I would like to plot each column of a matrix as a boxplot and then label the outliers in each boxplot as the row name they belong to in the matrix. To use an example:
vv=matrix(c(1,2,3,4,8,15,30),nrow=7,ncol=4,byrow=F)
rownames(vv)=c("one","two","three","four","five","six","seven")
boxplot(vv)
I would like to label the outlier in each plot (in this case 30) as the row name it belongs to, so in this case 30 belongs to row 7. Is there an easy way to do this? I have seen similar questions to this asked but none seemed to have worked the way I want it to.
There is a simple way. Note that b in Boxplot in following lines is a capital letter.
library(car)
Boxplot(y ~ x, id.method="y")
Or alternatively, you could use the "Boxplot" function from the {car} package which labels outliers for you.
See the following link: https://CRAN.R-project.org/package=car
In the example given it's a bit boring because they are all the same row. but here is the code:
bxpdat <- boxplot(vv)
text(bxpdat$group, # the x locations
bxpdat$out, # the y values
rownames(vv)[which(vv == bxpdat$out, arr.ind=TRUE)[, 1]], # the labels
pos = 4)
This picks the rownames that have values equal to the "out" list (i.e., the outliers) in the result of boxplot. Boxplot calls and returns the values from boxplot.stats. Take a look at:
str(bxpdat)
#DWin's solution works very well for a single boxplot, but will fail for anything with duplicate values, like the dataset I have created:
#Create data
set.seed(1)
basenums <- c(1,2,3,4,8,15,30)
vv=matrix(c(basenums, sample(basenums), 1-basenums,
c(0, 29, 30, 31, 32, 33, 60)),nrow=7,ncol=4,byrow=F)
dimnames(vv)=list(c("one","two","three","four","five","six","seven"), 1:4)
On this dataset, #DWin's solution gives:
Which is false, because in the 4th example, it is not possible for the minimum and maximum to be in the same row.
This solution is monstrous (and I hope can be simplified), but effective.
#Reshape data
vv_dat <- as.data.frame(vv)
vv_dat$row <- row.names(vv_dat)
library(reshape2)
new_vv <- melt(vv_dat, id.vars="row")
#Get boxplot data
bxpdat <- as.data.frame(boxplot(value~variable, data=new_vv)[c("out", "group")])
#Get matches with boxplot data
text_guide <- do.call(rbind, apply(bxpdat, 1,
function(x) new_vv[new_vv$value==x[1]&new_vv$variable==x[2], ]))
#Add labels
with(text_guide, text(x=as.numeric(variable)+0.2, y=value, labels=row))
Or you can simply run the code from this blog post:
source("https://raw.githubusercontent.com/talgalili/R-code-snippets/master/boxplot.with.outlier.label.r") # Load the function
set.seed(6484)
y <- rnorm(20)
x1 <- sample(letters[1:2], 20,T)
lab_y <- sample(letters, 20)
# plot a boxplot with interactions:
boxplot.with.outlier.label(y~x1, lab_y)
(which handles multiple outliers which are close to one another)
#sebastian-c
This is a slight modification of DWin solution that seem to work with more generality
bx1<-boxplot(pb,las=2,cex.axis=.8)
if(length(bx1$out)!=0){
## get the row of each outlier
out.rows<-sapply(1:length(bx1$out),function(i) which(vv[,bx1$group[i]]==bx1$out[i]))
text(bx1$group,bx1$out,
rownames(vv)[out.rows],
pos=4
)
}
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
I have a list of names sorted, like below:
ACVR2B
ADAM19
ADAM29
ADAM29
ADAMTS1
ADAMTS1
ADAMTS1
ADAMTS12
ADAMTS16
ADAMTS16
ADAMTS16
ADAMTS17
ADAMTS17
ADAMTSL1
ADCY10
would like to plot them as a histogram. It is very easy when these are values but with characters how can I do it in R or in open office?
Thank you
Try plotting the result of table(). The function table() computes the cross-tabulation frequency, which is exactly what you want.
set.seed(42)
x <- sample(letters, 100, replace = TRUE)
plot(table(x))
To plot the sorted values, try this:
z <- sort(table(x))
plot(z, xaxt="n", type="h")
axis(1, at=seq_along(z), names(z))
Given to what Andrie suggested: I did this:
Letter<-read.table("letters", header=T)
x <- sample(Letters, replace = F)
plot(sort(table(x)))
but the things is when, I want to plot in a descending order with only top 10 I miss out on the labels.
Can anyone suggest how to fix it and get only top 10.
I would like to ask a follow-up question related to the answer given in this post [Gantt style time line plot (in base R) ] on Gantt plots in base r. I feel like this is worth a new question as I think these plots have a broad appeal. I'm also hoping that a new question would attract more attention. I also feel like I need more space than the comments of that question to be specific.
The following code was given by #digEmAll . It takes a dataframe with columns referring to a start time, end time, and grouping variable and turns that into a Gantt plot. I have modified #digEmAll 's function very slightly to get the bars/segments in the Gantt plot to be contiguous to one another rather than having a gap. Here it is:
plotGantt <- function(data, res.col='resources',
start.col='start', end.col='end', res.colors=rainbow(30))
{
#slightly enlarge Y axis margin to make space for labels
op <- par('mar')
par(mar = op + c(0,1.2,0,0))
minval <- min(data[,start.col])
maxval <- max(data[,end.col])
res.colors <- rev(res.colors)
resources <- sort(unique(data[,res.col]),decreasing=T)
plot(c(minval,maxval),
c(0.5,length(resources)+0.5),
type='n', xlab='Duration',ylab=NA,yaxt='n' )
axis(side=2,at=1:length(resources),labels=resources,las=1)
for(i in 1:length(resources))
{
yTop <- i+0.5
yBottom <- i-0.5
subset <- data[data[,res.col] == resources[i],]
for(r in 1:nrow(subset))
{
color <- res.colors[((i-1)%%length(res.colors))+1]
start <- subset[r,start.col]
end <- subset[r,end.col]
rect(start,yBottom,end,yTop,col=color)
}
}
par(op) # reset the plotting margins
}
Here are some sample data. You will notice that I have four groups 1-4. However, not all dataframes have all four groups. Some only have two, some only have 3.
mydf1 <- data.frame(startyear=2000:2009, endyear=2001:2010, group=c(1,1,1,1,2,2,2,1,1,1))
mydf2 <- data.frame(startyear=2000:2009, endyear=2001:2010, group=c(1,1,2,2,3,4,3,2,1,1))
mydf3 <- data.frame(startyear=2000:2009, endyear=2001:2010, group=c(4,4,4,4,4,4,3,2,3,3))
mydf4 <- data.frame(startyear=2000:2009, endyear=2001:2010, group=c(1,1,1,2,3,3,3,2,1,1))
Here I run the above function, but specify four colors for plotting:
plotGantt(mydf1, res.col='group', start.col='startyear', end.col='endyear',
res.colors=c('red','orange','yellow','gray99'))
plotGantt(mydf2, res.col='group', start.col='startyear', end.col='endyear',
res.colors=c('red','orange','yellow','gray99'))
plotGantt(mydf3, res.col='group', start.col='startyear', end.col='endyear',
res.colors=c('red','orange','yellow','gray99'))
plotGantt(mydf4, res.col='group', start.col='startyear', end.col='endyear',
res.colors=c('red','orange','yellow','gray99'))
These are the plots:
What I would like to do is modify the function so that:
1) it will plot on the y-axis all four groups regardless of whether they actually appear in the data or not.
2) Have the same color associated with each group for every plot regardless of how many groups there are. As you can see, mydf2 has four groups and all four colors are plotted (1-red, 2-orange, 3-yellow, 4-gray). These colors are actually plotted with the same groups for mydf3 as that only contains groups 2,3,4 and the colors are picked in reverse order. However mydf1 and mydf4 have different colors plotted for each group as they do not have any group 4's. Gray is still the first color chosen but now it is used for the lowest occurring group (group2 in mydf1 and group3 in mydf3).
It appears to me that the main thing I need to work on is the vector 'resources' inside the function, and have that not just contain the unique groups but all. When I try manually overriding to make sure it contains all the groups, e.g. doing something as simple as resources <-as.factor(1:4) then I get an error:
'Error in rect(start, yBottom, end, yTop, col = color) : cannot mix zero-length and non-zero- length coordinates'
Presumably the for loop does not know how to plot data that do not exist for groups that don't exist.
I hope that this is a replicable/readable question and it's clear what I'm trying to do.
EDIT: I realize that to solve the color problem, I could just specify the colors for the 3 groups that exist in each of these sample dfs. However, my intention is to use this plot as an output to a function whereby it wouldn't be known ahead of time if all of the groups exist for a particular df.
I slightly modified your function to account for NA in start and end dates :
plotGantt <- function(data, res.col='resources',
start.col='start', end.col='end', res.colors=rainbow(30))
{
#slightly enlarge Y axis margin to make space for labels
op <- par('mar')
par(mar = op + c(0,1.2,0,0))
minval <- min(data[,start.col],na.rm=T)
maxval <- max(data[,end.col],na.rm=T)
res.colors <- rev(res.colors)
resources <- sort(unique(data[,res.col]),decreasing=T)
plot(c(minval,maxval),
c(0.5,length(resources)+0.5),
type='n', xlab='Duration',ylab=NA,yaxt='n' )
axis(side=2,at=1:length(resources),labels=resources,las=1)
for(i in 1:length(resources))
{
yTop <- i+0.5
yBottom <- i-0.5
subset <- data[data[,res.col] == resources[i],]
for(r in 1:nrow(subset))
{
color <- res.colors[((i-1)%%length(res.colors))+1]
start <- subset[r,start.col]
end <- subset[r,end.col]
rect(start,yBottom,end,yTop,col=color)
}
}
par(mar=op) # reset the plotting margins
invisible()
}
In this way, if you simply append all your possible group values to your data you'll get them printed on the y axis. e.g. :
mydf1 <- data.frame(startyear=2000:2009, endyear=2001:2010,
group=c(1,1,1,1,2,2,2,1,1,1))
# add all the group values you want to print with NA dates
mydf1 <- rbind(mydf1,data.frame(startyear=NA,endyear=NA,group=1:4))
plotGantt(mydf1, res.col='group', start.col='startyear', end.col='endyear',
res.colors=c('red','orange','yellow','gray99'))
About the colors, at the moment the ordered res.colors are applied to the sorted groups; so the 1st color in res.colors is applied to 1st (sorted) group and so on...
I want to draw boxplots in R and add names to outliers. So far I found this solution.
The function there provides all the functionality I need, but it scrambles incorrectly the labels. In the following example, it marks the outlier as "u" instead of "o":
library(plyr)
library(TeachingDemos)
source("http://www.r-statistics.com/wp-content/uploads/2011/01/boxplot-with-outlier-label-r.txt") # Load the function
set.seed(1500)
y <- rnorm(20)
x1 <- sample(letters[1:2], 20,T)
lab_y <- sample(letters, 20)
# plot a boxplot with interactions:
boxplot.with.outlier.label(y~x1, lab_y)
Do you know of any solution? The ggplot2 library is super nice, but provides no such functionality (as far as I know). My alternative is to use the text() function and extract the outlier information from the boxplot object. However, like this the labels may overlap.
Thanks a lot :-)
I took a look at this with debug(boxplot.with.outlier.label), and ... it turns out there's a bug in the function.
The error occurs on line 125, where the data.frame DATA is constructed from x,y and label_name.
Previously x and y have been reordered, while lab_y hasn't been. When the supplied value of x (your x1) isn't itself already in order, you'll get the kind of jumbling you experienced.
As an immediate fix, you can pre-order the x values like this (or do something more elegant)
df <- data.frame(y, x1, lab_y, stringsAsFactors=FALSE)
df <- df[order(df$x1), ]
# Needed since lab_y is not searched for in data (though it probably should be)
lab_y <- df$lab_y
boxplot.with.outlier.label(y~x1, lab_y, data=df)
The intelligent point label placement is a separate issue discussed here or here. There's no ultimate and ideal solution so you just have to pick one there.
So you would overplot the normal boxplot with labels, as follows:
set.seed(1501)
y <- c(4, 0, 7, -5, rnorm(16))
x1 <- c("a", "a", "b", "b", sample(letters[1:2], 16, T))
lab_y <- sample(letters, 20)
bx <- boxplot(y~x1)
out_lab <- c()
for (i in seq(bx$out)) {
out_lab[i] <- lab_y[which(y == bx$out[i])[1]]
}
identify(bx$group, bx$out, labels = out_lab, cex = 0.7)
Then, during the identify() is running, you just click to position where you want the label,
as described here. When finished, you just press "STOP".
Note that each outlier can have more than one label! In my solution, I just simply picked the first!!
PS: I feel ashamed for the for loop, but don't know how to vectorize it - feel free to post improvement.
EDIT: inspired by the Federico's link now I see it can be done much easier! Just these 2 commands:
boxplot(y~x1)
identify(as.integer(as.factor(x1)), y, labels = lab_y, cex = 0.7)
Anyone knows how to take advantage of ggplot or lattice in doing survival analysis? It would be nice to do a trellis or facet-like survival graphs.
So in the end I played around and sort of found a solution for a Kaplan-Meier plot. I apologize for the messy code in taking the list elements into a dataframe, but I couldnt figure out another way.
Note: It only works with two levels of strata. If anyone know how I can use x<-length(stratum) to do this please let me know (in Stata I could append to a macro-unsure how this works in R).
ggkm<-function(time,event,stratum) {
m2s<-Surv(time,as.numeric(event))
fit <- survfit(m2s ~ stratum)
f$time <- fit$time
f$surv <- fit$surv
f$strata <- c(rep(names(fit$strata[1]),fit$strata[1]),
rep(names(fit$strata[2]),fit$strata[2]))
f$upper <- fit$upper
f$lower <- fit$lower
r <- ggplot (f, aes(x=time, y=surv, fill=strata, group=strata))
+geom_line()+geom_ribbon(aes(ymin=lower,ymax=upper),alpha=0.3)
return(r)
}
I have been using the following code in lattice. The first function draws KM-curves for one group and would typically be used as the panel.group function, while the second adds the log-rank test p-value for the entire panel:
km.panel <- function(x,y,type,mark.time=T,...){
na.part <- is.na(x)|is.na(y)
x <- x[!na.part]
y <- y[!na.part]
if (length(x)==0) return()
fit <- survfit(Surv(x,y)~1)
if (mark.time){
cens <- which(fit$time %in% x[y==0])
panel.xyplot(fit$time[cens], fit$surv[cens], type="p",...)
}
panel.xyplot(c(0,fit$time), c(1,fit$surv),type="s",...)
}
logrank.panel <- function(x,y,subscripts,groups,...){
lr <- survdiff(Surv(x,y)~groups[subscripts])
otmp <- lr$obs
etmp <- lr$exp
df <- (sum(1 * (etmp > 0))) - 1
p <- 1 - pchisq(lr$chisq, df)
p.text <- paste("p=", signif(p, 2))
grid.text(p.text, 0.95, 0.05, just=c("right","bottom"))
panel.superpose(x=x,y=y,subscripts=subscripts,groups=groups,...)
}
The censoring indicator has to be 0-1 for this code to work. The usage would be along the following lines:
library(survival)
library(lattice)
library(grid)
data(colon) #built-in example data set
xyplot(status~time, data=colon, groups=rx, panel.groups=km.panel, panel=logrank.panel)
If you just use 'panel=panel.superpose' then you won't get the p-value.
I started out following almost exactly the approach you use in your updated answer. But the thing that's irritating about the survfit is that it only marks the changes, not each tick - e.g., it will give you 0 - 100%, 3 - 88% instead of 0 - 100%, 1 - 100%, 2 - 100%, 3 - 88%. If you feed that into ggplot, your lines will slope from 0 to 3, rather than remaining flat and dropping straight down at 3. That might be fine depending on your application and assumptions, but it's not the classic KM plot. This is how I handled the varying numbers of strata:
groupvec <- c()
for(i in seq_along(x$strata)){
groupvec <- append(groupvec, rep(x = names(x$strata[i]), times = x$strata[i]))
}
f$strata <- groupvec
For what it's worth, this is how I ended up doing it - but this isn't really a KM plot, either, because I'm not calculating out the KM estimate per se (although I have no censoring, so this is equivalent... I believe).
survcurv <- function(surv.time, group = NA) {
#Must be able to coerce surv.time and group to vectors
if(!is.vector(as.vector(surv.time)) | !is.vector(as.vector(group))) {stop("surv.time and group must be coercible to vectors.")}
#Make sure that the surv.time is numeric
if(!is.numeric(surv.time)) {stop("Survival times must be numeric.")}
#Group can be just about anything, but must be the same length as surv.time
if(length(surv.time) != length(group)) {stop("The vectors passed to the surv.time and group arguments must be of equal length.")}
#What is the maximum number of ticks recorded?
max.time <- max(surv.time)
#What is the number of groups in the data?
n.groups <- length(unique(group))
#Use the number of ticks (plus one for t = 0) times the number of groups to
#create an empty skeleton of the results.
curves <- data.frame(tick = rep(0:max.time, n.groups), group = NA, surv.prop = NA)
#Add the group names - R will reuse the vector so that equal numbers of rows
#are labeled with each group.
curves$group <- unique(group)
#For each row, calculate the number of survivors in group[i] at tick[i]
for(i in seq_len(nrow(curves))){
curves$surv.prop[i] <- sum(surv.time[group %in% curves$group[i]] > curves$tick[i]) /
length(surv.time[group %in% curves$group[i]])
}
#Return the results, ordered by group and tick - easier for humans to read.
return(curves[order(curves$group, curves$tick), ])
}