I have a simple data set with two columns of data- K and SwStr.
K = c(.259, .215, .224, .223, .262, .233)
SwStr = c(.130, .117, .117, .114, .113, .111)
I plotted the data using:
plot(res$K, res$SwStr)
I want to plot the result of a linear model, using SwStr to predict K. I try to do that using:
graphic<-lm(K~SwStr-1, data=res)
P=predict(graphic)
plot(res$K, res$SwStr)
lines(P, lty="dashed", col="green", lwd=3)
But when I do this, I don't get any line plotted. What am I doing wrong?
(1) You are inverting the axes of the original plot. If you want SwStr on the x axis and K on the y axis you need
plot(res$SwStr, res$K)
or
with(res,plot(K~SwStr))
If you check the actual values of the plotted points on the graph, this might be obvious (especially if K and SwStr have different magnitudes) ...
For lm fits you can also use abline(graphic,...)
edit: (2) You also have to realize that predict gives just the predicted y values, not the x values. So you want something like this:
K=c(.259, .215, .224, .223, .262, .233)
SwStr=c(.130, .117, .117, .114, .113, .111)
g <- lm(K~SwStr-1)
par(las=1,bty="l") ## my favourites
plot(K~SwStr)
P <- predict(g)
lines(SwStr,P)
Depending on the situation, you may also want to use the newdata argument to predict to specify a set of evenly spaced x values ...
Related
I have fit a LOESS local regression to some data and I want to be able to find the X value associated with a given Y value.
plot(cars, main = "Stopping Distance versus Speed")
car_loess <- loess(cars$dist~cars$speed,span=.5)
lines(1:50, predict(car_loess,data.frame(speed=1:50)))
I was hoping that I could use teh inverse.predict function from the chemCal package, but that does not work for LOESS objects.
Does anyone have any idea how I might be able to do this calibrationa in a better way than predicticting Y values from a long vector of X values and looking through the resulting fitted Y for the Y value of interest and taking its corresponding X value?
Practically speaking in the above example, let's say I wanted to find the speed at which the stopping distance is 15.
Thanks!
The predicted line that you added to the plot is not quite right. Use code like this instead:
# plot the loess line
lines(cars$speed, car_loess$fitted, col="red")
You can use the approx() function to get a linear approximation from the loess line at a give y value. It works just fine for the example that you give:
# define a given y value at which you wish to approximate x from the loess line
givenY <- 15
estX <- approx(x=car_loess$fitted, y=car_loess$x, xout=givenY)$y
# add corresponding lines to the plot
abline(h=givenY, lty=2)
abline(v=estX, lty=2)
But, with a loess fit, there may be more than one x for a given y. The approach I am suggesting does not provide you with ALL of the x values for the given y. For example ...
# example with non-monotonic x-y relation
y <- c(1:20, 19:1, 2:20)
x <- seq(y)
plot(x, y)
fit <- loess(y ~ x)
# plot the loess line
lines(x, fit$fitted, col="red")
# define a given y value at which you wish to approximate x from the loess line
givenY <- 15
estX <- approx(x=fit$fitted, y=fit$x, xout=givenY)$y
# add corresponding lines to the plot
abline(h=givenY, lty=2)
abline(v=estX, lty=2)
I'm trying to plot some data with 2d density contours using ggplot2 in R.
I'm getting one slightly odd result.
First I set up my ggplot object:
p <- ggplot(data, aes(x=Distance,y=Rate, colour = Company))
I then plot this with geom_points and geom_density2d. I want geom_density2d to be weighted based on the organisation's size (OrgSize variable). However when I add OrgSize as a weighting variable nothing changes in the plot:
This:
p+geom_point()+geom_density2d()
Gives an identical plot to this:
p+geom_point()+geom_density2d(aes(weight = OrgSize))
However, if I do the same with a loess line using geom_smooth, the weighting does make a clear difference.
This:
p+geom_point()+geom_smooth()
Gives a different plot to this:
p+geom_point()+geom_smooth(aes(weight=OrgSize))
I was wondering if I'm using density2d inappropriately, should I instead be using contour and supplying OrgSize as the 'height'? If so then why does geom_density2d accept a weighting factor?
Code below:
require(ggplot2)
Company <- c("One","One","One","One","One","Two","Two","Two","Two","Two")
Store <- c(1,2,3,4,5,6,7,8,9,10)
Distance <- c(1.5,1.6,1.8,5.8,4.2,4.3,6.5,4.9,7.4,7.2)
Rate <- c(0.1,0.3,0.2,0.4,0.4,0.5,0.6,0.7,0.8,0.9)
OrgSize <- c(500,1000,200,300,1500,800,50,1000,75,800)
data <- data.frame(Company,Store,Distance,Rate,OrgSize)
p <- ggplot(data, aes(x=Distance,y=Rate))
# Difference is apparent between these two
p+geom_point()+geom_smooth()
p+geom_point()+geom_smooth(aes(weight = OrgSize))
# Difference is not apparent between these two
p+geom_point()+geom_density2d()
p+geom_point()+geom_density2d(aes(weight = OrgSize))
geom_density2d is "accepting" the weight parameter, but then not passing to MASS::kde2d, since that function has no weights. As a consequence, you will need to use a different 2d-density method.
(I realize my answer is not addressing why the help page says that geom_density2d "understands" the weight argument, but when I have tried to calculate weighted 2D-KDEs, I have needed to use other packages besides MASS. Maybe this is a TODO that #hadley put in the help page that then got overlooked?)
I have checked my references, it seems to me that to fit a dataset with x and y, many tutorial need to first plot the x and y, then the fitted line is plot. The normal procedure is like below:
## Calculate the fitted line
smoothingSpline = smooth.spline(tree_number[2:100], jaccard[1:99], spar=0.35)
plot(tree_number[2:100],jaccard[1:99]) #plot the data points
lines(smoothingSpline) # add the fitted spline line.
However, I do not want to plot the tree_number and jaccard, but rather, I only want to plot the fitted spline line in the plot, how should I do?
You can use the associcated plot function:
plot(smoothingSpline, type="l")
Or you can extract the x and y values explicitly and plot them
plot(smoothingSpline$x, smoothingSpline$y, type="l")
Why not just plot(smoothingSpline, type = "l")? That should allow you to add the fitted spline line without having to first plot the data points.
I have a probability density function in a plot called ph that i derived from two samples of data, by the help of a user of stackoverflow, in this way
few <-read.table('outcome.dat',head=TRUE)
many<-read.table('alldata.dat',head=TRUE)
mh <- hist(many$G,breaks=seq(0,1.,by=0.03), plot=FALSE)
fh <- hist(few$G, breaks=mh$breaks, plot=FALSE)
ph <- fh
ph$density <- fh$counts/(mh$counts+0.001)
plot(ph,freq=FALSE,col="blue")
I would like to fit the best curve of the plot of ph, but i can't find a working method.
how can i do this? I have to extract the vaule from ph and then works on they? or there is same function that works on
plot(ph,freq=FALSE,col="blue")
directly?
Assuming you mean that you want to perform a curve fit to the data in ph, then something along the lines of
nls(FUN, cbind(ph$counts, ph$mids),...) may work. You need to know what sort of function 'FUN' you think the histogram data should fit, e.g. normal distribution. Read the help file on nls() to learn how to set up starting "guess" values for the coefficients in FUN.
If you simply want to overlay a curve onto the histogram, then smoo<-spline(ph$mids,ph$counts);
lines(smoo$x,smoo$y)
will come close to doing that. You may have to adjust the x and/or y scaling.
Do you want a density function?
x = rnorm(1000)
hist(x, breaks = 30, freq = FALSE)
lines(density(x), col = "red")
How to plot the density of a single column dataset as dots? For example
x <- c(1:40)
On the same plot using the same scale of the x-axis and y-axis, how to add another data set as line format which represent the density of another data that represents the equation of
y = exp(-x)
to the plot?
The equation is corrected to be y = exp(-x).
So, by doing plot(density(x)) or plot(density(y)), I got two separated figures. How to add them in the same axis and using dots for x, smoothed line for y?
You can add a line to a plot with the lines() function. Your code, modified to do what you asked for, is the following:
x <- 1:40
y <- exp(-x)
plot(density(x), type = "p")
lines(density(y))
Note that we specified the plot to give us points with the type parameter and then added the density curve for y with lines. The help pages for ?plot, ?par, ?lines would be some insightful reading. Also, check out the R Graph Gallery to view some more sophisticated graphs that generally have the source code attached to them.