I am attempting to carry out lasso regression using the lars package but can not seem to get the lars bit to work. I have inputted code:
diabetes<-read.table("diabetes.txt", header=TRUE)
diabetes
library(lars)
diabetes.lasso = lars(diabetes$x, diabetes$y, type = "lasso")
However, I get an error message of :
Error in rep(1, n) : invalid 'times' argument.
I have tried entering it like this:
diabetes<-read.table("diabetes.txt", header=TRUE)
library(lars)
data(diabetes)
diabetes.lasso = lars(age+sex+bmi+map+td+ldl+hdl+tch+ltg+glu, y, type = "lasso")
But then I get the error message:
'Error in lars(age+sex + bmi + map + td + ldl + hdl + tch + ltg + glu, y, type = "lasso") :
object 'age' not found'
Where am I going wrong?
EDIT: Data - as below but with another 5 columns.
ldl hdl tch ltg glu
1 -0.034820763 -0.043400846 -0.002592262 0.019908421 -0.017646125
2 -0.019163340 0.074411564 -0.039493383 -0.068329744 -0.092204050
3 -0.034194466 -0.032355932 -0.002592262 0.002863771 -0.025930339
4 0.024990593 -0.036037570 0.034308859 0.022692023 -0.009361911
5 0.015596140 0.008142084 -0.002592262 -0.031991445 -0.046640874
I think some of the confusion may have to do with the fact that the diabetes data set that comes with the lars package has an unusual structure.
library(lars)
data(diabetes)
sapply(diabetes,class)
## x y x2
## "AsIs" "numeric" "AsIs"
sapply(diabetes,dim)
## $x
## [1] 442 10
##
## $y
## NULL
##
## $x2
## [1] 442 64
In other words, diabetes is a data frame containing "columns" which are themselves matrices. In this case, with(diabetes,lars(x,y,type="lasso")) or lars(diabetes$x,diabetes$y,type="lasso") work fine. (But just lars(x,y,type="lasso") won't, because R doesn't know to look for the x and y variables within the diabetes data frame.)
However, if you are reading in your own data, you'll have to separate the response variable and the predictor matrix yourself, something like
X <- as.matrix(mydiabetes[names(mydiabetes)!="y",])
mydiabetes.lasso = lars(X, mydiabetes$y, type = "lasso")
Or you might be able to use
X <- model.matrix(y~.,data=mydiabetes)
lars::lars does not appear to have a formula interface, which means you cannot use the formula specification for the column names (and furthermore it does not accept a "data=" argument). For more information on this and other "data mining" topics, you might want to get a copy of the classic text: "Elements of Statistical Learning". Try this:
# this obviously assumes require(lars) and data(diabetes) have been executed.
> diabetes.lasso = with( diabetes, lars(x, y, type = "lasso"))
> summary(diabetes.lasso)
LARS/LASSO
Call: lars(x = x, y = y, type = "lasso")
Df Rss Cp
0 1 2621009 453.7263
1 2 2510465 418.0322
2 3 1700369 143.8012
3 4 1527165 86.7411
4 5 1365734 33.6957
5 6 1324118 21.5052
6 7 1308932 18.3270
7 8 1275355 8.8775
8 9 1270233 9.1311
9 10 1269390 10.8435
10 11 1264977 11.3390
11 10 1264765 9.2668
12 11 1263983 11.0000
Related
Applying lmer() function across all columns in dataframe. I have made a list of variables and used lapply. Below is the code:
varlist=names(Genus_abundance)[5:ncol(Genus_abundance)]
lapply(varlist, function(x){lmer(substitute(i ~ Status + (1|Match), list(i=as.name(x), data=Genus_abundance, na.action = na.exclude)))})
However, I keep getting this error:
Error in eval(predvars, data, env) : object 'Acetatifactor' not found
I have checked and Acetatifactor is in the Genus_abundance dataframe.
Bit stuck about where its going wrong
EDIT:
Added a working example:
set.seed(43)
n <- 6
dat <- data.frame(id=1:n, Status=rep(LETTERS[1:2], n/2), age= sample(18:90, n, replace=TRUE), match=1:n, Acetatifactor=runif(n), Acutalibacter=runif(n), Adlercreutzia=runif(n))
head(dat)
id Status age match Acetatifactor Acutalibacter Adlercreutzia
1 1 A 49 1 0.1861022 0.1364904 0.8626298
2 2 B 31 2 0.7297301 0.8246794 0.3169752
3 3 A 23 3 0.4118721 0.5923042 0.2592606
4 4 B 64 4 0.4140497 0.7943970 0.7422665
5 5 A 60 5 0.4803101 0.7690324 0.7473611
6 6 B 79 6 0.4274945 0.9180564 0.9179040
lapply(varlist,
function(x){lmer(substitute(i ~ status + (1|match), list(i=as.name(x))),
data=dd)
})
The specific problem here is misplaced parentheses. You should close the substitute(..., list(i=as.name(x))) with three close-parentheses so that the whole chunk is properly understood as the first argument to lme4.
More generally I agree with #Kat in the comments that this is a good place to look. Since your arguments are already strings (not symbols) you don't really need all of the substitute() business and could use
fit_fun <- function(v) {
lmer(reformulate(c("status", "(1|match)"), response = v),
data = dd, na.action = na.exclude)
}
lapply(varlist, fit_fun)
Or you could use refit to fit the first column, then update the fit with each of the next columns. For large models this is much more efficient.
m1 <- lmer(resp1 ~ status + (1|match), ...)
m_other <- lapply(dd[-(1:3)], refit, object = m1)
c(list(m1), m_other)
I run a CV Lasso with the cv.gamlr function in R. I can get the coefficients for the lambdas that correspond to the “1se” or “min” criterion.
set.seed(123)
lasso<-cv.gamlr(x = X, y = Y, family ='binomial')
coef(lasso,select = "1se")
coef(lasso,select = "min")
But what if I want to obtain the coefficients for a specific lambda, stored in the lasso$gamlr$lambda vector? Is it possible to obtain them?
For example, to get the coefficients for the first lambda in the model... Something like this:
lambda_100<- lasso$gamlr$lambda[100]
coef(lasso,select = lambda_100)
Of course, this sends the following error:
Error in match.arg(select) : 'arg' must be NULL or a character vector
Thanks :)
The coefficients are stored under lasso$gamlr$beta, in your example, you can access them like this:
library(gamlr)
x = matrix(runif(500),ncol=5)
y = rnorm(100)
cvfit <- cv.gamlr(x, y, gamma=1)
dim(cvfit$gamlr$beta)
[1] 5 100
length(cvfit$gamlr$lambda)
[1] 100
cvfit$gamlr$lambda[100]
seg100
0.00125315
cvfit$gamlr$beta[,drop=FALSE,100]
5 x 1 sparse Matrix of class "dgCMatrix"
seg100
1 0.12960060
2 -0.16406246
3 -0.46566731
4 0.08197053
5 -0.54170494
Or if you prefer it in a vector:
cvfit$gamlr$beta[,100]
1 2 3 4 5
0.12960060 -0.16406246 -0.46566731 0.08197053 -0.54170494
I must do a Lasso regression with the package glmnetand I have problems to generate my x model.matrix
My data.frame: 108 observations, Y response variable, 24 predictors, here is an overview:
CONVENTIONAL_HUmin CONVENTIONAL_HUmean CONVENTIONAL_HUstd CONVENTIONAL_HUmax
1 37.9400539686119 63.4903779286635 11.7592095845857 85.2375439991287
2 23.8400539686119 80.5903779286635 15.0592095845857 125.837543999129
3 19.3035945249441 73.2764716205565 12.8816244173147 130.24141901586
CONVENTIONAL_HUQ1 CONVENTIONAL_HUQ2 CONVENTIONAL_HUQ3 HISTO_Skewness HISTO_Kurtosis
1 54.9938390994964 65.4873070322704 72.8863025473031 -0.203420585259268 2.25208159159488
2 70.8938390994964 80.3873070322704 91.4863025473031 -0.117420585259268 2.91208159159488
3 64.4689755423307 73.8666609177099 81.7351818199415 -0.0908104900456161 2.8751327713366
HISTO_ExcessKurtosis HISTO_Entropy_log10 HISTO_Entropy_log2 HISTO_Energy...Uniformity.
1 -0.751917020142877 0.701345471328916 2.32782599847774 0.219781577333287
2 -0.0887170201428774 0.793345471328916 2.63782599847774 0.184781577333287
3 -0.127231561113029 0.738530858918985 2.45445652190669 0.206887426065656
GLZLM_SZE GLZLM_LZE GLZLM_LGZE GLZLM_HGZE GLZLM_SZLGE
1 0.366581916604228 35.7249100350856 8.7285612359045e-05 11497.6407737833 3.22615226279017e-05
2 0.693581916604228 984.424910035086 8.5685612359045e-05 11697.6407737833 5.98615226279017e-05
3 0.622711792823853 1103.10288991619 8.5573088970709e-05 11571.7421733917 5.33303855950858e-05
GLZLM_SZHGE GLZLM_LZLGE GLZLM_LZHGE GLZLM_GLNU GLZLM_ZLNU
1 4164.91570215061 0.00314512237564268 405585.990838764 2.66964898745512 2.47759091065361
2 8064.91570215061 0.0835651223756427 11581585.9908388 12.9796489874551 38.5375909106536
3 7295.45317481887 0.0949686480587339 12926109.9421091 15.0930512668698 37.6083347285291
GLZLM_ZP Y
1 0.219643444043173 1
2 0.112643444043173 0
3 0.104031438564764 0
My code for the model.matrix
x=model.matrix(Y~.,data=data.det)
It générale a very large model.matrix with 244728 elements ! It seems that it has duplicated a hundred times each predictor of the 24 !
Here's an overview of the data.matrix:
(Intercept) CONVENTIONAL_HUmin-10.5599460313881
CONVENTIONAL_HUmin-117.359946031388 CONVENTIONAL_HUmin-13.0599460313881
CONVENTIONAL_HUmin-154.359946031388 CONVENTIONAL_HUmin-17.6599460313881
CONVENTIONAL_HUmin-18.3599460313881 CONVENTIONAL_HUmin-2.87994603138811
CONVENTIONAL_HUmin-21.281710504529 CONVENTIONAL_HUmin-28.3599460313881
CONVENTIONAL_HUmin-3.44994603138811 CONVENTIONAL_HUmin-3.89640547505594
CONVENTIONAL_HUmin-67.0599460313881 CONVENTIONAL_HUmin-682.359946031388
CONVENTIONAL_HUmin-9.08171050452898 CONVENTIONAL_HUmin1.04428949547101
CONVENTIONAL_HUmin1.63928949547101 CONVENTIONAL_HUmin10.8400539686119
CONVENTIONAL_HUmin10.968289495471 CONVENTIONAL_HUmin11.5400539686119
CONVENTIONAL_HUmin11.618289495471 CONVENTIONAL_HUmin11.6400539686119
CONVENTIONAL_HUmin12.518289495471 CONVENTIONAL_HUmin12.5400539686119
CONVENTIONAL_HUmin13.4400539686119 CONVENTIONAL_HUmin13.6400539686119
CONVENTIONAL_HUmin13.7400539686119 CONVENTIONAL_HUmin13.818289495471
CONVENTIONAL_HUmin14.5400539686119 CONVENTIONAL_HUmin14.6693017607572
CONVENTIONAL_HUmin14.8400539686119 CONVENTIONAL_HUmin16.9400539686119
CONVENTIONAL_HUmin17.0400539686119 CONVENTIONAL_HUmin17.618289495471
CONVENTIONAL_HUmin18.2400539686119 CONVENTIONAL_HUmin18.8400539686119
CONVENTIONAL_HUmin19.3035945249441 CONVENTIONAL_HUmin20.0400539686119
CONVENTIONAL_HUmin20.818289495471 CONVENTIONAL_HUmin21.0400539686119
CONVENTIONAL_HUmin21.118289495471 CONVENTIONAL_HUmin21.3400539686119
CONVENTIONAL_HUmin21.5400539686119 CONVENTIONAL_HUmin21.9400539686119
...
attr(,"contrasts")$CONVENTIONAL_HUmin
[1] "contr.treatment"
Not convenient at all because I end up with much more predictors in the input x for Lasso Regression which makes hazardous selection of the predictors even more present
Have you an idea of the source of the dysfunction ? any suggestion to fix that ?
Try this, you want a matrix not a model matrix...
# make a matrix of your predictors minus your outcome
x <- as.matrix(data.detect[-25])
# put the y column in a vector
y <- data.detect$Y
# run it
fit.lasso <- glmnet(x, y, family = "binomial", alpha = 1)
This question already has an answer here:
r predict function returning too many values [closed]
(1 answer)
Closed 6 years ago.
Assume I have have fit a regression model with multiple predictor variables in R, like in the following toy example:
n <- 20
x <- rnorm(n)
y <- rnorm(n)
z <- x + y + rnorm(n)
m <- lm(z ~ x + y + I(y^2))
Now I have new date, consisting of x and y values, and I want to predict the corresponding z values:
x.new <- rnorm(5)
y.new <- rnorm(5)
Question: How should I best call predict to apply the fitted model to the new data?
Here are a few things I tried, which do not work:
Attempt 1. Trying to use the x.new and y.new as the columns of a new data frame:
> predict(m, data=data.frame(x=x.new, y=y.new))
1 2 3 4 5 6 7
-0.0157090 1.1667958 -1.3797101 0.1185750 0.7786496 1.7666232 -0.6692865
8 9 10 11 12 13 14
1.9720532 0.3514206 1.1677019 0.6441418 -2.3010431 -0.3228424 -0.2181511
15 16 17 18 19 20
-0.8883275 0.4549592 -1.0377040 0.1750522 -2.4542843 1.2250101
This gave 20 values instead of 5, so cannot be right.
Attempt 2: Maybe predict got confused because the y^2 values were not supplied? Try to use model.frame to provide data in the correct form.
> predict(m, model.frame(~ x.new + y.new + I(y.new^2)))
1 2 3 4 5 6 7
-0.0157090 1.1667958 -1.3797101 0.1185750 0.7786496 1.7666232 -0.6692865
8 9 10 11 12 13 14
1.9720532 0.3514206 1.1677019 0.6441418 -2.3010431 -0.3228424 -0.2181511
15 16 17 18 19 20
-0.8883275 0.4549592 -1.0377040 0.1750522 -2.4542843 1.2250101
Warning message:
'newdata' had 5 rows but variables found have 20 rows
Again, this results in 20 values (plus a warning), so cannot be right.
The parameter is newdata (not data) when telling predict what to predict for.
predict(m, newdata = data.frame(x = x.new, y = y.new))
Following up from this question (see for reproducible data frame) I want to run MCMCGLMM n times, where n is the number of randomisations. I have tried to construct a loop which runs all the chains, and saves them (to retrieve the posterior distributions of the randomised variable later) but I am encountering problems.
This is what the data frame looks like (when n = 5, hence R1-R5), A = response variable, L and V are random effect variables, B is a fixed effect, R1-R5 are random assignments of L with structure of V maintained:
ID L B V A R1 R2 R3 R4 R5
1 1_1_1 1 1 1 11.1 6 19 21 1 31
2 1_1_1 1 1 1 6.9 6 19 21 1 31
3 1_1_4 1 1 4 7.7 2 24 8 22 22
4 1_1_4 1 1 4 10.5 2 24 8 22 22
5 1_1_5 1 1 5 8.5 11 27 14 17 22
6 1_1_7 1 1 7 11.2 5 24 13 18 25
I can create the names I want to assign to my chains, and the names of the variable that changes with each run of the MCMC chain (R1-Rn):
n = 5
Rs = as.vector(rep(NA,n))
for(i in 1:n){
Rs[i] = paste("R",i, sep = "")
}
Rs
Output:
> Rs
[1] "R1" "R2" "R3" "R4" "R5"
I then tried this loop to produce 5 chains:
for(i in 1:n){
chains[i] = MCMCglmm(A ~1 + B,
random = as.formula(paste0("~" ,Rs[i], " + Vial")),
rcov = ~units,
nitt = 500,
thin = 2,
burnin = 50,
prior = prior2,
family = "gaussian",
start = list(QUASI = FALSE),
data = df)
}
Thanks Roland for helping to get the random effect to call properly, previously I was getting an error Error in buildZ(rmodel.terms[r] ... object Rs[i] not found- fixed by as.formula
But this stores all of the data in chains and seemingly only the $Sol components, but I need to be able to access the values within the VCV, specifically the posterior distributions of the R variables (e.g. summary(chainR1$VCV))
In summary: It seems I am making a mistake in how I assign the chain names, does anyone have a suggestion of how to do this, and save the posterior distributions or even the whole chain?
Using assign was a key point:
n = 10 #Number of chains to run
chainVCVdf = matrix(rep(NA, times = ((nitt-burnin)/thin)*n), ncol = n)
colnames(chainVCVdf)=c(rep("X", times = n))
for(i in 1:n){
assign("chainX",paste0("chain",Rs[i]))
chainX = MCMCglmm(A ~1 + B,
random = as.formula(paste0("~" ,Rs[i], " + V")),
rcov = ~units,
nitt = nitt,
thin = thin,
burnin = burnin,
prior = prior1,
family = "gaussian",
start = list(QUASI = FALSE),
data = df)
assign("chainVCV", chainX$VCV[,1])
chainVCVdf[,i]=(chainVCV)
colnames(chainVCVdf)[i] = colnames(chainX$VCV)[1]
}
It then became possible to build a matrix of the VCV component that I am interested in (namely the randomised L assignment in columns R1-Rn)
It seems as though you want to run a number of different MCMCglmm formulas in a loop. #Roland has helped you found the solution to this (although I personally would create the formulas prior to the loop). #Roland also points out that in order to save the results of each model, you should save them in a list - rather than a chain as you are currently doing. You could also save each model as an .RData file, as seen in the end of the question. To formalize an answer to this question I would perform this in the following way:
Rs = paste0("~R", 1:5, " + V") ## Create all model formulae
chainNames = paste0("chainR", 1:5) ## Names for each model
chains = list() ## Initialize list
## Loop over models
for(i in 1:length(Rs)){
chains[[i]] = MCMCglmm(A ~1 + B,
random = formula(Rs[i]),
rcov = ~units,
nitt = 500,
thin = 2,
burnin = 50,
prior = prior2,
family = "gaussian",
start = list(QUASI = FALSE),
data = df)
}
names(chains) = chainNames ## Name each model
save(chains, "chainsR1-R5.Rdata") ## Save all model output
A side note, paste0 is the same as paste, but with the argument sep="" by default