when using the "curve" function in R, how do you suppress/stop the plot from showing up? For example, this code always plots the curve
my_curve = curve(x)
Is there a parameter to do this or should I being using a different function? I just want the x y points as a dataframe from the curve.
curve() is from the graphics library and is unhandy for generating lists.
Just try using:
x = seq(from, to, length.out = n)
y = function(x)
If you stick to the curve function, the closest to a solution I know is adding dev.off() after the curve() statement!
Here's a way to take advantage of the part of curve that you want without generating a plot.
I made a copy of the curve function (just type curve in the console); called it by a new name (curve2); and commented out the four lines at the end starting with if (isTRUE(add)). When it's called and assigned, I had a list with two vectors—x and y. No plot.
I have 23 different groups,each of them consists of from 7 to 20 individual samples (totally approximately 350-400 observations) with their own x,y & z coordinates. I'd like to produce 3D plot based on the data i have by means of plot3d function of rgl R package. It's not a big deal in general. The problem, that i'd like to make each one from the mentioned above 23 groups to be easy distinguishable on the 3D plot. I tried to use different colors for each group, but unfortunately it's not possible to find a 23 well recognizable by human eyes colors. I was thinking about pch parameter like in the plot function of base R library. But, again, as i can see there is not such option in the plot3d function. Besides, i have to explain, that there are too much points in my data set and adding the labels to each point (e.g. with text3d rgl function) is not a good idea (they will overlap with each other and give in result some kind of a mess on the 3D plot). Is there way to figure out it (i gues it's very common problem)? Thank you in advance!
Below is code of some toy example for better explanation:
# generate data
prefix=rep("ID",69)
suffix=rep(1:23,3)
suffix_2=as.character(suffix[order(suffix)])
titles_1=paste(prefix,suffix,sep="_")
titles_2=titles_1[order(titles_1)]
x=1:69
y=x+20
z=x+50
df=data.frame(titles_2,x,y,z)
# load rgl library
library('rgl')
# make 3D plot
plot3d(x,y,z)
If you like living on the bleeding edge, there's a new function rgl::pch3d() that draws symbols using the same codes as points() does
in base graphics. It's in rgl 0.95.1475, available on R-forge (and within a few hours on Github; see How do I install the latest version of rgl?). It's not completely working with rglwidget() yet.
The example code
open3d()
i <- 0:25; x <- i %% 5; y <- rep(0, 26); z <- i %/% 5
pch3d(x, y, z, pch = i, bg = "green")
text3d(x, y, z + 0.3, i)
pch3d(x + 5, y, z, pch = LETTERS[i+1])
text3d(x + 5, y, z + 0.3, i+65)
produces this display (after some resizing and rotation):
It's not perfect, but how about using letters a-w to distinguish the groups?
with(df,plot3d(x,y,z))
with(df,text3d(x,y,z,texts=letters[titles_2]))
Because i'm going to use the 3D plot for publication purposes i used this solution for now. It's not pretended to be the best one.
# generate data
prefix=rep("ID",69)
suffix=rep(1:23,3)
suffix_2=as.character(suffix[order(suffix)])
titles_1=paste(prefix,suffix,sep="_")
titles_2=titles_1[order(titles_1)]
x=1:69
y=x+20
z=x+50
df=data.frame(titles_2,x,y,z)
# load rgl library
library('rgl')
# load randomcoloR library
library(randomcoloR)
# create a custom palette
palette <- distinctColorPalette(23)
palette(palette)
# make 3D plot
plot3d(x,y,z,size = 10,col=suffix[order(suffix)])
I'm new to R. Previously, I've been able to overlay 2 separate plots that were of the same kind, p1 and p2, using plot (p1); plot (p2, add=T).
I'm struggling with the definition of factors when overlaying a barplot with a point plot showing all individual points.
I can individually plot the barplot as I want it. The point plot looks like I want it, but I realize I'm using an incorrect definition of phase as numerical to force R plot to display each value, rather than default to a boxplot (like when I use plot(my.df$cond, my.df$val).
Any tips on defining my variable types correctly or whether I'm using the correct barplot and plot functions, would be greatly appreciated. Thank you so much.
shpad <- c(1,2,5,6,1,2,5,6,1,2,5,6,1,2,5,6)
my.df <- data.frame(val=c(0.0738,0.0518,0.002,0.0397,0.1452,0.1152,0.1774,0.0658,0.0218,0.0497,-0.0296,0.0653,0.0848,0.1296,0.1416,0.0923,
phase=c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4),
sub=c(1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4),
cond=c("NsNm", "NsNm", "NsNm", "NsNm", "NsLm", "NsLm", "NsLm", "NsLm", "LsNm", "LsNm", "LsNm", "LsNm", "LsLm", "LsLm", "LsLm", "LsLm"))
avg <-tapply(my.df$val, my.df$phase, mean)
barplot(avg, border=NA, names.arg=c("NsNm", "NsLm", "LsNm", "LsLm"),col=c("blue","darkblue","red", "darkred"),ylab = "score",ylim=c(-0.03,0.25))
plot(my.df$phase, my.df$val, type="p", ylim=c(-0.03,0.25), ylab = "score", pch=shpad)
tl;dr: problem is that if instead of the last line, I have plot(my.df$phase, my.df$val, type="p", ylim=c(-0.03,0.25), ylab = "score", pch=shpad, add=T), the formats are incongruent.
Alright, so, I've tried for a bit to accomplish what you wanted, but the best I could do with the base plotting system is this:
Which is accomplished purely by your lines of code above except for the last line, which I replaced with
points(my.df$phase,my.df$val,type="p",pch=shpad)
However, I think you can do much better, if you want to keep the same kind of plot, using the ggplot2 library. Using this code:
library('ggplot2')
new.df <- data.frame(avg,phase=levels(factor(phase)))
ggplot(new.df) +
geom_bar(stat="identity",aes(x=levels(phase),y=avg, fill=c("NsNm","NsLm","LsNm","LsLm")))+
geom_point(aes(x=my.df$phase,y=my.df$val,shape=factor(shpad))) +
scale_x_discrete(name="Type",labels=c("NsNm","NsLm","LsNm","LsLm")) +
ylab("Score")
you can make this chart:
I didn't adjust the coloring and the point types and the legend titles (not sure how important they are, but those can be fiddled with). However, you can see this probably produces the result you were aiming for.
I am very new to R and would like to draw a line graph. I have got as far as importing my data into R and don't really know where to go next! I've searched the internet for examples of how to plot a line graph, but can't find anything that explains why the various commands are being used (which I think that I need to learn what is going on). Can anyone recommend any such tutorials/instructions that are aimed at the beginner?
Probably complicating the matter further, the line graph I'd like to draw doesn't have evenly spaced data points on the x-axis (0.19, 0.31 and 0.36). I'd like to reflect this in the plot, but have no idea how to program this.
Thanks in advance for everyone's help!
There are many ways to plot in R. One is with Base r commands, like this
x <- c(0.19, 0.31, 0.36)
y <- c(1,2,3)
plot(x,y,type = "l")
Look online for plot examples for ggplot and lattice graphs.
I suggest to report some data and and example of the code you are dealing with. It helps the community to examine your problem.
Anyway, if I got your problem correctly, R deals automatically with NAs, that are not reported in the plot. You can have a line graph with type = "l" ("l" stands for "line") in the plot() function.
x <- rnorm(100)
plot(x, type = "l")
So, I am able to use the plot() function in R to graph different functions. However, I am finding that the graphs in R do not typically show the entire curve of the function. Let me use an example:
fun <- function(x){
x^3 + 2*x^2 + 3*x + 4
}
plot(fun)
However, when I plot the same function using the Desmos Graphing Calculator it shows all four quadrants of the Cartesian graph whereas R is only showing one:
My question: How can I modify RPlot to show all four quadrants as opposed to just one as in the above case?
I think you can do this just by extending the default range (which is [0,1]):
plot(fun,from=-5,to=5,ylim=c(-8,8),col="red")
grid()
abline(v=0,h=0,lty=2)
I've added a few frills to make it look a little more like the desired plot. Adding a point on the y axis is easy; adding the x-intercept is not quite so easy.
points(0,fun(0),pch=16)
points(Re(polyroot(c(4,3,2,1))[2]),0,pch=16)