I am running the same regression with small alterations of x variables several times. My aim is after having determined the fit and significance of each variable for this linear regression model to view all all major plots. Instead of having to create each plot one by one, I want a function to loop through my variables (x1...xn) from the following list.
fit <-lm( y ~ x1 + x2 +... xn))
The plots I want to create for all x are
1) 'x versus y' for all x in the function above
2) 'x versus predicted y
3) x versus residuals
4) x versus time, where time is not a variable used in the regression but provided in the dataframe the data comes from.
I know how to access the coefficients from fit, however I am not able to use the coefficient names from the summary and reuse them in a function for creating the plots, as the names are characters.
I hope my question has been clearly described and hasn't been asked already.
Thanks!
Create some mock data
dat <- data.frame(x1=rnorm(100), x2=rnorm(100,4,5), x3=rnorm(100,8,27),
x4=rnorm(100,-6,0.1), t=(1:100)+runif(100,-2,2))
dat <- transform(dat, y=x1+4*x2+3.6*x3+4.7*x4+rnorm(100,3,50))
Make the fit
fit <- lm(y~x1+x2+x3+x4, data=dat)
Compute the predicted values
dat$yhat <- predict(fit)
Compute the residuals
dat$resid <- residuals(fit)
Get a vector of the variable names
vars <- names(coef(fit))[-1]
A plot can be made using this character representation of the name if you use it to build a string version of a formula and translate that. The four plots are below, and the are wrapped in a loop over all the vars. Additionally, this is surrounded by setting ask to TRUE so that you get a chance to see each plot. Alternatively you arrange multiple plots on the screen, or write them all to files to review later.
opar <- par(ask=TRUE)
for (v in vars) {
plot(as.formula(paste("y~",v)), data=dat)
plot(as.formula(paste("yhat~",v)), data=dat)
plot(as.formula(paste("resid~",v)), data=dat)
plot(as.formula(paste("t~",v)), data=dat)
}
par(opar)
The coefficients are stored in the fit objects as you say, but you can access them generically in a function by referring to them this way:
x <- 1:10
y <- x*3 + rnorm(1)
plot(x,y)
fit <- lm(y~x)
fit$coefficient[1] # intercept
fit$coefficient[2] # slope
str(fit) # a lot of info, but you can see how the fit is stored
My guess is when you say you know how to access the coefficients you are getting them from summary(fit) which is a bit harder to access than taking them directly from the fit. By using fit$coeff[1] etc you don't have to have the name of the variable in your function.
Three options to directly answer what I think was the question: How to access the coefficients using character arguments:
x <- 1:10
y <- x*3 + rnorm(1)
fit <- lm(y~x)
# 1
fit$coefficient["x"]
# 2
coefname <- "x"
fit$coefficient[coefname]
#3
coef(fit)[coefname]
If the question was how to plot the various functions then you should supply a sufficiently complex construction (in R) to allow demonstration of methods with a well-specified set of objects.
Related
I'm confused by the coefficients produced by the output of lm
Here's a copy of the data I'm working with
(postprocessed.csv)
"","time","value"
"1",1,2.61066016308988
"2",2,3.41246054742996
"3",3,3.8608767964033
"4",4,4.28686048552237
"5",5,4.4923132964825
"6",6,4.50557049744317
"7",7,4.50944447661246
"8",8,4.51097373134893
"9",9,4.48788748823809
"10",10,4.34603985656981
"11",11,4.28677073671406
"12",12,4.20065901625172
"13",13,4.02514194962519
"14",14,3.91360194972916
"15",15,3.85865748409081
"16",16,3.81318053258601
"17",17,3.70380706527433
"18",18,3.61552922363713
"19",19,3.61405310598722
"20",20,3.64591327503384
"21",21,3.70234435835577
"22",22,3.73503970503372
"23",23,3.81003078640584
"24",24,3.88201196162666
"25",25,3.89872518158949
"26",26,3.97432743542362
"27",27,4.2523675144599
"28",28,4.34654855854847
"29",29,4.49276038902684
"30",30,4.67830892029687
"31",31,4.91896819673664
"32",32,5.04350767355202
"33",33,5.09073406942046
"34",34,5.18510849382162
"35",35,5.18353176529036
"36",36,5.2210776270173
"37",37,5.22643491929207
"38",38,5.11137006553725
"39",39,5.01052467981257
"40",40,5.0361056705898
"41",41,5.18149486951409
"42",42,5.36334869132276
"43",43,5.43053620818444
"44",44,5.60001072279525
I have fitted a 4th order polynomial to this data using the following script:
library(ggplot2)
library(matrixStats)
library(forecast)
df_input <- read.csv("postprocessed.csv")
x <- df_input$time
y <- df_input$value
df <- data.frame(x, y)
poly4model <- lm(y~poly(x, degree=4), data=df)
v <- seq(30, 40)
vv <- poly4model$coefficients[1] +
poly4model$coefficients[2] * v +
poly4model$coefficients[3] * (v ^ 2) +
poly4model$coefficients[4] * (v ^ 3) +
poly4model$coefficients[5] * (v ^ 4)
pdf("postprocessed.pdf")
plot(df)
lines(v, vv, col="red", pch=20, lw=3)
dev.off()
I initially tried using the predict function to do this, but couldn't get that to work, so resorted to implementing this "workaround" using some new vectors v and vv to store the data for the line in the region I am trying to plot.
Ultimatly, I am trying to do this:
Fit a 4th order polynomial to the data
Plot the 4th order polynomial over the range of data in one color
Plot the 4th order polynomial over the range from the last value to the last value + 10 (prediction) in a different color
At the moment I am fairly sure using v and vv to do this is not "the best way", however I would have thought it should work. What is happening is that I get very large values.
Here is a screenshot from Desmos. I copied and pasted the same coefficients as shown by typing poly4model$coefficients into the console. However, something must have gone wrong because this function is nothing like the data.
I think I've provided enough info to be able to run this short script. However I will add the pdf as well.
It is easiest to use the predict function to create your line. To do that, you pass the model and a data frame with the desired independent variables to the predict function.
x <- df_input$time
y <- df_input$value
df <- data.frame(x, y)
poly4model <- lm(y~poly(x, degree=4), data=df)
v <- seq(30, 40)
#Notice the column in the dataframe is the same variable name
# as the variable in the model!
predict(poly4model, data.frame(x=v))
plot(df)
lines(v, predict(poly4model, data.frame(x=seq(30, 40))), col="red", pch=20, lw=3)
NOTE
The function poly "Returns or evaluates orthogonal polynomials of degree 1 to degree over the specified set of points x: these are all orthogonal to the constant polynomial of degree 0." To return the "normal" polynomial coefficients one needs to use the "raw=TRUE" option in the function.
poly4model <- lm(y~poly(x, degree=4, raw=TRUE), data=df)
Now your equation above will work.
I have a small data base (txt file).
I want to obtain an exponential regression in R.
The commands that I am using are:
regression <- read.delim("C:/Users/david/OneDrive/Desktop/regression.txt")
View(regression)
source('~/.active-rstudio-document', echo=TRUE)
m <- nls(DelSqRho ~ (1-exp(-a*(d-b)**2)), data=regression, start=list(a=1, b=1))
y_est<-predict(m,regression$d)
plot(x,y)
lines(x,y_est)
summary(m)
But, when I run it, I get an error:
Error in nls(DelSqRho ~ (1 - exp(-a * (d - b)^2)), data = regression, :
step factor 0.000488281 reduced below 'minFactor' of 0.000976562
and I do not know how to solve it, how to obtain the exponwential regression, please, any hint?
nls is quite sensitive to the values of the starting parameters and so you want to choose values that give a reasonable fit to the data (minpack.lm::nlsLM can be a bit more forgiving).
You can plot the curve at your starting values of a=1 and b=1 and see that it doesn't do a great job of capturing the curve.
regression <- read.delim("regression.txt")
with(regression, plot(d, DelSqRho, ylim=c(-3, 1)))
xs <- seq(min(regression$d), max(regression$d), length=100)
a <- 1; b <- 1; ys <- 1 - exp(-a* (xs - b)**2)
lines(xs, ys)
One way to get starting values is by rearranging the objective function.
y = 1 - exp(-a*(x-b)**2) can be rearranged as log(1/(1-y)) = ab^2 - 2abx + ax^2 (here y must be less than one). Linear regression can then be used to get an estimate of a and b.
start_m <- lm(log(1/(1-DelSqRho)) ~ poly(d, 2, raw=TRUE), regression)
unname(a <- coef(start_m)[3]) # as `a` is aligned with the quadratic term
# [1] -0.2345953
unname(b <- sqrt(coef(start_m)[1]/coef(start_m)[3]))
# [1] 2.933345
(Sometimes it is not possible to rearrange the data in this way and you can try to get a rough idea of the parameters by plotting the curves at various starting parameters. nls2 can also do a brute force search or grid search over starting parameters.)
We can now try to estimate the nls model at these parameters:
m <- nls(DelSqRho ~ 1-exp(-a*(d-b)**2), data=regression, start=list(a=a, b=b))
coef(m)
# a b
# -0.2379078 2.8868374
And plot the results:
# note that `newdata` must be a named list or data frame
# in which to look for variables with which to predict.
y_est <- predict(m, newdata=data.frame(d=xs))
with(regression, plot(d, DelSqRho))
lines(xs, y_est, col="red", lwd=2)
The fit isn't great and is perhaps suggestive that a more flexible model is required.
In R, I have a dataset of (x, y) points that is constantly being updated via simulation (values are appended to the end of the dataset).
I would like to compute the slope (via a linear model) of the line created by the data using only the last 10 listed datapoints.
The confusion here arises from the fact that the data are changing, and so I suspect a loop may be needed to iterate over the indices of the datapoints.
In R, one usually does something like
linreg <- lm(y ~ x, data = d) # set up linear model
summary.linreg <- summary(linreg) # output summary of model
beta1 <- coef(summary.linreg)[2] # extract slope
The change that is needed in my case is in linreg, specifically
linreg <- lm(y[?] ~ x[?], data = d) # subset response and predictor
For a non-changing dataset of 10 x-y points, one simply does [?] = [1:10] and the problem is solved. In my case though, I am at a standstill as to the best way to proceed efficiently.
Any thoughts?
No, don't subset inside the formula. Subset the data.frame. Inside your loop, after each database update, do this:
linreg <- lm(y ~ x, data = tail(d, 10))
If you want to loop over a data.frame rows, do this:
linreg <- lm(y ~ x, data = d[i:(i+9),])
If your data.frame is large and you only need the slope, you should use the more low-level function lm.fit for better performance. There might also be packages that provide functions for rolling regression.
Good day,
I have tried to figure this out, but I really can't!! I'll supply an example of my data in R:
x <- c(36,71,106,142,175,210,246,288,357)
y <- c(19.6,20.9,19.8,21.2,17.6,23.6,20.4,18.9,17.2)
table <- data.frame(x,y)
library(nlmrt)
curve <- "y~ a + b*exp(-0.01*x) + (c*x)"
ones <- list(a=1, b=1, c=1)
Then I use wrapnls to fit the curve and to find a solution:
solve <- wrapnls(curve, data=table, start=ones, trace=FALSE)
This is all fine and works for me. Then, using the following, I obtain a prediction of y for each of the x values:
predict(solve)
But how do I find the prediction of y for new x values? For instance:
new_x <- c(10, 30, 50, 70)
I have tried:
predict(solve, new_x)
predict(solve, 10)
It just gives the same output as:
predict(solve)
I really hope someone can help! I know if I use the values of 'solve' for parameters a, b, and c and substitute them into the curve formula with the desired x value that I would be able to this, but I'm wondering if there is a simpler option. Also, without plotting the data first.
Predict requires the new data to be a data.frame with column names that match the variable names used in your model (whether your model has one or many variables). All you need to do is use
predict(solve, data.frame(x=new_x))
# [1] 18.30066 19.21600 19.88409 20.34973
And that will give you a prediction for just those 4 values. It's somewhat unfortunate that any mistakes in specifying the new data results in the fitted values for the original model being returned. An error message probably would have been more useful, but oh well.
I ran a multiple regression with several continuous predictors, a few of which came out significant, and I'd like to create a scatterplot or scatter-like plot of my DV against one of the predictors, including a "regression line". How can I do this?
My plot looks like this
D = my.data; plot( D$probCategorySame, D$posttestScore )
If it were simple regression, I could add a regression line like this:
lmSimple <- lm( posttestScore ~ probCategorySame, data=D )
abline( lmSimple )
But my actual model is like this:
lmMultiple <- lm( posttestScore ~ pretestScore + probCategorySame + probDataRelated + practiceAccuracy + practiceNumTrials, data=D )
I would like to add a regression line that reflects the coefficient and intercept from the actual model instead of the simplified one. I think I'd be happy to assume mean values for all other predictors in order to do this, although I'm ready to hear advice to the contrary.
This might make no difference, but I'll mention just in case, the situation is complicated slightly by the fact that I probably will not want to plot the original data. Instead, I'd like to plot mean values of the DV for binned values of the predictor, like so:
D[,'probCSBinned'] = cut( my.data$probCategorySame, as.numeric( seq( 0,1,0.04 ) ), include.lowest=TRUE, right=FALSE, labels=FALSE )
D = aggregate( posttestScore~probCSBinned, data=D, FUN=mean )
plot( D$probCSBinned, D$posttestScore )
Just because it happens to look much cleaner for my data when I do it this way.
To plot the individual terms in a linear or generalised linear model (ie, fit with lm or glm), use termplot. No need for binning or other manipulation.
# plot everything on one page
par(mfrow=c(2,3))
termplot(lmMultiple)
# plot individual term
par(mfrow=c(1,1))
termplot(lmMultiple, terms="preTestScore")
You need to create a vector of x-values in the domain of your plot and predict their corresponding y-values from your model. To do this, you need to inject this vector into a dataframe comprised of variables that match those in your model. You stated that you are OK with keeping the other variables fixed at their mean values, so I have used that approach in my solution. Whether or not the x-values you are predicting are actually legal given the other values in your plot should probably be something you consider when setting this up.
Without sample data I can't be sure this will work exactly for you, so I apologize if there are any bugs below, but this should at least illustrate the approach.
# Setup
xmin = 0; xmax=10 # domain of your plot
D = my.data
plot( D$probCategorySame, D$posttestScore, xlim=c(xmin,xmax) )
lmMultiple <- lm( posttestScore ~ pretestScore + probCategorySame + probDataRelated + practiceAccuracy + practiceNumTrials, data=D )
# create a dummy dataframe where all variables = their mean value for each record
# except the variable we want to plot, which will vary incrementally over the
# domain of the plot. We need this object to get the predicted values we
# want to plot.
N=1e4
means = colMeans(D)
dummyDF = t(as.data.frame(means))
for(i in 2:N){dummyDF=rbind(dummyDF,means)} # There's probably a more elegant way to do this.
xv=seq(xmin,xmax, length.out=N)
dummyDF$probCSBinned = xv
# if this gives you a warning about "Coercing LHS to list," use bracket syntax:
#dummyDF[,k] = xv # where k is the column index of the variable `posttestScore`
# Getting and plotting predictions over our dummy data.
yv=predict(lmMultiple, newdata=subset(dummyDF, select=c(-posttestScore)))
lines(xv, yv)
Look at the Predict.Plot function in the TeachingDemos package for one option to plot one predictor vs. the response at a given value of the other predictors.