I have a simple effect plot in R using the effects package. Here is an example:
require(effects)
A <- rnorm(100)
B <- rnorm(100)
C <- factor(rep(c("right", "left"), 50))
eff <- effect("B:C", lm(A~B*C))
plot(eff)
I want to change the strip text of the panels (in this case C:left and C:right) to any text (eg. text:left). There is an unelegant way to achieve this by changing the data using:
names(eff$x)[names(eff$x)=="C"] <- "text"
I don't want to change the data though. Also changing the element name restricts me from using spaces or other special symbols. Using the function to change the strip names in plot.eff
plot(..., strip = function(...) strip.default(..., strip.names = c(factor.names, TRUE)))
gives me an error though ("formal argument "strip" matched by multiple actual arguments") since this seems to be used by the xyplot too.
Is there a way to change the strip text without modifying the data? Any help would be highly appreciated!
Related
My problem is multifaceted.
I would like to plot multiple columns saved in a data frame. Those columns do not have an x variable but would essentially be 1 to 101 consistent for all. I have seen that I can transfer them into long format but most ggplot options require an X. I tried zoo which does what I want it to, but the x-label is all jumbled and I am not aware of how to fix it. (Example of data below, and plot)
df <- zoo(HIP_131_Y0_LC_walk1[1:9])
plot(df)
I have multiple data frames saved in a list so ultimately would like to run a function and apply to all. The zoo function solves step one but I am not able to apply to all the data frames in the list.
graph<-lapply(myfiles,function(x) zoo(x) )
print(graph)
Ideally I would like to also mark minimum and maximum, which I am aware can be done with ggplot but not zoo.
Thank you so much for your help in advance
Assuming that the problem is overlapped panel names there are numerous solutions to this:
abbreviate the names using abbreviate. We show this for plot.zoo and autoplot.zoo .
put the panel name in the upper left. We show this for plot.zoo using a custom panel.
Use a header on each panel. We show this using xyplot.zoo and using ggplot.
The examples below use the test input in the Note at the end. (Next time please provide a complete example including all input in reproducible form.)
The first two examples below abbreviates the panel names and using plot.zoo and autoplot.zoo (which uses ggplot2). The third example uses xyplot.zoo (which uses lattice). This automatically uses headers and is probably the easiest solution.
library(zoo)
plot(z, ylab = abbreviate(names(z), 8))
library(ggplot2)
zz <- setNames(z, abbreviate(names(z), 8))
autoplot(zz)
library (lattice)
xyplot(z)
(click on plots to see expanded; continued after plots)
This fourth example puts the panel names in the upper left of the panel themselves using plot.zoo with a custom panel.
pnl <- function(x, y, ..., pf = parent.frame()) {
legend("topleft", names(z)[pf$panel.number], bty = "n", inset = -0.1)
lines(x, y)
}
plot(z, panel = pnl, ylab = "")
(click on plot to see it expanded)
We can also get headers with autoplot.zoo similar to in lattice above.
library(ggplot2)
autoplot(z, facets = ~ Series, col = I("black")) +
theme(legend.position = "none")
(click to expand; continued after graphics)
List
If you have a list of vectors L (see Note at end for a reproducible example of such a list) then this will produce a zoo object:
do.call("merge", lapply(L, zoo))
Note
Test input used above.
library(zoo)
set.seed(123)
nms <- paste0(head(state.name, 9), "XYZ") # long names
m <- matrix(rnorm(101*9), 101, dimnames = list(NULL, nms))
z <- zoo(m)
L <- split(m, col(m)) # test list using m in Note
t = table(iris$Species)
pie(t, labels=rownames(t))
This draws a simple pie. I want that the labels are a little bit more away from the pie. I checked the par() docu but I think I don't understand it completly and I missed the option for that.
This question is explicite about R's own pie() and not related to any other extern R package.
I don't think you can really do this with the pie function. If you look at View(pie) you'll see that the labels are drawn using the text function. This means that they are not really axis labels, and that par has little effect on them. You could try to do stuff by using the arguments of the text function (i.e. pos = 2, offset = 1) but this will affect all labels in the exact same way and results in warnings. To me it seems that the only way is the stupid way by adding some spaces before/ after labels. ie:
t = table(iris$Species)
nms = rownames(t)
# spaces needed after the labels
nms[2] = paste0(nms[2], strrep(' ', 7))
# spaces needed before the labels
nms[c(1, 3)] = paste0(strrep(' ', 7), nms[c(1, 3)])
pie(t, labels = nms)
If you want to a better solution, you could rewrite the pie function to be a bit more flexible or use a different package.
With plot_ly(), I am able to add what I want in the tool tip, but I don't manage to get rid of the default value. Is there way to do this ?
In the example below, for the first point, the tooltip is "250 company1". I would like to get only "company1". I have a solution using ggplot2 and then ggplotly() with the tooltip option, but I would rather remain with plotly only.
require(plotly)
seq <- 1:10
name <- c(paste0("company",1:10))
value <- c(250,125,50,40,40,30,20,20,10,10)
d <- data.frame(seq,name,value)
plot_ly(data=d,x=seq,y=value,text=name)
You need the hoverinfo parameter for plot_ly:
seq <- 1:10
name <- c(paste0("company",1:10))
value <- c(250,125,50,40,40,30,20,20,10,10)
d <- data.frame(seq,name,value)
plot_ly(data=d,x=seq,y=value,text=name,hoverinfo="text")
More reading: plotly R chart attribute reference
Gopala, I asked a similar question and found the answer useful. Perhaps it will help you too.
How can I remove the size line in the hoverinfo of a Plotly chart in R?
I have several subjects for which I need to generate a plot, as I have many subjects I'd like to have several plots in one page rather than one figure for subject.
Here it is what I have done so far:
Read txt file with subjects name
subjs <- scan ("ListSubjs.txt", what = "")
Create a list to hold plot objects
pltList <- list()
for(s in 1:length(subjs))
{
setwd(file.path("C:/Users/", subjs[[s]])) #load subj directory
ifile=paste("Co","data.txt",sep="",collapse=NULL) #Read subj file
dat = read.table(ifile)
dat <- unlist(dat, use.names = FALSE) #make dat usable for ggplot2
df <- data.frame(dat)
pltList[[s]]<- print(ggplot( df, aes(x=dat)) + #save each plot with unique name
geom_histogram(binwidth=.01, colour="cyan", fill="cyan") +
geom_vline(aes(xintercept=0), # Ignore NA values for mean
color="red", linetype="dashed", size=1)+
xlab(paste("Co_data", subjs[[s]] , sep=" ",collapse=NULL)))
}
At this point I can display the single plots for example by
print (pltList[1]) #will print first plot
print(pltList[2]) # will print second plot
I d like to have a solution by which several plots are displayed in the same page, I 've tried something along the lines of previous posts but I don't manage to make it work
for example:
for (p in seq(length(pltList))) {
do.call("grid.arrange", pltList[[p]])
}
gives me the following error
Error in arrangeGrob(..., as.table = as.table, clip = clip, main = main, :
input must be grobs!
I can use more basic graphing features, but I d like to achieve this by using ggplot. Many thanks for consideration
Matilde
Your error comes from indexing a list with [[:
consider
pl = list(qplot(1,1), qplot(2,2))
pl[[1]] returns the first plot, but do.call expects a list of arguments. You could do it with, do.call(grid.arrange, pl[1]) (no error), but that's probably not what you want (it arranges one plot on the page, there's little point in doing that). Presumably you wanted all plots,
grid.arrange(grobs = pl)
or, equivalently,
do.call(grid.arrange, pl)
If you want a selection of this list, use [,
grid.arrange(grobs = pl[1:2])
do.call(grid.arrange, pl[1:2])
Further parameters can be passed trivially with the first syntax; with do.call care must be taken to make sure the list is in the correct form,
grid.arrange(grobs = pl[1:2], ncol=3, top=textGrob("title"))
do.call(grid.arrange, c(pl[1:2], list(ncol=3, top=textGrob("title"))))
library(gridExtra) # for grid.arrange
library(grid)
grid.arrange(pltList[[1]], pltList[[2]], pltList[[3]], pltList[[4]], ncol = 2, main = "Whatever") # say you have 4 plots
OR,
do.call(grid.arrange,pltList)
I wish I had enough reputation to comment instead of answer, but anyway you can use the following solution to get it work.
I would do exactly what you did to get the pltList, then use the multiplot function from this recipe. Note that you will need to specify the number of columns. For example, if you want to plot all plots in the list into two columns, you can do this:
print(multiplot(plotlist=pltList, cols=2))
I have found that the beanplot is the best way to represent my data. I want to look at multiple beanplots together to visualize my data. Each of my plots contains 3 variables, so each one looks something like what would be generated by this code:
library(beanplot)
a <- rnorm(100)
b <- rnorm(100)
c <- rnorm(100)
beanplot(a, b ,c ,ylim = c(-4, 4), main = "Beanplot",
col = c("#CAB2D6", "#33A02C", "#B2DF8A"), border = "#CAB2D6")
(Would have just included an image but my reputation score is not high enough, sorry)
I have 421 of these that I want to put into one long PDF (EDIT: One plot per page is fine, this was just poor wording on my part). The approach I have taken was to first generate the beanplots in a for loop and store them in a list at each iteration. Then I will use the multiplot function (from the R Cookbook page on multiplot) to display all of my plots on one long column so I can begin my analysis.
The problem is that the beanplot function does not appear to be set up to assign plot objects as a variable. Example:
library(beanplot)
a <- rnorm(100)
b <- rnorm(100)
plot1 <- beanplot(a, b, ylim = c(-5,5), main = "Beanplot",
col = c("#CAB2D6", "#33A02C", "#B2DF8A"), border = "#CAB2D6")
plot1
If you then type plot1 into the R console, you will get back two of the plot parameters but not the plot itself. This means that when I store the plots in the list, I am unable to graph them with multiplot. It will simply return the plot parameters and a blank plot.
This behavior does not seem to be the case with qplot for example which will return a plot when you recall the stored plot. Example:
library(ggplot2)
a <- rnorm(100)
b <- rnorm(100)
plot2 <- qplot(a,b)
plot2
There is no equivalent to the beanplot that I know of in ggplot. Is there some sort of workaround I can use for this issue?
Thank you.
You can simply open a PDF device with pdf() and keep the default parameter onefile=TRUE. Then call all your beanplot()s, one after the other. They will all be in one PDF document, each one on a separate page. See here.