I want to exam which variable impacts most on the outcome, in my data, which is the stock yield. My data is like below.
And my code is also attached.
library(randomForest)
require(data.table)
data = fread("C:/stockcrazy.csv")
PEratio <- data$offeringPE/data$industryPE
data_update <- data.frame(data,PEratio)
train <- data_update[1:47,]
test <- data_update[48:57,]
For the above subset data set train and test, I am not sure if I need to do a cross validation on this data. And I don't know how to do it.
data.model <- randomForest(yield ~ offerings + offerprice + PEratio + count + bingo
+ purchase , data=train, importance=TRUE)
par(mfrow = c(1, 1))
varImpPlot(data.model, n.var = 6, main = "Random Forests: Top 6 Important Variables")
importance(data.model)
plot(data.model)
model.pred <- predict(data.model, newdata=test)
model.pred
d <- data.frame(test,model.pred)
I am sure not sure if the result of IncMSE is good or bad. Can someone interpret this?
Additionally, I found the predicted values of the test data is not a good prediction of the real data. So how can I improve this?
Let's see. Let's start with %IncMSE:
I found this really good answer on cross validated about %IncMSE which I quote:
if a predictor is important in your current model, then assigning
other values for that predictor randomly but 'realistically' (i.e.:
permuting this predictor's values over your dataset), should have a
negative influence on prediction, i.e.: using the same model to
predict from data that is the same except for the one variable, should
give worse predictions.
So, you take a predictive measure (MSE) with the original dataset and
then with the 'permuted' dataset, and you compare them somehow. One
way, particularly since we expect the original MSE to always be
smaller, the difference can be taken. Finally, for making the values
comparable over variables, these are scaled.
This means that in your case the most important variable is purchase i.e. when the variable purchase was permuted (i.e. the order of the values randomly changed) the resulting model was 12% worse than having the variable in its original order in terms of calculating the mean square error. The MSE was 12% higher using a permuted purchase variable meaning that the this variable is the most important. Variable importance is just a measure of how important your predictor variables were in the model you used. In your case purchase was the most important and P/E ratio was the least (for those 6 variables). This is not something you can interpret as good or bad because it doesn't show you how well the model fits unseen data. I hope this is clear now.
For the cross-validation:
You do not need to do a cross validation during the training phase because it happens automatically. Approximately, 2/3 of the records are used for the creation of a tree and the 1/3 that is left out (out-of-bag data) is used to assess the tree afterwards (the R squared for the tree is computed using the oob data)
As for the improvement of the model:
By showing just the 10 first lines of the predicted and the actual values of yield, you cannot make a safe decision on whether the model is good or bad. What you need is a test of fitness. The most common one is the R squared. It is simplistic but for comparing models and getting a first opinion about your model it does its job. This is calculated by the model for every tree that you make and can be accessed by data.model$rsq. This ranges from 0 to 1 with 1 being the perfect model and 0 showing really poor fit ( it can sometimes even take negative values which shows a bad fit). If your rsq is bad then you can try the following to improve your model although it is not certain that you will get the results you wish for:
Calibrate your trees in a different way. Change the number of trees grown and prune the trees by specifying a big nodesize number. (here you use the default 500 trees and a nodesize of 5 which might overfit your model.)
Increase the number of variables if possible.
Choose a different model. There are cases were a random Forest would not work well
Related
Still fairly new to GLM and a bit confused about how to establish my model.
About my project:
I sampled the microbiome (and measured a diversity index value = Shannon) from the root system of a sample of 9 trees (=tree1_cat).
In each tree I sampled fine and thick roots (=rootpart) and each tree was sampled four times (=days) over the course of one season. Thus I have a nested design but have to keep time in mind for autocorrelation. Also not all values are present, thus I have a few missing values). So far I have tried and tested the following:
Model <- gls(Shannon ~ tree1_cat/rootpart + tree1_cat + days,
na.action = na.omit, data = psL.meta,
correlation = corAR1(form =~ 1|days),
weights = varIdent(form= ~ 1|days))
Furthermore I've tried to get more insight and used anova(Model) to get the p-values of those factors. Am I allowed to use those p-values? Also I've used emmeans(Model, specs = pairwise ~ rootpart)for pairwise comparisons but since rootpart was entered as nested factor it only gives me the paired interactions.
It all works, but I am not sure, whether this is the right model! Any help would be highly appreciated!
It would be helpful to know your scientific question, but let's suppose you're interested in differences in Shannon diversity between fine and thick roots and in time trends. A model you could use would be:
library(lmerTest)
lmer(Shannon ~ rootpart*days + (rootpart*days|tree1_cat), data = ...)
The fixed-effect component rootpart*days can be expanded into 1 + rootpart + days + rootpart:days (where 1 signifies the intercept)
intercept: SD in fine roots on day 0 (hopefully that's the beginning of the season)
rootpart: difference between fine and thick roots on day 0
days: change per day in SD in fine roots (slope)
rootpart:days difference in slope between thick roots and fine roots
The random-effect component (rootpart*days|tree1_cat) measures how all four of these effects vary across trees, and their correlations (e.g. do trees with a larger-than-average difference between fine and thick roots on day 0 also have a larger-than-average change over time in fine root SD?)
This 'maximal' random effects model is almost certainly too complex for your data; a rough rule of thumb says you should have 10-20 data points per parameter estimated, the fixed-effect model takes 4 parameters. A full model with 4 random effects requires the estimate of a 4×4 covariance matrix, which has (4*5)/2 = 10 parameters all by itself. I might just try (1+days|tree1_cat) (random slopes) or (rootpart|tree_cat) (among-tree difference in fine vs. thick differences), with a bias towards allowing for the variation in the effect that is your primary interest (e.g. if your primary question is about fine vs. thick then go with (rootpart|tree_cat).
I probably wouldn't worry about autocorrelation at all, nor heteroscedasticity by day (your varIdent(~1|days) term) unless those patterns are very strongly evident in the data.
If you want to allow for autocorrelation you'll need to fit the model with nlme::lme or glmmTMB (lmer still doesn't have machinery for autocorrelation models); something like
library(nlme)
lme(Shannon ~ rootpart*days,
random = ~days|tree1_cat,
data = ...,
correlation = corCAR1(form = ~days|tree1_cat)
)
You need to use corCAR1 (continuous-time autoregressive order-1) rather than the more common corAR1 for unevenly sampled data. Be aware that lme is more finicky/worse at dealing with singular models, so you may discover you need to simplify your model before you can actually get this model to run.
I'm working with random forest models in R as a part of an independent research project. I have fit my random forest model and generated the overall importance of each predictor to the models accuracy. However, in order to interpret my results in a research paper, I need to understand whether the variables have a positive or negative impact on the response variable.
Is there a way to produce this information from a random forest model? I.e. I expect age to have a positive impact on the likelihood a surgical complication occurs, but existence of osteoarthritis not so much.
Code:
surgery.bagComp = randomForest(complication~ahrq_ccs+age+asa_status+bmi+baseline_cancer+baseline_cvd+baseline_dementia+baseline_diabetes+baseline_digestive+baseline_osteoart+baseline_psych+baseline_pulmonary,data=surgery,mtry=2,importance=T,cutoff=c(0.90,0.10)) #The cutoff is the probability for each group selection, probs of 10% or higher are classified as 'Complication' occurring
surgery.bagComp #Get stats for random forest model
imp=as.data.frame(importance(surgery.bagComp)) #Analyze the importance of each variable in the model
imp = cbind(vars=rownames(imp), imp)
imp = imp[order(imp$MeanDecreaseAccuracy),]
imp$vars = factor(imp$vars, levels=imp$vars)
dotchart(imp$MeanDecreaseAccuracy, imp$vars,
xlim=c(0,max(imp$MeanDecreaseAccuracy)), pch=16,xlab = "Mean Decrease Accuracy",main = "Complications - Variable Importance Plot",color="black")
Importance Plot:
Any suggestions/areas of research anyone can suggest would be greatly appreciated.
In order to interpret my results in a research paper, I need to understand whether the variables have a positive or negative impact on the response variable.
You need to be perform "feature impact" analysis, not "feature importance" analysis.
Algorithmically, it's about traversing decision tree data structures and observing what was the impact of each split on the prediction outcome. For example, consider the split "age <= 40". Does the left branch (condition evaluates to true) carry lower likelihood than the right branch (condition evaluates to false)?
Feature importances may give you a hint which features to look for, but it cannot be "transformed" to feature impacts.
You might find the following articles helpful: WHY did your model predict THAT? (Part 1 of 2) and WHY did your model predict THAT? (Part 2 of 2).
I'm using the package glmnet, I need to run several LASSO analysis for the calibration of a large number of variables (%reflectance for each wavelength throughout the spectrum) against one dependent variable. I have a couple of doubts on the procedure and on the results I wish to solve. I show my provisional code below:
First I split my data in training (70% of n) and testing sets.
smp_size <- floor(0.70 * nrow(mydata))
set.seed(123)
train_ind <- sample(seq_len(nrow(mydata)), size = smp_size)
train <- mydata[train_ind, ]
test <- mydata[-train_ind, ]
Then I separate the target trait (y) and the independent variables (x) for each set as follows:
vars.train <- train[3:2153]
vars.test <- test[3:2153]
x.train <- data.matrix(vars.train)
x.test <- data.matrix(vars.test)
y.train <- train$X1
y.test <- test$X1
Afterwords, I run a cross-validated LASSO model for the training set and extract and writte the non-zero coefficients for lambdamin. This is because one of my concerns here is to note which variables (wavebands of the reflectance spectrum) are selected by the model.
install.packages("glmnet")
library(glmnet)
cv.lasso.1 <- cv.glmnet(y=y.train, x= x.train, family="gaussian", nfolds =
5, standardize=TRUE, alpha=1)
coef(cv.lasso.1,s=cv.lasso.1$lambda.min) # Using lambda min.
(cv.lasso.1)
install.packages("broom")
library(broom)
c <- tidy(coef(cv.lasso.1, s="lambda.min"))
write.csv(c, file = "results")
Finally, I use the function “predict” and apply the object “cv.lasso1” (the model obtained previously) to the variables of the testing set (x.2) in order to get the prediction of the variable and I run the correlation between the predicted and the actual values of Y for the testing set.
predict.1.2 <- predict(cv.lasso.1, newx=x.2, type = "response", s =
"lambda.min")
cor.test(x=c(predict.1.2), y=c(y.2))
This is a simplified code and had no problem so far, the point is that I would like to make a loop (of one hundred repetitions) of the whole code and get the non-zero coefficients of the cross-validated model as well as the correlation coefficient of the predicted vs actual values (for the testing set) for each repetition. I've tried but couldn't get any clear results. Can someone give me some hint?
thanks!
In general, running repeated analyses of the same type over and over on the same data can be tricky. And in your case, may not be necessary the way in which you have outlined it.
If you are trying to find the variables most predictive, you can use PCA, Principal Component Analysis to select variables with the most variation within the a variable AND between variables, but it does not consider your outcome at all, so if you have poor model design it will pick the least correlated data in your repository but it may not be predictive. So you should be very aware of all variables in the set. This would be a way of reducing the dimensionality in your data for a linear or logistic regression of some sort.
You can read about it here
yourPCA <- prcomp(yourData,
center = TRUE,
scale. = TRUE)
Scaling and centering are essential to making these models work right, by removing the distance between your various variables setting means to 0 and standard deviations to 1. Unless you know what you are doing, I would leave those as they are. And if you have skewed or kurtotic data, you might need to address this prior to PCA. Run this ONLY on your predictors...keep your target/outcome variable out of the data set.
If you have a classification problem you are looking to resolve with much data, try an LDA, Linear Discriminant Analysis which looks to reduce variables by optimizing the variance of each predictor with respect to the OUTCOME variable...it specifically considers your outcome.
require(MASS)
yourLDA =r <- lda(formula = outcome ~ .,
data = yourdata)
You can also set the prior probabilities in LDA if you know what a global probability for each class is, or you can leave it out, and R/ lda will assign the probabilities of the actual classes from a training set. You can read about that here:
LDA from MASS package
So this gets you headed in the right direction for reducing the complexity of data via feature selection in a computationally solid method. In looking to build the most robust model via repeated model building, this is known as crossvalidation. There is a cv.glm method in boot package which can help you get this taken care of in a safe way.
You can use the following as a rough guide:
require(boot)
yourCVGLM<- cv.glmnet(y = outcomeVariable, x = allPredictorVariables, family="gaussian", K=100) .
Here K=100 specifies that you are creating 100 randomly sampled models from your current data OBSERVATIONS not variables.
So the process is two fold, reduce variables using one of the two methods above, then use cross validation to build a single model from repeated trials without cumbersome loops!
Read about cv.glm here
Try starting on page 41, but look over the whole thing. The repeated sampling you are after is called booting and it is powerful and available in many different model types.
Not as much code and you might hope for, but pointing you in a decent direction.
I am using R to do some evaluations for two different forecasting models. The basic idea of the evaluation is do the comparison of Pearson correlation and it corresponding p-value using the function of cor.() . The graph below shows the final result of the correlation coefficient and its p-value.
we suggestion that model which has lower correlation coefficient with corresponding lower p-value(less 0,05) is better(or, higher correlation coefficient but with pretty high corresponding p-value).
so , in this case, overall, we would say that the model1 is better than model2.
but the question here is, is there any other specific statistic method to quantify the comparison?
Thanks a lot !!!
Assuming you're working with time series data since you called out a "forecast". I think what you're really looking for is backtesting of your forecast model. From Ruey S. Tsay's "An Introduction to Analysis of Financial Data with R", you might want to take a look at his backtest.R function.
backtest(m1,rt,orig,h,xre=NULL,fixed=NULL,inc.mean=TRUE)
# m1: is a time-series model object
# orig: is the starting forecast origin
# rt: the time series
# xre: the independent variables
# h: forecast horizon
# fixed: parameter constriant
# inc.mean: flag for constant term of the model.
Backtesting allows you to see how well your models perform on past data and Tsay's backtest.R provides RMSE and Mean-Absolute-Error which will give you another perspective outside of correlation. Caution depending on the size of your data and complexity of your model, this can be a very slow running test.
To compare models you'll normally look at RMSE which is essentially the standard deviation of the error of your model. Those two are directly comparable and smaller is better.
An even better alternative is to set up training, testing, and validation sets before you build your models. If you train two models on the same training / test data you can compare them against your validation set (which has never been seen by your models) to get a more accurate measurement of your model's performance measures.
One final alternative, if you have a "cost" associated with an inaccurate forecast, apply those costs to your predictions and add them up. If one model performs poorly on a more expensive segment of data, you may want to avoid using it.
As a side-note, your interpretation of a p value as less is better leaves a little to be [desired] quite right.
P values address only one question: how likely are your data, assuming a true null hypothesis? It does not measure support for the alternative hypothesis.
New to R.
Looking to limit the range of values that can be predicted.
df.Train <- data.frame(S=c(1,2,2,2,1),L=c(1,2,3,3,1),M=c(400,450,400,700,795),V=c(423,400,555,600,800),G=c(4,3.2,2,2.7,3.4), stringsAsFactors=FALSE)
m.Train <- lm(G~S+L+M+V,data=df.Train)
df.Test <- data.frame(S=c(1,2,1,2,1),L=c(1,2,3,1,1),M=c(400,450,500,800,795),V=c(423,475,555,600,555), stringsAsFactors=FALSE)
round(predict(m.Train, df.Test, type="response"),digits=1)
#seq(0,4,.1) #Predicted values should fall in this range
I've experimented with the predict() options but no luck.
Is there an option in predict? Should I be limiting it in the model?
Thank you
There are ways to transform your response variable, G in this occasion but there needs to be a good reason to do this. For example, if you want the output to be probabilities between 0 and 1 and your response variable is binary (0,1) then you need a logistic regression.
It all comes down to what data you have and whether a model / transformation of the response variable would be appropriate. In your example you do not specify what the data is and therefore we cannot say anything about which model or which transformation to use.
Setting the above on the side, if you really care about the prediction and do not care about the model or the transformation (but why wouldn't you care?) it looks like your data could use a quasipossion generalised linear model which might provide the output you need:
df.Train <- data.frame(S=c(1,2,2,2,1),L=c(1,2,3,3,1),M=c(400,450,400,700,795),V=c(423,400,555,600,800),G=c(4,3.2,2,2.7,3.4), stringsAsFactors=FALSE)
m.Train <- glm(G~S+L+M+V,data=df.Train, family=quasipoisson)
df.Test <- data.frame(S=c(1,2,1,2,1),L=c(1,2,3,1,1),M=c(400,450,500,800,795),V=c(423,475,555,600,555), stringsAsFactors=FALSE)
> predict(m.Train, df.Test, type="response")
1 2 3 4 5
4.000000 2.840834 3.062754 3.615447 4.573276
#probably not as good as you want
The model is using a log link by default which ensures the values will be positive. There is no guarantee that the model will not predict values greater than 4 but since you fed it values of less than 4 (your G variable) then chances are that most of the predictions will follow that distribution (like in this example). You might then need to consider how to treat predictions that go above 4.
In general you should consider carefully which model to choose and which response transformation. The poison model above for example is usually used for count data. However, you should never manipulate predictions on your own so if you choose the lm model in the end make sure you use the predictions it gives.
EDIT
It looks like in your case a non-linear regression might be what you need. The problem using a linear model like lm is that predictions can be greater than the max of the observed cases and less than the min of the observed cases. In which case doing a linear regression might not be appropriate. There are algorithms that will never predict a value greater than the max or less than the min. Such a case might be better suited in your case. One of these algorithms is the k-nearest neighbor for example:
library(FNN)
> knn.reg(df.Train[1:4], test=df.Test[1:4], y=df.Train[5], k=3)
Prediction:
[1] 3.066667 3.066667 3.066667 2.700000 3.100000
As you can see the predictions will never go above 4. That said knn is a local solution algorithm so again you need to research whether this is a good approach or not for your problem and your data. In terms of predictions though it definitely confirms your conditions. Knn is a very easy to understand algorithm that relies on distances between points to calculate predictions.
Hope it helps :)