I'd like to use R to generate two categorical variables (such as eye color and hair color, for instance) where I can specify the degree to which these two variables are associated. It doesn't really matter to me which levels of eye color would be associated with which levels of hair color, but just being able to specify an overall association, such as by specifying the odds ratio, is a requirement. Also, I know there are ways to do this for two normally distributed continuous variables using, for example, the mvtnorm package, so I could take that route and then choose cut points to make the variables categorical after the fact, but I don't want to do it that way if I can avoid it. Any help would be greatly appreciated!
Edit: apologies for not being clearer from the start, but what I'm really asking I suppose is whether or not there's a function anybody knows of in some R package that will do this in one or two lines.
If you can specify the odds ratios (and you also need to specify the baseline odds), you just convert them to probabilities and use runif().
Edit (I misunderstood the question): Take a look at the bindata package.
If you like, here is a function I wrote that you can use to generate such data without the package. It is rather clunky; it's intended to be self-explanatory rather than elegant or fast.
odds.to.probs <- function(odds){
probs <- odds / (odds+1)
return(probs)
}
get.correlated.binary.data <- function(N, odds.x.eq.0, odds.y.eq.0.x.eq.0,
odds.ratio){
odds.y.eq.0.x.eq.1 <- odds.y.eq.0.x.eq.0*odds.ratio
prob.x.eq.0 <- odds.to.probs(odds.x.eq.0)
prob.y.eq.0.x.eq.0 <- odds.to.probs(odds.y.eq.0.x.eq.0)
prob.y.eq.0.x.eq.1 <- odds.to.probs(odds.y.eq.0.x.eq.1)
x <- ifelse(runif(N)<=prob.x.eq.0, 0, 1)
y <- rep(NA, N)
y <- ifelse(x==0, ifelse(runif(sum(x))<=prob.y.eq.0.x.eq.0, 0, 1), y)
y <- ifelse(x==1, ifelse(runif( (N-sum(x)) )<=prob.y.eq.0.x.eq.1, 0, 1), y)
dat <- data.frame(x=x, y=y)
return(dat)
}
> set.seed(9)
> dat <- get.correlated.binary.data(30, 3, 1.5, -.03)
> table(dat)
y
x 0 1
0 10 13
1 0 7
Related
I've got the following code that generates a random data set with a graph of the following,
x1=abs(rnorm(200))
x2=abs(rnorm(200))-7*x1^2
plot(x1,x2)
My goal is to separate the data so that the first 100 points are blue and the remaining 100 points are red in a data.frame. So I have two quick questions,
1) How do I separate the data so as I move along x1 the first 100 points are blue and the other are red? I've added an image below for clarification, mind my artistic talent with the snipping tool.
2) If after the colours are assigned, is a simple z=data.frame(x1,x2, colours) enough to get the data into a dataset so that I may run the data using some basic machine learning tools, such as SVM, Bagging and Boosting?
Cheers for the help.
set.seed(42)
dat <- data.frame(x1 = abs(rnorm(200)))
dat$x2 <- abs(rnorm(200)) - 7*dat$x1^2
dat$col <- ifelse(rank(dat$x1) <= 100, "blue", "red")
plot(x2 ~ x1, data = dat, col = col)
# also: plot(dat$x1, dat$x2, col = dat$col)
The "first 100" is subjective depending on your needs and the context of the data. One might also want the euclidean distance from origin (pythagorean), manhattan distance, or some other valuation. Or x1 <= mean(x1) or x1 <= median(x1). Lots of ways, this is just one way, where we use ifelse to differentiate/assign.
I have a measurment of which should fit an hysteresis. For visualisation purpose I would like to plot a line approximating the hysteresis to help explain this pattern.
I created an example in the following image using the code below.
I would like to have an output similar to the green curve - however I don't have this data directly available, and I don't care whether it is pointy.
However most smoothing functions such as smooth.spline which I plotted in blue - allow no loops. The closest I can find is from the bezier library - plotted in red. Not nicely visible here but it produces a loop, however it fits poorly (and gives some warnings and takes quite some time).
Can you suggest a method?
set.seed(12345)
up <- seq(0,1,length.out=100)^3
down <- sqrt(seq(1,0,length.out=100))
x <- c(seq(0,1,length.out=length(up)),
seq(1,0, length.out=length(down)))
data <- data.frame(x=x, y=c(up,down),
measuredx=x + rnorm(length(x))*0.01,
measuredy=c(up,down) + rnorm(length(up)+length(down))*0.03)
with(data,plot(measuredx,measuredy, type = "p"))
with(data,lines(x,y, col='green'))
sp <- with(data,smooth.spline(measuredx, measuredy))
with(sp, lines(x,y, col="blue"))
library(bezier)
bf <- bezierCurveFit(as.matrix(data[,c(1,3)]))
lines(bezier(t=seq(0, 1, length=500), p=bf$p), col="red", cex=0.25)
UPDATE
As it turns out my actual problem is slightly different I ask another question to reflect my actual issue in the question: How to fit a smooth hysteresis in a poorly distributed data set?
set.seed(12345)
up <- seq(0,1,length.out=100)^3
down <- sqrt(seq(1,0,length.out=100))
x <- c(seq(0,1,length.out=length(up)),
seq(1,0, length.out=length(down)))
data <- data.frame(x=x, y=c(up,down),
measuredx=x + rnorm(length(x))*0.01,
measuredy=c(up,down) + rnorm(length(up)+length(down))*0.03)
Instead of smoothing data$measuredy directly over data$measuredx, do two separate smoothing, by smoothing each against a time stamp variable. Then combine the fitted values from two smoothing. This is a general way for smoothing a closed curve or a loop. (See also Q & A: Smoothing Continuous 2D Points)
t <- seq_len(nrow(data) + 1)
xs <- smooth.spline(t, c(data$measuredx, data$measuredx[1]))$y
ys <- smooth.spline(t, c(data$measuredy, data$measuredy[1]))$y
with(data, plot(measuredx, measuredy))
lines(xs, ys)
c(data$measuredx, data$measuredx[1]) for example is just to ensure that the last value in the vector agrees with the first, so that it completes a cycle.
The curve is not really closed at the bottom left corner, because smooth.spline is doing smoothing not interpolation, so even if we have ensure that data vector completes a cycle, the fitted one may not be a closed one. A practical workaround is to use weighted regression, imposing heavy weight on this spot to make it closed.
t <- seq_len(nrow(data) + 1)
w <- rep(1, length(t)) ## initially identical weight everywhere
w[c(1, length(w))] <- 100000 ## give heavy weight
xs <- smooth.spline(t, c(data$measuredx, data$measuredx[1]), w)$y
ys <- smooth.spline(t, c(data$measuredy, data$measuredy[1]), w)$y
with(data, plot(measuredx, measuredy), col = 8)
lines(xs, ys, lwd = 2)
I am trying to generate a series of numbers to simulate a Levy Walk in R. Currently I am using the following code:
alpha=2
n=1000
x=rep(0,n)
y=rep(0,n)
for (i in 2:n){
theta=runif(1)*2*pi
f=runif(1)^(-1/alpha)
x[i]=x[i-1]+f*cos(theta)
y[i]=y[i-1]+f*sin(theta)
}
The code is working as expected and I am able to generate the numbers according to my requirements. The figure below shows on such Levy Walk:
The following histogram confirms that the numbers generated (i.e. f) actually belong to a power law:
My question is as follows:
The step lengths generated (i.e. f) are quite large. Haw can I modify the code so that the step lengths only fall within some bound [fmin, fmax]?
P.S. I have intentionally not vectorized the code.
Try using this:
f=runif(1, fmax^(-alpha), fmin^(-alpha))^(-1/alpha)
Note that you need 0 < fmin < fmax.
BTW, you can vectorize your code like this:
theta <- runif(n-1)*2*pi
f <- runif(n-1, fmax^(-alpha), fmin^(-alpha))^(-1/alpha)
x <- c(0, cumsum(f*cos(theta)))
y <- c(0, cumsum(f*sin(theta)))
Just for precision, what you're simmulating here is a Lévy flight. For it to be a Lévy walk, you should allow the particle to "walk" from the beginning to the end of each flight (with a for, for example). If you plot your resulting simmulation with plot(x, y, type = "o") you will see that there are no positions within flights (no walking) using your code.
library(ggplot2)
library(gridExtra)
alpha= 5
n= 1000
x= rep(0,n)
y= rep(0,n)
fmin= 1
fmax= n
for (i in 2:n){
theta= runif(n-1)*2*pi
f= runif(n-1, fmax^(-alpha), fmin^(-alpha))^(-1/alpha)
x= c(0, cumsum(f*cos(theta)))
y= c(0, cumsum(f*sin(theta)))
}
ggplot(data.frame(x=x, y=y), aes(x, y))+geom_point()+geom_path()
I really need your R skills here. Been working with this plot for several days now. I'm a R newbie, so that might explain it.
I have sequence coverage data for chromosomes (basically a value for each position along the length of every chromosome, making the length of the vectors many millions). I want to make a nice coverage plot of my reads. This is what I got so far:
Looks alright, but I'm missing y-labels so I can tell which chromosome it is, and also I've been having trouble modifying the x-axis, so it ends where the coverage ends. Additionally, my own data is much much bigger, making this plot in particular take extremely long time. Which is why I tried this HilbertVis plotLongVector. It works but I can't figure out how to modify it, the x-axis, the labels, how to make the y-axis logged, and the vectors all get the same length on the plot even though they are not equally long.
source("http://bioconductor.org/biocLite.R")
biocLite("HilbertVis")
library(HilbertVis)
chr1 <- abs(makeRandomTestData(len=1.3e+07))
chr2 <- abs(makeRandomTestData(len=1e+07))
par(mfcol=c(8, 1), mar=c(1, 1, 1, 1), ylog=T)
# 1st way of trying with some code I found on stackoverflow
# Chr1
plotCoverage <- function(chr1, start, end) { # Defines coverage plotting function.
plot.new()
plot.window(c(start, length(chr1)), c(0, 10))
axis(1, labels=F)
axis(4)
lines(start:end, log(chr1[start:end]), type="l")
}
plotCoverage(chr1, start=1, end=length(chr1)) # Plots coverage result.
# Chr2
plotCoverage <- function(chr2, start, end) { # Defines coverage plotting function.
plot.new()
plot.window(c(start, length(chr1)), c(0, 10))
axis(1, labels=F)
axis(4)
lines(start:end, log(chr2[start:end]), type="l")
}
plotCoverage(chr2, start=1, end=length(chr2)) # Plots coverage result.
# 2nd way of trying with plotLongVector
plotLongVector(chr1, bty="n", ylab="Chr1") # ylab doesn't work
plotLongVector(chr2, bty="n")
Then I have another vector called genes that are of special interest. They are about the same length as the chromosome-vectors but in my data they contain more zeroes than values.
genes_chr1 <- abs(makeRandomTestData(len=1.3e+07))
genes_chr2 <- abs(makeRandomTestData(len=1e+07))
These gene vectors I would like plotted as a red dot under the chromosomes! Basically, if the vector has a value there (>0), it is presented as a dot (or line) under the long vector plot. This I have not idea how to add! But it seems fairly straightforward.
Please help me! Thank you so much.
DISCLAIMER: Please do not simply copy and paste this code to run off the entire positions of your chromosome. Please sample positions (for example, as #Gx1sptDTDa shows) and plot those. Otherwise you'd probably get a huge black filled rectangle after many many hours, if your computer survives the drain.
Using ggplot2, this is really easily achieved using geom_area. Here, I've generated some random data for three chromosomes with 300 positions, just to show an example. You can build up on this, I hope.
# construct a test data with 3 chromosomes and 100 positions
# and random coverage between 0 and 500
set.seed(45)
chr <- rep(paste0("chr", 1:3), each=100)
pos <- rep(1:100, 3)
cov <- sample(0:500, 300)
df <- data.frame(chr, pos, cov)
require(ggplot2)
p <- ggplot(data = df, aes(x=pos, y=cov)) + geom_area(aes(fill=chr))
p + facet_wrap(~ chr, ncol=1)
You could use the ggplot2 package.
I'm not sure what exactly you want, but here's what I did:
This has 7000 random data points (about double the amount of genes on Chromosome 1 in reality). I used alpha to show dense areas (not many here, as it's random data).
library(ggplot2)
Chr1_cov <- sample(1.3e+07,7000)
Chr1 <- data.frame(Cov=Chr1_cov,fil=1)
pl <- qplot(Cov,fil,data=Chr1,geom="pointrange",ymin=0,ymax=1.1,xlab="Chromosome 1",ylab="-",alpha=I(1/50))
print(pl)
And that's it. This ran in less than a second. ggplot2 has a humongous amount of settings, so just try some out. Use facets to create multiple graphs.
The code beneath is for a sort of moving average, and then plotting the output of that. It is not a real moving average, as a real moving average would have (almost) the same amount of data points as the original - it will only make the data smoother. This code, however, takes an average for every n points. It will of course run quite a bit faster, but you will loose a lot of detailed information.
VeryLongVector <- sample(500,1e+07,replace=TRUE)
movAv <- function(vector,n){
chops <- as.integer(length(vector)/n)
count <- 0
pos <- 0
Cov <-0
pos[1:chops] <- 0
Cov[1:chops] <- 0
for(c in 1:chops){
tmpcount <- count + n
tmppos <- median(count:tmpcount)
tmpCov <- mean(vector[count:tmpcount])
pos[c] <- tmppos
Cov[c] <- tmpCov
count <- count + n
}
result <- data.frame(pos=pos,cov=Cov)
return(result)
}
Chr1 <- movAv(VeryLongVector,10000)
qplot(pos,cov,data=Chr1,geom="line")
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), ])
}