I am trying to compute robust/cluster standard errors after using mlogit() to fit a Multinomial Logit (MNL) in a Discrete Choice problem. Unfortunately, I suspect I am having problems with it because I am using data in long format (this is a must in my case), and getting the error #Error in ef/X : non-conformable arrays after sandwich::vcovHC( , "HC0").
The Data
For illustration, please gently consider the following data. It represents data from 5 individuals (id_ind ) that choose among 3 alternatives (altern). Each of the five individuals chose three times; hence we have 15 choice situations (id_choice). Each alternative is represented by two generic attributes (x1 and x2), and the choices are registered in y (1 if selected, 0 otherwise).
df <- read.table(header = TRUE, text = "
id_ind id_choice altern x1 x2 y
1 1 1 1 1.586788801 0.11887832 1
2 1 1 2 -0.937965347 1.15742493 0
3 1 1 3 -0.511504401 -1.90667519 0
4 1 2 1 1.079365680 -0.37267925 0
5 1 2 2 -0.009203032 1.65150370 1
6 1 2 3 0.870474033 -0.82558651 0
7 1 3 1 -0.638604013 -0.09459502 0
8 1 3 2 -0.071679538 1.56879334 0
9 1 3 3 0.398263302 1.45735788 1
10 2 4 1 0.291413453 -0.09107974 0
11 2 4 2 1.632831160 0.92925495 0
12 2 4 3 -1.193272276 0.77092623 1
13 2 5 1 1.967624379 -0.16373709 1
14 2 5 2 -0.479859282 -0.67042130 0
15 2 5 3 1.109780885 0.60348187 0
16 2 6 1 -0.025834772 -0.44004183 0
17 2 6 2 -1.255129594 1.10928280 0
18 2 6 3 1.309493274 1.84247199 1
19 3 7 1 1.593558740 -0.08952151 0
20 3 7 2 1.778701074 1.44483791 1
21 3 7 3 0.643191170 -0.24761157 0
22 3 8 1 1.738820924 -0.96793288 0
23 3 8 2 -1.151429915 -0.08581901 0
24 3 8 3 0.606695064 1.06524268 1
25 3 9 1 0.673866953 -0.26136206 0
26 3 9 2 1.176959443 0.85005871 1
27 3 9 3 -1.568225496 -0.40002252 0
28 4 10 1 0.516456176 -1.02081089 1
29 4 10 2 -1.752854918 -1.71728381 0
30 4 10 3 -1.176101700 -1.60213536 0
31 4 11 1 -1.497779616 -1.66301234 0
32 4 11 2 -0.931117325 1.50128532 1
33 4 11 3 -0.455543630 -0.64370825 0
34 4 12 1 0.894843784 -0.69859139 0
35 4 12 2 -0.354902281 1.02834859 0
36 4 12 3 1.283785176 -1.18923098 1
37 5 13 1 -1.293772990 -0.73491317 0
38 5 13 2 0.748091387 0.07453705 1
39 5 13 3 -0.463585127 0.64802031 0
40 5 14 1 -1.946438667 1.35776140 0
41 5 14 2 -0.470448172 -0.61326604 1
42 5 14 3 1.478763383 -0.66490028 0
43 5 15 1 0.588240775 0.84448489 1
44 5 15 2 1.131731049 -1.51323232 0
45 5 15 3 0.212145247 -1.01804594 0
")
The problem
Consequently, we can fit an MNL using mlogit() and extract their robust variance-covariance as follows:
library(mlogit)
library(sandwich)
mo <- mlogit(formula = y ~ x1 + x2|0 ,
method ="nr",
data = df,
idx = c("id_choice", "altern"))
sandwich::vcovHC(mo, "HC0")
#Error in ef/X : non-conformable arrays
As we can see there is an error produced by sandwich::vcovHC, which says that ef/X is non-conformable. Where X <- model.matrix(x) and ef <- estfun(x, ...). After looking through the source code on the mirror on GitHub I spot the problem which comes from the fact that, given that the data is in long format, ef has dimensions 15 x 2 and X has 45 x 2.
My workaround
Given that the show must continue, I am computing the robust and cluster standard errors manually using some functions that I borrow from sandwich and I adjusted to accommodate the Stata's output.
> Robust Standard Errors
These lines are inspired on the sandwich::meat() function.
psi<- estfun(mo)
k <- NCOL(psi)
n <- NROW(psi)
rval <- (n/(n-1))* crossprod(as.matrix(psi))
vcov(mo) %*% rval %*% vcov(mo)
# x1 x2
# x1 0.23050261 0.09840356
# x2 0.09840356 0.12765662
Stata Equivalent
qui clogit y x1 x2 ,group(id_choice) r
mat li e(V)
symmetric e(V)[2,2]
y: y:
x1 x2
y:x1 .23050262
y:x2 .09840356 .12765662
> Clustered Standard Errors
Here, given that each individual answers 3 questions is highly likely that there is some degree of correlation among individuals; hence cluster corrections should be preferred in such situations. Below I compute the cluster correction in this case and I show the equivalence with the Stata output of clogit , cluster().
id_ind_collapsed <- df$id_ind[!duplicated(mo$model$idx$id_choice,)]
psi_2 <- rowsum(psi, group = id_ind_collapsed )
k_cluster <- NCOL(psi_2)
n_cluster <- NROW(psi_2)
rval_cluster <- (n_cluster/(n_cluster-1))* crossprod(as.matrix(psi_2))
vcov(mo) %*% rval_cluster %*% vcov(mo)
# x1 x2
# x1 0.1766707 0.1007703
# x2 0.1007703 0.1180004
Stata equivalent
qui clogit y x1 x2 ,group(id_choice) cluster(id_ind)
symmetric e(V)[2,2]
y: y:
x1 x2
y:x1 .17667075
y:x2 .1007703 .11800038
The Question:
I would like to accommodate my computations within the sandwich ecosystem, meaning not computing the matrices manually but actually using the sandwich functions. Is it possible to make it work with models in long format like the one described here? For example, providing the meat and bread objects directly to perform the computations? Thanks in advance.
PS: I noted that there is a dedicated bread function in sandwich for mlogit, but I could not spot something like meat for mlogit, but anyways I am probably missing something here...
Why vcovHC does not work for mlogit
The class of HC covariance estimators can just be applied in models with a single linear predictor where the score function aka estimating function is the product of so-called "working residuals" and a regressor matrix. This is explained in some detail in the Zeileis (2006) paper (see Equation 7), provided as vignette("sandwich-OOP", package = "sandwich") in the package. The ?vcovHC also pointed to this but did not explain it very well. I have improved this in the documentation at http://sandwich.R-Forge.R-project.org/reference/vcovHC.html now:
The function meatHC is the real work horse for estimating the meat of HC sandwich estimators - the default vcovHC method is a wrapper calling sandwich and bread. See Zeileis (2006) for more implementation details. The theoretical background, exemplified for the linear regression model, is described below and in Zeileis (2004). Analogous formulas are employed for other types of models, provided that they depend on a single linear predictor and the estimating functions can be represented as a product of “working residual” and regressor vector (Zeileis 2006, Equation 7).
This means that vcovHC() is not applicable to multinomial logit models as they generally use separate linear predictors for the separate response categories. Similarly, two-part or hurdle models etc. are not supported.
Basic "robust" sandwich covariance
Generally, for computing the basic Eicker-Huber-White sandwich covariance matrix estimator, the best strategy is to use the sandwich() function and not the vcovHC() function. The former works for any model with estfun() and bread() methods.
For linear models sandwich(..., adjust = FALSE) (default) and sandwich(..., adjust = TRUE) correspond to HC0 and HC1, respectively. In a model with n observations and k regression coefficients the former standardizes with 1/n and the latter with 1/(n-k).
Stata, however, divides by 1/(n-1) in logit models, see:
Different Robust Standard Errors of Logit Regression in Stata and R. To the best of my knowledge there is no clear theoretical reason for using specifically one or the other adjustment. And already in moderately large samples, this makes no difference anyway.
Remark: The adjustment with 1/(n-1) is not directly available in sandwich() as an option. However, coincidentally, it is the default in vcovCL() without specifying a cluster variable (i.e., treating each observation as a separate cluster). So this is a convenient "trick" if you want to get exactly the same results as Stata.
Clustered covariance
This can be computed "as usual" via vcovCL(..., cluster = ...). For mlogit models you just have to consider that the cluster variable just needs to be provided once (as opposed to stacked several times in long format).
Replicating Stata results
With the data and model from your post:
vcovCL(mo)
## x1 x2
## x1 0.23050261 0.09840356
## x2 0.09840356 0.12765662
vcovCL(mo, cluster = df$id_choice[1:15])
## x1 x2
## x1 0.1766707 0.1007703
## x2 0.1007703 0.1180004
I would like to conduct a linear regression that will have three steps: 1) Running the regression on all data points 2) Taking out the 10 outiers as found by using the absolute distanse value of rstandard 3) Running the regression again on the new data frame.
I know how to do it manually but these is very awkwarding. Is there a way to do it automatically? Can it be done for taking out columns as well?
Here is my toy data frame and code (I'll take out 2 top outliers):
df <- read.table(text = "userid target birds wolfs
222 1 9 7
444 1 8 4
234 0 2 8
543 1 2 3
678 1 8 3
987 0 1 2
294 1 7 16
608 0 1 5
123 1 17 7
321 1 8 7
226 0 2 7
556 0 20 3
334 1 6 3
225 0 1 1
999 0 3 11
987 0 30 1 ",header = TRUE)
model<- lm(target~ birds+ wolfs,data=df)
rstandard <- abs(rstandard(model))
df<-cbind(df,rstandard)
g<-subset(df,rstandard > sort(unique(rstandard),decreasing=T)[3])
g
userid target birds wolfs rstandard
4 543 1 2 3 1.189858
13 334 1 6 3 1.122579
modelNew<- lm(target~ birds+ wolfs,data=df[-c(4,13),])
I don't see how you could do this without estimating two models, the first to identify the most influential cases and the second on the data without those cases. You could simplify your code and avoid cluttering the workspace, however, by doing it all in one shot, with the subsetting process embedded in the call to estimate the "final" model. Here's code that does this for the example you gave:
model <- lm(target ~ birds + wolfs,
data = df[-(as.numeric(names(sort(abs(rstandard(lm(target ~ birds + wolfs, data=df))), decreasing=TRUE)))[1:2]),])
Here, the initial model, evaluation of influence, and ensuing subsetting of the data are all built into the code that comes after the first data =.
Also, note that the resulting model will differ from the one your code produced. That's because your g did not correctly identify the two most influential cases, as you can see if you just eyeball the results of abs(rstandard(lm(target ~ birds + wolfs, data=df))). I think it has to do with your use of unique(), which seems unnecessary, but I'm not sure.
I'm trying to plot a logistic regression in R, for a continuous independent variable and a dichotomous dependent variable. I have very limited experience with R, but my professor has asked me to add this graph to a paper I'm writing, and he said R would probably be the best way to create it. Anyway, I'm sure there are tons of mistakes here, but this is the sort of this previously suggested on StackOverflow:
ggplot(vvv, aes(x = vvv$V1, y=vvv$V2)) + geom_point() + stat_smooth(method="glm", family="binomial", se=FALSE)
curve(predict(ggg, data.frame(V1=x), type="response"), add=TRUE)
where vvv is the name of my csv file (31 obs. of 2 variables), V1 is the continuous variable, and V2 is the dichotomous one. Also, ggg (List of 30?) is the following:
ggg<- glm(formula = vvv$V2 ~ vvv$V1, family = "binomial", data = vvv)
The ggplot function produces a graph of my data points, but no logistic regression curve. The curve function results in the following error:
"Error in curve(predict(ggg, data.frame(V1 = x), type = "resp"), add = TRUE) : 'expr' did not evaluate to an object of length 'n'
In addition: Warning message:'newdata' had 101 rows but variables found have 31 rows"
I'm not sure what the problem is, and I'm having trouble finding resources for this specific issue. Can anybody help? It would be greatly appreciated :)
Edit: Thanks to anyone who responded! My data, vvv, is the following, where the percent was the initial probability for presence/absence of a species in a specific area, and the 1 and 0 indicate whether or not a species ended up being observed.:
V1 V2
1 95.00% 1
2 95.00% 0
3 95.00% 1
4 92.00% 1
5 92.00% 1
6 92.00% 1
7 92.00% 1
8 92.00% 1
9 92.00% 1
10 92.00% 1
11 85.00% 1
12 85.00% 1
13 85.00% 1
14 85.00% 1
15 85.00% 1
16 80.00% 1
17 80.00% 0
18 77.00% 1
19 77.00% 1
20 75.00% 0
21 70.00% 1
22 70.00% 0
23 70.00% 0
24 70.00% 1
25 70.00% 0
26 69.00% 1
27 65.00% 0
28 60.00% 1
29 50.00% 1
30 35.00% 0
31 25.00% 0
As #MrFlick commented, V1 is probably a factor. So, first you have to change it to numeric class. This just substitutes "%" for nothing and divides by 100, so you will have proportions as numeric class:
vvv$V1<-as.numeric(sub("%","",vvv$V1))/100
Doing this, you can use your own code and you will have a plot for a logistic regression:
ggplot(vvv, aes(x = vvv$V1, y=vvv$V2)) + geom_point() + stat_smooth(method="glm", family="binomial", se=F)
This should print not only the points, but also the logistic regression curve. I don't understand what is the point of using curves. From what I could understand from your question, this is enough for what you need.
I'm trying to build a neural net with the neuralnet package and I'm having some trouble with it. I've been successful with the nnet package but no luck with the neuralnet one. I have read the whole documentation package and can't find the solution, or maybe I'm not able to spot it.
The training command I'm using is
nn<-neuralnet(V15 ~ V1 + V2 + V3 + V4 + V5 + V6 + V7 + V8 + V9 + V10 + V11 + V12 + V13 + V14,data=test.matrix,lifesign="full",lifesign.step=100,hidden=8)
and for prediction
result<- compute(nn,data.matrix)$net.result
The training takes a whole lot longer than the nnet training. I have tried using the same algorithm as nnet (backpropagation instead of resilent backpropagation) and nothing, changed the activation function too (and the linear.output=F) and pretty much everything else, and the result didn't improved. Predicted values are all the same. I don't understand why the nnet works for me, while the neuralnet one doesn't.
I could really use some help, my (lack of) understanding of both things (neural nets and R) it's probably the cause, but can't find why.
My dataset is from UCI. I want to use the neural network for a binary classification. A sample of the data would be:
25,Private,226802,11th,7,Never-married,Machine-op-inspct,Own-child,Black,Male,0,0,40,United-States,<=50K.
38,Private,89814,HS-grad,9,Married-civ-spouse,Farming-fishing,Husband,White,Male,0,0,50,United-States,<=50K.
28,Local-gov,336951,Assoc-acdm,12,Married-civ-spouse,Protective-serv,Husband,White,Male,0,0,40,United-States,>50K.
44,Private,160323,Some-college,10,Married-civ-spouse,Machine-op-inspct,Husband,Black,Male,7688,0,40,United-States,>50K.
18,?,103497,Some-college,10,Never-married,NA,Own-child,White,Female,0,0,30,United-States,<=50K.
34,Private,198693,10th,6,Never-married,Other-service,Not-in-family,White,Male,0,0,30,United-States,<=50K.
29,?,227026,HS-grad,9,Never-married,?,Unmarried,Black,Male,0,0,40,United-States,<=50K.
63,Self-emp-not-inc,104626,Prof-school,15,Married-civ-spouse,Prof-specialty,Husband,White,Male,3103,0,32,United-States,>50K.
24,Private,369667,Some-college,10,Never-married,Other-service,Unmarried,White,Female,0,0,40,United-States,<=50K.
55,Private,104996,7th-8th,4,Married-civ-spouse,Craft-repair,Husband,White,Male,0,0,10,United-States,<=50K.
65,Private,184454,HS-grad,9,Married-civ-spouse,Machine-op-inspct,Husband,White,Male,6418,0,40,United-States,>50K.
36,Federal-gov,212465,Bachelors,13,Married-civ-spouse,Adm-clerical,Husband,White,Male,0,0,40,United-States,<=50K.
26,Private,82091,HS-grad,9,Never-married,Adm-clerical,Not-in-family,White,Female,0,0,39,United-States,<=50K.
Converted into a matrix, with the factors as numerical values:
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15
39 7 77516 10 13 5 1 2 5 2 2174 0 40 39 0
50 6 83311 10 13 3 4 1 5 2 0 0 13 39 0
38 4 215646 12 9 1 6 2 5 2 0 0 40 39 0
53 4 234721 2 7 3 6 1 3 2 0 0 40 39 0
28 4 338409 10 13 3 10 6 3 1 0 0 40 5 0
37 4 284582 13 14 3 4 6 5 1 0 0 40 39 0
49 4 160187 7 5 4 8 2 3 1 0 0 16 23 0
52 6 209642 12 9 3 4 1 5 2 0 0 45 39 1
31 4 45781 13 14 5 10 2 5 1 14084 0 50 39 1
42 4 159449 10 13 3 4 1 5 2 5178 0 40 39 1
37 4 280464 16 10 3 4 1 3 2 0 0 80 39 1
30 7 141297 10 13 3 10 1 2 2 0 0 40 19 1
23 4 122272 10 13 5 1 4 5 1 0 0 30 39 0
Summary of the predicted values:
V1
Min. :0.2446871
1st Qu.:0.2446871
Median :0.2446871
Mean :0.2451587
3rd Qu.:0.2446871
Max. :1.0000000
Value of the Wilcoxon-Mann-Whitney test (area under the curve) shows that the prediction performance is virtualy the same as a random.
performance(predneural,"auc")#y.values
[1] 0.5013319126
The first reason to consider when you get weird results with neural networks is normalization. Your data must be normalized, otherwise, yes, the training will result in skewed NN which will produce the same outcome all the time, it is a common symptom.
Looking at your data set, there are values >>1 which means they are all treated by NN essentially the same. The reason for it is that the traditionally used response functions are (almost) constant outside some range around 0.
Always normalize your data before feeding it into a neural network.
Similar to the answer from #sashkello, I faced a similar issue earlier when my data was not properly normalized. Once I normalized the data everything ran correctly.
Recently, I faced this issue again and after debugging, I found that there can be another reason for neural networks giving the same output. If you have a neural network that has a weight decay term such as that in the RSNNS package, make sure that your decay term is not so large that all weights go to essentially 0.
I was using the caret package for in R. Initially, I was using a decay hyperparameter = 0.01. When I looked at the diagnostics, I saw that the RMSE was being calculated for each fold (of cross validation), but the Rsquared was always NA. In this case all predictions were coming out to the same value.
Once I reduced the decay to a much lower value (1E-5 and lower), I got the expected results.
I hope this helps.
I'm adding this here for anyone who might have the same problem I had.
if any of the above didn't work and you're using TensorFlow with a custom training loop. make sure to set training=True as in:
predictions = model(inputs, training=True)
If anyone has the same problem, I solved it by using the parameter rep when defining the neural network. It seems that the training of the network is done only once if you don't set this parameter and that leads to the network returning a vector of identical values (or values which are very similar, e.g. 0.99872 and 0.97891).
I believe that the problem could also be in the default value of err.fct parameter, which I set to ce for binary classification.
This is the code which led to normal results:
library(neuralnet)
model <- neuralnet(formula = allow ~ .,
data = podaci_train,
linear.output = FALSE,
err.fct = "ce",
hidden = 4,
rep = 3)
predictions <- compute(model, subset(podaci_test, select = -c(allow)))
predictions <- predictions$net.result
max(predictions)
min(predictions)
This is the output (maximum probability - 94.57%, minimum probability - 0.01%):
[1] 0.9456731
[1] 0.0009583263
The usage of rep leads to some weird behavior in RStudio when plotting, because there are multiple models in different training iterations. Therefore, if you don't want to make your environment crash from too much plotting, use an additional parameter:
plot(model, rep = 'best')
I'm a PhD student of genetics and I am trying do association analysis of some genetic data using linear regression. In the table below I'm regressing each 'trait' against each 'SNP' There is also a interaction term include as 'var'
I've only used R for 2 weeks and I don't have any programming background so please explain any help provided as I want to understand.
This is a sample of my data:
Sample ID var trait 1 trait 2 trait 3 SNP1 SNP2 SNP3
77856517 2 188 3 2 1 0 0
375689755 8 17 -1 -1 1 -1 -1
392513415 8 28 14 4 1 1 1
393612038 8 85 14 6 1 1 0
401623551 8 152 11 -1 1 0 0
348466144 7 -74 11 6 1 0 0
77852806 4 81 16 6 1 1 0
440614343 8 -93 8 0 0 1 0
77853193 5 3 6 5 1 1 1
and this is the code I've been using for a single regression:
result1 <-lm(trait1~SNP1+var+SNP1*var, na.action=na.exclude)
I want to run a loop where every trait is tested against each SNP.
I've been trying to modify codes I've found online but I always run into some error that I don't understand how to solve.
Thank you for any and all help.
Personally I don't find the problem so easy. Specially for an R novice.
Here a solution based on creating dynamically the regression formula.
The idea is to use paste function to create different formula terms, y~ x + var + x * var then coercing the result string tp a formula using as.formula. Here y and x are the formula dynamic terms: y in c(trait1,trai2,..) and x in c(SNP1,SNP2,...). Of course here I use lapply to loop.
lapply(1:3,function(i){
y <- paste0('trait',i)
x <- paste0('SNP',i)
factor1 <- x
factor2 <- 'var'
factor3 <- paste(x,'var',sep='*')
listfactor <- c(factor1,factor2,factor3)
form <- as.formula(paste(y, "~",paste(listfactor,collapse="+")))
lm(formula = form, data = dat)
})
I hope someone come with easier solution, ore more R-ish one:)
EDIT
Thanks to #DWin comment , we can simplify the formula to just y~x*var since it means y is modeled by x,var and x*var
So the code above will be simplified to :
lapply(1:3,function(i){
y <- paste0('trait',i)
x <- paste0('SNP',i)
LHS <- paste(x,'var',sep='*')
form <- as.formula(paste(y, "~",LHS)
lm(formula = form, data = dat)
})