I have plotted five graphs and a legend. The graphs work just fine, however the legens disappears without an error.
My preview in RStudio looks like this
When I zoom in, the area where the legend should be is blank.
I use the following code:
opar <- par (no.readonly = TRUE)
par (mfrow = c(3, 2))
library(deSolve)
# Plot A
LotVmod <- function (Time, State, Pars) {
with(as.list(c(State, Pars)), {
dx = (b*x) - (b*x*x/K) - (y*(x^k/(x^k+C^k)*(l*x/(1+l*h*x))))
dy = (y*e*(x^k/(x^k+C^k)*(l*x/(1+l*h*x)))) - (m*y)
return(list(c(dx, dy)))
})
}
Pars <- c(b = 1.080, e = 2.200, K = 130.000, k = 20.000, l = 2.000,
h = 0.030, C = 2.900, m = 0.050)
State <- c(x = 0.25, y = 2.75)
Time <- seq(1, 9, by = 1)
out <- as.data.frame(ode(func = LotVmod, y = State, parms = Pars, times = Time))
matplot(out[,-1], type = "l", xlim = c(1, 9), ylim = c(0, 45),
xlab = "time",
ylab = "population",
main = "Compartment A")
mtext ( "Coefficient of Variance 4.96", cex = 0.8 )
x <- c(# Validation data)
y <- c(# Validation data)
lines (Time, x, type="l", lty=1, lwd=2.5, col="black")
lines (Time, y, type="l", lty=1, lwd=2.5, col="red")
# Legend
plot.new()
legend("center", c(expression (italic ("F. occidentalis")*" observed"),
expression (italic ("M. pygmaeus")*" observed"),
expression (italic ("F. occidentalis")*" simulated"),
expression (italic ("M. pygmaeus")*" simulated")),
lty = c(1, 1, 1, 2),
col = c(1, 2, 1, 2),
lwd = c(2.5, 2.5, 1, 1),
box.lwd = 0, bty = "n")
# Plot C to F = same as A
par(opar)
My output doesn't give an error. I have used the exact same code before without any trouble, thus I restarted R, removed all objects, cleared all plots and restarted both RStudio and my computer.
Try to add xpd=TRUE in your legend statement. I.e.
legend("center", c(expression (italic ("F. occidentalis")*" observed"),
expression (italic ("M. pygmaeus")*" observed"),
expression (italic ("F. occidentalis")*" simulated"),
expression (italic ("M. pygmaeus")*" simulated")),
lty = c(1, 1, 1, 2),
col = c(1, 2, 1, 2),
lwd = c(2.5, 2.5, 1, 1),
box.lwd = 0, bty = "n", xpd=TRUE)
By default, the legend is cut off by the plotting region. This xpd parameter enables plotting outside the plot region. See e.g. ?par for more on xpd.
This is due to how the plot canvas is set up and how rescaling that device works. The way you do it, you add the legend in the plotting region of the top right plot. The plotting region is however not the complete device, but only the part inside the space formed by the axes. If you rescale, that plotting region will be rescaled as well. The margins around the plotting region don't change size though, so zooming in makes your plotting region so small that it doesn't fit the legend any longer. It is hidden by the margins around the plotting region.
For that reason AEBilgrau is very right you need to add xpd = TRUE. This allows the legend to extend outside of the plotting region, so it doesn't disappear behind the margins when resizing the plotting device.
Related
I am plotting correlation coefficients (values = 0.0:1.0) for two isotopes measured in each individual from two populations. I would like to have a fixed aspect-ratio for my scatter-plot so that the x- and y-axis are exactly the same size no matter the graphics device. Suggestions?
This is my first plot in R, any comments on refinements to my code is appreciated? Finally, is it worth investing in learning the basic plotting techniques or should I jump right to ggplot2 or lattice?
My plot script:
## Create dataset
WW_corr <-
structure(list(South_N15 = c(0.7976495, 0.1796725, 0.5338347,
0.4103769, 0.7447027, 0.5080296, 0.7566544, 0.7432026, 0.8927161
), South_C13 = c(0.76706752, 0.02320767, 0.88429902, 0.36648357,
0.73840937, 0.0523504, 0.52145159, 0.50707858, 0.51874445), North_N15 = c(0.7483608,
0.4294148, 0.9283554, 0.8831571, 0.5056481, 0.1945943, 0.8492716,
0.5759033, 0.7483608), North_C13 = c(0.08114805, 0.47268136,
0.94975596, 0.06023815, 0.33652839, 0.53055943, 0.30228833, 0.8864435,
0.08114805)), .Names = c("South_N15", "South_C13", "North_N15",
"North_C13"), row.names = c(NA, -9L), class = "data.frame")
opar <- par()
## Plot results
par(oma = c(1, 0, 0, 0), mar = c(4, 5, 2, 2))
plot(1,1,xlim=c(0:1.0), ylim=c(0:1.0), type="n", las=1, bty="n", main = NULL,
ylab=expression(paste("Correlation Coefficient (r) for ", delta ^{15},"N ",
"\u0028","\u2030","\u0029")),
xlab=expression(paste("Correlation Coefficient (r) for ", delta ^{13},"C ",
"\u0028","\u2030","\u0029")))
points(WW_corr$South_N15, WW_corr$South_C13, pch = 23, cex = 1.25,
bg ="antiquewhite4", col = "antiquewhite4")
points(WW_corr$North_N15, WW_corr$North_C13, pch = 15, cex = 1.25,
bg ="black")
axis(1, at = seq(0, 1.0, by = 0.1), labels = F, tick = TRUE, tck = -0.01)
axis(2, at = seq(0, 1.0, by = 0.1), labels = F, tick = TRUE, tck = -0.01)
abline(h=.86, v=.86, col = "gray60", lty = 2)
legend("topleft", c("North", "South"), pch = c(15, 23),
col = c("black", "antiquewhite4"), pt.bg = c("black", "antiquewhite4"),
horiz=TRUE, bty = "n")
par(opar)
par(pty="s")
plot(...)
sets the plot type to be square, which will do the job (I think) in your case because your x and y ranges are the same. Fairly well hidden option documented in ?par.
Using asp=1 as a parameter to plot will get interpreted by the low-level plot.window call and should give you a unitary aspect ratio. There is the potential that a call using ylim and xlim could conflict with an aspect ratio scpecification and the asp should "prevail". That's a very impressive first R graph, by the away. And an excellent question construction. High marks.
The one jarring note was your use of the construction xlim=c(0:1.0). Since xlim expects a two element vector, I would have expected xlim=c(0,1). Fewer keystrokes and less subject to error in the future if you changed to a different set of limits, since the ":" operator would give you unexpected results if you tried that with "0:2.5".
I was wondering if it is possible to seperate two plots from eachother (both should be on the same plot, using double Y axis). So the double plot should be split into two but without actually plotting them seperate - par(mfrow(1,2)).
I was trying to imitate it with layout plot, or with latticeExtra, ggplot but no success.
I have two different dataset one for the exchange rate one for the logaritmic returns.
par(mar=c(4,4,3,4))
plot(rates$EURHUF~rates$Date, type="l", ylab="Rate", main="EUR/HUF", xlab="Time")
par(new=TRUE)
plot(reteslog$EURHUF~rateslog$Date, type="l", xaxt="n", yaxt="n", ylab="", xlab="", col="red")
axis(side=4)
mtext("Log return", side=4, line=3)
legend("topleft", c("EUR/HUF Rates","EUR/HUF Logreturns"), col=c("black", "red"), lty=c(1,1))
So far I am here, I just don't know how to seperate them or scale them (maybe using margin, or layout?)
Thank you very much guys for helping
I have a solution to this that isn't too outlandish, and is entirely in base, which is nice. For it to work, you just need to be able to force all of your data onto the same scale, which usually isn't a hassle.
The idea is that once your data is on the same scale, you can plot it all normally, and then add in custom axes that show the respective scales of the different data.
set.seed(1986)
d01 <- sample(x = 1:20,
size = 200,
replace = TRUE)
d02 <- sample(x = 31:45,
size = 200,
replace = TRUE)
# pdf(file = "<some/path/to/image.pdf>",
# width = 4L,
# height = 4L) # plot to a pdf
jpeg(file = "<some/path/to/image.jpeg>") # plot to a jpeg
par(mar=c(3.5, 3.5, 2, 3.5)) # parameters to make things prettier
par(mgp=c(2.2, 1, 0)) # parameters to make things prettier
plot(x = 0,
y = 0,
type = "n",
xlim = c(1, 200),
ylim = c(1, 50),
xlab = "Label 01!",
ylab = "Label 02!",
axes = FALSE,
frame.plot = TRUE)
points(d01,
pch = 1,
col = "blue") # data 01
points(d02,
pch = 2,
col = "red") # data 02
mtext("Label 03!",
side = 4,
line = 2) # your extra y axis label
xticks <- seq(from = 0,
to = 200,
by = 50) # tick mark labels
xtickpositions <- seq(from = 0,
to = 200,
by = 50) # tick mark positions on the x axis
axis(side = 1,
at = xtickpositions,
labels = xticks,
col.axis="black",
las = 2,
lwd = 0,
lwd.ticks = 1,
tck = -0.025) # add your tick marks
y01ticks <- seq(from = 0,
to = 1,
by = 0.1) # tick mark labels
y01tickpositions <- seq(from = 0,
to = 50,
by = 5) # tick mark positions on the y01 axis
axis(side = 2,
at = y01tickpositions,
labels = y01ticks,
las = 2,
lwd = 0,
lwd.ticks = 1,
tck = -0.025) # add your tick marks
y02ticks <- seq(from = 0,
to = 50,
by = 5L) # tick mark labels
y02tickpositions <- seq(from = 0,
to = 50,
by = 5) # tick mark positions on the y02 axis
axis(side = 4,
at = y02tickpositions,
labels = y02ticks,
las = 2,
lwd = 0,
lwd.ticks = 1,
tck = -0.025) # add your tick marks
dev.off() # close plotting device
A few notes:
Sizing for this plot was originally set for a pdf, which unfortunately cannot be uploaded here, however that device call is included as commented out code above. You can always play with parameters to find out what works best for you.
It can be advantageous to plot all of your axis labels with mtext().
Including simple example data in your original post is often much more helpful than the exact data you're working with. As of me writing this, I don't really know what your data looks like because I don't have access to those objects.
I've inherited this R code that plots a simple line graph. However, it does it so that the y axis values are plotted downwards below 0 (plots it in the 4th quadrant with 0 at the top and +3600 at the bottom). I want to plot the data right-side up (1st quadrant) so the y axis data goes from 0 up to +3600 at the top like a typical grade-school plot.
I've tried ylim = rev(y) but it returns an error...
I've also tried flipping the seq() command but no luck there.
list.vlevel = numeric(9) # placeholder
plot(
rep(0, length(list.vlevel)),
seq(1, length(list.vlevel)),
type = "n",
xlim = biaslim,
axes = F,
main = paste(list.var.bias[vv], list.score.bias[vv]),
xlab = "",
ylab = ""
)
abline(h = seq(1, length(list.vlevel)),
lty = 3,
col = 8)
axis(2,
labels = list.vlevel,
at = seq(length(list.vlevel), 1, -1),
las = 1)
axis(1)
box()
legend(
x = min(biasarray.var.runhour),
y = length(list.vlevel),
legend = expname,
lty = 3,
lwd = 3,
col = expcol
)
for (exp in seq(length(expname), 1, -1)) {
lines(
biasarray.var.runhour[exp, ],
seq(length(list.vlevel), 1, -1),
col = expcol[exp],
lwd = 3,
lty = 3
)
}
abline(v = 0, lty = 3)
The plot should end up in the first quadrant with yaxis values increasing from 0 upwards to +###.
The axis(2, ...) line draws the y axis. You can see that is the labels follow a descending sequence: seq(length(list.vlevel), 1, -1). seq(1, length(list.vlevel))
Similarly, inside lines(), probably you need to make the same change from seq(length(list.vlevel), 1, -1) to ``seq(1, length(list.vlevel))`
That's as much as we can tell with the info you've provided - can't run any of yoru code without values for all the constants you use, e.g., biasarray.var.runhour, list.var.bias, vv, etc.
I am plotting correlation coefficients (values = 0.0:1.0) for two isotopes measured in each individual from two populations. I would like to have a fixed aspect-ratio for my scatter-plot so that the x- and y-axis are exactly the same size no matter the graphics device. Suggestions?
This is my first plot in R, any comments on refinements to my code is appreciated? Finally, is it worth investing in learning the basic plotting techniques or should I jump right to ggplot2 or lattice?
My plot script:
## Create dataset
WW_corr <-
structure(list(South_N15 = c(0.7976495, 0.1796725, 0.5338347,
0.4103769, 0.7447027, 0.5080296, 0.7566544, 0.7432026, 0.8927161
), South_C13 = c(0.76706752, 0.02320767, 0.88429902, 0.36648357,
0.73840937, 0.0523504, 0.52145159, 0.50707858, 0.51874445), North_N15 = c(0.7483608,
0.4294148, 0.9283554, 0.8831571, 0.5056481, 0.1945943, 0.8492716,
0.5759033, 0.7483608), North_C13 = c(0.08114805, 0.47268136,
0.94975596, 0.06023815, 0.33652839, 0.53055943, 0.30228833, 0.8864435,
0.08114805)), .Names = c("South_N15", "South_C13", "North_N15",
"North_C13"), row.names = c(NA, -9L), class = "data.frame")
opar <- par()
## Plot results
par(oma = c(1, 0, 0, 0), mar = c(4, 5, 2, 2))
plot(1,1,xlim=c(0:1.0), ylim=c(0:1.0), type="n", las=1, bty="n", main = NULL,
ylab=expression(paste("Correlation Coefficient (r) for ", delta ^{15},"N ",
"\u0028","\u2030","\u0029")),
xlab=expression(paste("Correlation Coefficient (r) for ", delta ^{13},"C ",
"\u0028","\u2030","\u0029")))
points(WW_corr$South_N15, WW_corr$South_C13, pch = 23, cex = 1.25,
bg ="antiquewhite4", col = "antiquewhite4")
points(WW_corr$North_N15, WW_corr$North_C13, pch = 15, cex = 1.25,
bg ="black")
axis(1, at = seq(0, 1.0, by = 0.1), labels = F, tick = TRUE, tck = -0.01)
axis(2, at = seq(0, 1.0, by = 0.1), labels = F, tick = TRUE, tck = -0.01)
abline(h=.86, v=.86, col = "gray60", lty = 2)
legend("topleft", c("North", "South"), pch = c(15, 23),
col = c("black", "antiquewhite4"), pt.bg = c("black", "antiquewhite4"),
horiz=TRUE, bty = "n")
par(opar)
par(pty="s")
plot(...)
sets the plot type to be square, which will do the job (I think) in your case because your x and y ranges are the same. Fairly well hidden option documented in ?par.
Using asp=1 as a parameter to plot will get interpreted by the low-level plot.window call and should give you a unitary aspect ratio. There is the potential that a call using ylim and xlim could conflict with an aspect ratio scpecification and the asp should "prevail". That's a very impressive first R graph, by the away. And an excellent question construction. High marks.
The one jarring note was your use of the construction xlim=c(0:1.0). Since xlim expects a two element vector, I would have expected xlim=c(0,1). Fewer keystrokes and less subject to error in the future if you changed to a different set of limits, since the ":" operator would give you unexpected results if you tried that with "0:2.5".
I am plotting correlation coefficients (values = 0.0:1.0) for two isotopes measured in each individual from two populations. I would like to have a fixed aspect-ratio for my scatter-plot so that the x- and y-axis are exactly the same size no matter the graphics device. Suggestions?
This is my first plot in R, any comments on refinements to my code is appreciated? Finally, is it worth investing in learning the basic plotting techniques or should I jump right to ggplot2 or lattice?
My plot script:
## Create dataset
WW_corr <-
structure(list(South_N15 = c(0.7976495, 0.1796725, 0.5338347,
0.4103769, 0.7447027, 0.5080296, 0.7566544, 0.7432026, 0.8927161
), South_C13 = c(0.76706752, 0.02320767, 0.88429902, 0.36648357,
0.73840937, 0.0523504, 0.52145159, 0.50707858, 0.51874445), North_N15 = c(0.7483608,
0.4294148, 0.9283554, 0.8831571, 0.5056481, 0.1945943, 0.8492716,
0.5759033, 0.7483608), North_C13 = c(0.08114805, 0.47268136,
0.94975596, 0.06023815, 0.33652839, 0.53055943, 0.30228833, 0.8864435,
0.08114805)), .Names = c("South_N15", "South_C13", "North_N15",
"North_C13"), row.names = c(NA, -9L), class = "data.frame")
opar <- par()
## Plot results
par(oma = c(1, 0, 0, 0), mar = c(4, 5, 2, 2))
plot(1,1,xlim=c(0:1.0), ylim=c(0:1.0), type="n", las=1, bty="n", main = NULL,
ylab=expression(paste("Correlation Coefficient (r) for ", delta ^{15},"N ",
"\u0028","\u2030","\u0029")),
xlab=expression(paste("Correlation Coefficient (r) for ", delta ^{13},"C ",
"\u0028","\u2030","\u0029")))
points(WW_corr$South_N15, WW_corr$South_C13, pch = 23, cex = 1.25,
bg ="antiquewhite4", col = "antiquewhite4")
points(WW_corr$North_N15, WW_corr$North_C13, pch = 15, cex = 1.25,
bg ="black")
axis(1, at = seq(0, 1.0, by = 0.1), labels = F, tick = TRUE, tck = -0.01)
axis(2, at = seq(0, 1.0, by = 0.1), labels = F, tick = TRUE, tck = -0.01)
abline(h=.86, v=.86, col = "gray60", lty = 2)
legend("topleft", c("North", "South"), pch = c(15, 23),
col = c("black", "antiquewhite4"), pt.bg = c("black", "antiquewhite4"),
horiz=TRUE, bty = "n")
par(opar)
par(pty="s")
plot(...)
sets the plot type to be square, which will do the job (I think) in your case because your x and y ranges are the same. Fairly well hidden option documented in ?par.
Using asp=1 as a parameter to plot will get interpreted by the low-level plot.window call and should give you a unitary aspect ratio. There is the potential that a call using ylim and xlim could conflict with an aspect ratio scpecification and the asp should "prevail". That's a very impressive first R graph, by the away. And an excellent question construction. High marks.
The one jarring note was your use of the construction xlim=c(0:1.0). Since xlim expects a two element vector, I would have expected xlim=c(0,1). Fewer keystrokes and less subject to error in the future if you changed to a different set of limits, since the ":" operator would give you unexpected results if you tried that with "0:2.5".