I'm having trouble plotting 2 abline()s on a graph of the log10 Brain mass and log10 body mass. I'm following someone else's script but it just doesn't work for me. This is what I have:
This is the graph produced:
Why are the lines off there like that? Are the values I'm getting for the intercept and slope incorrect, or am I using the wrong ones? I've done other examples of this and its worked OK, but I've always ended up using the first model, never the second one so I'm not sure if I'm using the right values.
If you want to represent a linear regression of the log of the body mass compared to the log of the brain mass, the code is:
model <- lm(log10(brain)~log10(body))
then
abline(model$coefficients[2], model$coefficients[1])
When you don't know which parameter to enter in a function, use the help of that function. For abline(), the first parameter is the slope and the second one is the intercept.
Currently, your model use log10(brain), log10(body) and class.
If you want to assess the quality of your model, look at the residuals.
plot(model)
You can also just use the result of your lm like this:
model <- lm(log10(brain)~log10(body))
plot(log10(brain)~log10(body))
abline(model,col=2)
Related
I've fit a model using a 5 parameter logistic fit using the drm library. I apologize if this is a dumb question; I'm just getting started with r.
If dose in on my x-axis and response is on my y-axis, it is very easy to use this model to predict my response based on a given dose. You can either use the function PR or predict. However, I want to estimate a dose for a given response. I can't find a function to do this. For my assay, I fit my data to a standard curve and now I have measured a response from my unknowns. I would like to estimate concentration (dose) based on this response. I could fit the data in the opposite direction (flip x and y) but the fit differs slightly and that's not a very conventional strategy. If anyone has any suggestions I would greatly appreciate them. thank you
In case any one comes across this, the easiest way to do this is to use the ED function with the argument type = "absolute".
I have a time series of returns. I found out the optimal c(p,d,q) of an ARIMA(p,d,q) (I called it return.order) so that I finally got:
returns.arima <- prima(returns, order=returns.order)
This is not a lm file so I don't get the following diagnostic plots after computing plot(returns.arima):
but instead I get just the inverse unit roots:
So my questions are the following:
Is there a way to maybe convert my format to the lm one, so that I can produce this output easily?
If not, let's say I managed to get the residuals QQ-Plot through the following:
qqnorm(residuals.arima)
however, I didn't manage to find a way to report the outliers as in the first picture I posted.
Somebody knows how to do it?
What about Residual vs leverage, scale location and Cook's Distance Plots?
It's simple, but I can't remember how this procedure is called, hence I was not able to find the function to do so. I want to explore the effects and gradients of a simple lm() model by plotting the response of one variable at a time, the others being kept constant.
Can anybody tell me which function to use to do so? I seem to remember it's a function generating several plots, or something like this. It could be something akin to sensitivity analysis... Sorry for the beginner question.
Thank you in advance!
The car package has a lot of utilities for analyzing regression models. This sounds like a component+residual plot (or partial residuals plot).
library(car) # for avPlots(...)
fit <- lm(mpg~wt+hp+disp, mtcars)
crPlots(fit)
As noted in the comments, termplot(...) does basically the same thing.
I was using functional data analysis techniques to smooth a discrete dataset. I was using the fda package and the function smooth.basis. It gives me a fitted smooth object, yet now I need to get the inverse of it. The codes are something like the following:
basisobj<-create.bspline.basis(c(0,1),7)
fit=smooth.basis(argvals=Data$Time,y=Data$Score,fdParobj=basisobj)$fd
I wish to know what is the value of Data$Time when Data$Score is, say 20.
I did try the inverse.predict function but it could only give the result for lm function.
Thank you!
I have data points that represent a logarithmic function.
Is there an approach where I can just estimate the function that describes this data using R?
Thanks.
I assume that you mean that you have vectors y and x and you try do fit a function y(x)=Alog(x).
First of all, fitting log is a bad idea, because it doesn't behave well. Luckily we have x(y)=exp(y/A), so we can fit an exponential function which is much more convenient. We can do it using nonlinear least squares:
nls(x~exp(y/A),start=list(A=1.),algorithm="port")
where start is an initial guess for A. This approach is a numerical optimization, so it may fail.
The more stable way is to transform it to a linear function, log(x(y))=y/A and fit a straight line using lm:
lm(log(x)~y)
If I understand right you want to estimate a function given some (x,y) values of it. If yes check the following links.
Read about this:
http://en.wikipedia.org/wiki/Spline_%28mathematics%29
http://en.wikipedia.org/wiki/Polynomial_interpolation
http://en.wikipedia.org/wiki/Newton_polynomial
http://en.wikipedia.org/wiki/Lagrange_polynomial
Googled it:
http://www.stat.wisc.edu/~xie/smooth_spline_tutorial.html
http://stat.ethz.ch/R-manual/R-patched/library/stats/html/smooth.spline.html
http://www.image.ucar.edu/GSP/Software/Fields/Help/splint.html
I never used R so I am not sure if that works or not, but if you have Matlab i can explain you more.