- functions. For panel.histogram, this means not passing a formula and supplying an appropriate value to the breaks argument:
EDIT Proper percent values on y-axis and type of plots
xyplot(100*cdf~x,data=my.df,
panel=function(...){
panel.histogram(..., breaks = do.breaks(range(x), nint = 8),
type = "percent")
panel.xyplot(..., type = "s")
})
This answer is just a placeholder until a better answer comes.
The hist() function from the graphics package has an option called add. The following does what you want in the "classical" way:
plot( my.df$x, my.df$cdf * 100, type= "l" )
hist( my.df$x, add= T )