Related
I am doing quarterly analysis, for which I want to plot a graph. To maintain continuity on x axis I have turned quarters into factors. But then when I am using plot function and trying to color it red, the col argument is not working.
An example:
quarterly_analysis <- data.frame(Quarter = as.factor(c(2020.1,2020.2,2020.3,2020.4,2021.1,2021.2,2021.3,2021.4)),
AvgDefault = as.numeric(c(0.24,0.27,0.17,0.35,0.32,0.42,0.38,0.40)))
plot(quarterly_analysis, col="red")
But I am getting the graph in black color as shown below:
Converting it to a factor is not ideal to plot unless you have multiple values for each factor - it tries to plot a box plot-style plot. For example, with 10 observations in the same factor, the col = "red" color shows up as the fill:
set.seed(123)
fact_example <- data.frame(factvar = as.factor(rep(LETTERS[1:3], 10)),
numvar = runif(30))
plot(fact_example$factvar, fact_example$numvar,
col = "red")
With only one observation for each factor, this is not ideal because it is just showing you the line that the box plot would make.
You could use border = "red:
plot(quarterly_analysis$Quarter,
quarterly_analysis$AvgDefault, border="red")
Or if you want more flexibility, you can plot it numerically and do a little tweaking for more control (i.e., can change the pch, or make it a line graph):
# make numeric x values to plot
x_vals <- as.numeric(substr(quarterly_analysis$Quarter,1,4)) + rep(seq(0, 1, length.out = 4))
par(mfrow=c(1,3))
plot(x_vals,
quarterly_analysis$AvgDefault, col="red",
pch = 7, main = "Square Symbol", axes = FALSE)
axis(1, at = x_vals,
labels = quarterly_analysis$Quarter)
axis(2)
plot(x_vals,
quarterly_analysis$AvgDefault, col="red",
type = "l", main = "Line graph", axes = FALSE)
axis(1, at = x_vals,
labels = quarterly_analysis$Quarter)
axis(2)
plot(x_vals,
quarterly_analysis$AvgDefault, col="red",
type = "b", pch = 7, main = "Both", axes = FALSE)
axis(1, at = x_vals,
labels = quarterly_analysis$Quarter)
axis(2)
Data
set.seed(123)
quarterly_analysis <- data.frame(Quarter = as.factor(paste0(2019:2022,
rep(c(".1", ".2", ".3", ".4"),
each = 4))),
AvgDefault = runif(16))
quarterly_analysis <- quarterly_analysis[order(quarterly_analysis$Quarter),]
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 trying to plot in base R with the regular plot() fcn. However, when passing a vector of which pch to use, it will not plot the pch, it will only plot the number '1' instead of the shape of the pch I am calling.
Generating some data (my real data has over 400 rows for both the loads and meta objects:
loads <- data.frame(PC1 = c(11.32, 13.18, 12.82, 24.70), PC2 = c(-23.05, -24.71, -20.28, 10.09))
row.names(loads) <- c("100_A", "100_B", "100_C", "100_Orig")
meta <- data.frame(pch = c(17, 17, 17, 16), color = c("red", "red", "blue", "blue"))
row.names(meta) <- row.names(loads)
To plot:
x <- loads[, 1] ; y <- loads[, 2]
pch <- meta$pch
col <- meta$color
plot(x, y,
col = col, pch = pch, cex = 2, lwd = 4,
xlab = paste("PC1"), ylab = paste("PC2"))
Now, this will graph the correct color (red and blue) in the order I have them in the vector; the real issue becomes the plotting the pch. Instead of a circle (pch = 16) or a triangle (pch = 17) it's plotting a red or blue number 1 instead! I have included a pic of what my data is actually doing.
Thinking that the pch vector I am passing cannot have quotes around it, I have removed the quotes with the following code:
pch <- meta$pch
pch <-as.vector(noquote(pch))
class(pch)
[1] "character"
However, this generates the same results (getting a number 1 plotted). Interestingly, when use this code, it works fine. It turns all my colors to blue, and I get nice blue circles.
plot(x, y,
col = "blue, pch = 16, cex = 2, lwd = 4,
xlab = paste("PC1"), ylab = paste("PC2"))
This tells me that the plot function isn't recognizing my long vector composed of pch 16 and 17's mixed in.
Alternatively, if I use the rep function to generate my pch vector, a test shows it works fine. But I have over 400 rows. I cannot manually type rep for each pch. I will be here for eternity typing that out.
Any suggestions on what to do?????
Try defining the col as character and the pch as numeric like this:
plot(x, y,
col = as.character(col), pch = as.numeric(pch), cex = 2, lwd = 4,
xlab = paste("PC1"), ylab = paste("PC2"))
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 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.