I am trying to calculate a regression variable based on a range of variables in my data set. I would like the regression variable (ei: Threshold 1) to be calculated using a different variable set in each iteration of running the regression.
Aim to collected SSR values for each threshold range, and thus identify the ideal threshold based on the data.
Data (df) variables: Yield, Prec, Price, 0C, 1C, 2C, 3C, 4C, 5C, 6C, 7C, 8C, 9C, 10C
Each loop calculates "thresholds" by selecting a different "b" each time.
a <- df$0C
b <- df$1C
Threshold1 <- (a-b)
Threshold2 <- (b)
Where "b" would be changing in each loop, ranging from 1C to 9C.
Each individual threshold set (1 and 2) should be used to run a regression, and save the SSR for comparison with the subsequent regression utilizing thresholds based on a new "b" value (ranging from 1C TO 9C)
Regression:
reg <- lm(log(Yield)~Threshold1+Threshold2+log(Price)+prec+I(prec^2),data=df)
for each loop of the Regression, I vary the components of calculating thresholds in the following manner:
Current approach is centered around the following code:
df <- read.csv("Data.csv",header=TRUE)
names(df)
0C-9Cvarlist <- names(df)[9:19]
ssr.vec <- matrix(,21,1)
for(i in 1:length(varlist)){
a <- df$0C
b <- df$[i]
Threshold1 <- (a-b)
Threshold2 <- (b)
reg <- lm(log(Yield)~Threshold1+Threshold2+log(Price)+prec+I(prec^2),data=df)
r2 <- summary(reg)$r.squared
ssr.vec[i,] <- c(varlist,r2)
}
colnames(ssr.vec) <- c("varlist","r2")
I am failing to achieve the desired result with the above approach.
Thank you.
I can spot quite a few mistakes...
You need to add variables of interest (Threshold1 anf Threshold2) to the data in the regression. Also, I think that you need to select varlist[i] and not varlist to create your ssr.vec. You need 2 columns to your ssr.vec which is a matrix, so you should call it matrix. You also cannot use something like df$[i] to extract a column! Why is the matrix of length 21 ?! Change the column name to C0,..,C9 and not 0C,..,9C.
For future reference, solve the simple errors before asking question... and include error messages in your post!
This should do the job:
df <- read.csv("Data.csv",header=TRUE)
names(df)[8:19] = paste0("C",0:10)
varlist <- names(df)[9:19]
ssr.vec <- matrix(,21,2)
for(i in 1:length(varlist)){
a <- df$C0
b <- df[,i+9]
df$Threshold1 <- (a-b)
df$Threshold2 <- (b)
reg <- lm(log(Yield)~Threshold1+Threshold2+log(Price)+prec+I(prec^2),data=df)
r2 <- summary(reg)$r.squared
ssr.vec[i,] <- c(varlist[i],r2)
}
colnames(ssr.vec) <- c("varlist","r2")
Related
I have several models that I would like to compare their choices of important predictors over the same data set, Lasso being one of them. The data set I am using consists of census data with around a thousand variables that have been renamed to "x1", "x2" and so on for convenience sake (The original names are extremely long). I would like to report the top features then rename these variables with a shorter more concise name.
My attempt to solve this is by extracting the top variables in each iterated model, put it into a list, then finding the mean of the top variables in X amount of loops. However, my issue is I still find variability with the top 10 most used predictors and so I cannot manually alter the variable names as each run on the code chunk yields different results. I suspect this is because I have so many variables in my analysis and due to CV causing the creation of new models every bootstrap.
For the sake of a simple example I used mtcars and will look for the top 3 most common predictors due to only having 10 variables in this data set.
library(glmnet)
data("mtcars") # Base R Dataset
df <- mtcars
topvar <- list()
for (i in 1:100) {
# CV and Splitting
ind <- sample(nrow(df), nrow(df), replace = TRUE)
ind <- unique(ind)
train <- df[ind, ]
xtrain <- model.matrix(mpg~., train)[,-1]
ytrain <- df[ind, 1]
test <- df[-ind, ]
xtest <- model.matrix(mpg~., test)[,-1]
ytest <- df[-ind, 1]
# Create Model per Loop
model <- glmnet(xtrain, ytrain, alpha = 1, lambda = 0.2)
# Store Coeffecients per loop
coef_las <- coef(model, s = 0.2)[-1, ] # Remove intercept
# Store all nonzero Coefficients
topvar[[i]] <- coef_las[which(coef_las != 0)]
}
# Unlist
varimp <- unlist(topvar)
# Count all predictors
novar <- table(names(varimp))
# Find the mean of all variables
meanvar <- tapply(varimp, names(varimp), mean)
# Return top 3 repeated Coefs
repvar <- novar[order(novar, decreasing = TRUE)][1:3]
# Return mean of repeated Coefs
repvar.mean <- meanvar[names(repvar)]
repvar
Now if you were to rerun the code chunk above you would notice that the top 3 variables change and so if I had to rename these variables it would be difficult to do if they are not constant and changing every run. Any suggestions on how I could approach this?
You can use function set.seed() to ensure your sample will return the same sample each time. For example
set.seed(123)
When I add this to above code and then run twice, the following is returned both times:
wt carb hp
98 89 86
I'm trying to test the predictive power of a model by breaking the observations into 1/4th and 3/4th groups (test and train respectively), running a first-order regression with the independent variable train sample, using these coefficients to produce predicted values from the independent variable test sample, and then I would like to add new columns of these predicted values to the dependent variable test data for each iteration of the loop.
For context: TSIP500 is the full sample; iv is independent variable; dv is dependent variable, a max of 50 iterations is simply a test that isn't too large in quantity of iterations.
I was having trouble with the predict function so I did the equation manually. My code is below:
for(i in 1:50){
test_index <- sample(nrow(TSIP500iv), (1/4)*nrow(TSIP500iv), replace=FALSE)
train_500iv <- TSIP500[-test_index,"distance"]
test_500iv <- TSIP500[test_index,"distance"]
train_500dv <- TSIP500[-test_index,"percent_of_max"]
test_500dv <- TSIP500[test_index,"percent_of_max"]
reg_model <- lm(train_500dv~train_500iv)
int <- reg_model$coeff[1]
B1 <- reg_model$coeff[2]
predicted <- (int + B1*test_500iv)
predicted <- data.frame(predicted)
test_500dv <- data.frame(test_500dv)
test_500dv[,i] <- apply(predicted)
}
I've tried different approaches for the last line, but I always just get a singular column added. Any help would be tremendously appreciated.
for(i in 1:50){
test_index <- sample(nrow(TSIP500iv), (1/4)*nrow(TSIP500iv), replace=FALSE)
train_500iv <- TSIP500[-test_index,"distance"]
test_500iv <- TSIP500[test_index,"distance"]
train_500dv <- TSIP500[-test_index,"percent_of_max"]
test_500dv <- TSIP500[test_index,"percent_of_max"]
reg_model <- lm(train_500dv~train_500iv)
int <- reg_model$coeff[1]
B1 <- reg_model$coeff[2]
temp_results <- paste('pred',i,sep='_')
assign(temp_results, as.data.frame(int + B1*test_500iv))
test_500dv <- cbind(data.frame(test_500dv),temp_results)
}
I'm trying to implement my own linear regression likelihood ratio test.
The test is where you take the sum of squares of a reduced model and the sum of squares of a full model and compare it to the F statistic.
However, I am having some trouble implementing the function, especially when dealing with dummy variables.
This is the dataset I am working with and testing the function on.
Here is the code so far:
The function inputs are the setup matrix mat, the response matrix which has just one column, the indices (variables) being test, and the alpha value the test is at.
linear_regression_likelihood <- function(mat, response, indices, alpha) {
mat <- as.matrix(mat)
reduced <- mat[,c(1, indices)]
q <- 1 #set q = 1 just to test on data
p <- dim(mat)[2]
n <- dim(mat)[1]
f_stat <- qf(1-alpha, df1 = p-q, df2 = n-(p+1))
beta_hat_full <- qr.solve(t(mat)%*%mat)%*%t(mat)%*%response
y_hat_full <- mat%*%beta_hat_full
SSRes_full <- t(response - y_hat_full)%*%(response-y_hat_full)
beta_hat_red <- qr.solve(t(reduced)%*%reduced)%*%t(reduced)%*%response
y_hat_red <- reduced%*%beta_hat_red
SSRes_red <- t(response - y_hat_red)%*%(response-y_hat_red)
s_2 <- (t(response - mat%*%beta_hat_full)%*%(response - mat%*%beta_hat_full))/(n-p+1)
critical_value <- ((SSRes_red - SSRes_full)/(p-q))/s_2
print(critical_value)
if (critical_value > f_stat) {
return ("Reject H0")
}
else {
return ("Fail to Reject H0")
}
}
Here is the setup code, where I setup the matrix in the correct format. Data is the read in CSV file.
data <- data[, 2:5]
mat <- data[, 2:4]
response <- data[, 1]
library(ade4)
df <-data.frame(mat$x3)
dummy <- acm.disjonctif(df)
dummy
mat <- cbind(1, mat[1:2], dummy)
linear_regression_likelihood(mat, response, 2:3, 0.05)
This is the error I keep getting.
Error in solve.default(as.matrix(c)) : system is computationally singular: reciprocal condition number = 1.63035e-18
I know it has to do with taking the inverse of the matrix after it is multiplied, but the function is unable to do so. I thought it may be due to the dummy variables having too small of values, but I am not sure of any other way to include the dummy variables.
The test I am doing is to check whether the factor variable x3 has any affect on the response y. The actual answer which I verified using the built in functions states that we fail to reject the null hypothesis.
The error originates from line
beta_hat_full <- qr.solve(t(mat)%*%mat)%*%t(mat)%*%response
If you go through your function step-by-step you will see an error
Error in qr.solve(t(mat) %*% mat) : singular matrix 'a' in solve
The problem here is that your model matrix does not have full column rank, which translates to your regression coefficients not being unique. This is a result of the way you "dummyfied" x3. In order to ensure full rank, you need to remove one dummy column (or manually remove the intercept).
In the following example I remove the A column from dummy which means that resulting x3 coefficients measure the effect of a unit-change in B, C, and D against A.
# Read data
data <- read.csv("data_hw5.csv")
data <- data[, 2:5]
# Extract predictor and response data
mat <- data[, 2:4]
response <- data[, 1]
# Dummify categorical predictor x3
library(ade4)
df <-data.frame(mat$x3)
dummy <- acm.disjonctif(df)
dummy <- dummy[, -1] # Remove A to have A as baseline
mat <- cbind(1, mat[1:2], dummy)
# Apply linear_regression_likelihood
linear_regression_likelihood(mat, response, 2:3, 0.05);
# [,1]
#[1,] 8.291975
#[1] "Reject H0"
A note
The error could have been avoided if you had used base R's function model.matrix which ensures full rank when "dummyfying" categorical variables (model.matrix is also implicitly called in lm and glm to deal with categorical, i.e. factor variables).
Take a look at
mm <- model.matrix(y ~ x1 + x2 + x3, data = data)
which by default omits the first level of factor variable x3. mm is identical to mat after (correct) "dummification".
I want to check all the permutations and combinations of columns while selecting models in R. I have 8 columns in my data set and the below piece of code lets me check some of the models, but not all. Models like column 1+6, 1+2+5 will not be covered by this loop. Is there any better way to accomplish this?
best_model <- rep(0,3) #store the best model in this array
for(i in 1:8){
for(j in 1:8){
for(x in k){
diabetes_prediction <- knn(train = diabetes_training[, i:j], test = diabetes_test[, i:j], cl = diabetes_train_labels, k = x)
accuracy[x] <- 100 * sum(diabetes_test_labels == diabetes_prediction)/183
if( best_model[1] < accuracy[x] ){
best_model[1] = accuracy[x]
best_model[2] = i
best_model[3] = j
}
}
}
}
Well, this answer isn't complete, but maybe it'll get you started. You want to be able to subset by all possible subsets of columns. So instead of having i:j for some i and j, you want to be able to subset by c(1,6) or c(1,2,5), etc.
Using the sets package, you can for the power set (set of all subsets) of a set. That's the easy part. I'm new to R, so the hard part for me is understanding the difference between sets, lists, vectors, etc. I'm used to Mathematica, in which they're all the same.
library(sets)
my.set <- 1:8 # you want column indices from 1 to 8
my.power.set <- set_power(my.set) # this creates the set of all subsets of those indices
my.names <- c("a") #I don't know how to index into sets, so I created names (that are numbers, but of type characters)
for(i in 1:length(my.power.set)) {my.names[i] <- as.character(i)}
names(my.power.set) <- my.names
my.indices <- vector("list",length(my.power.set)-1)
for(i in 2:length(my.power.set)) {my.indices[i-1] <- as.vector(my.power.set[[my.names[i]]])} #this is the line I couldn't get to work
I wanted to create a list of lists called my.indices, so that my.indices[i] was a subset of {1,2,3,4,5,6,7,8} that could be used in place of where you have i:j. Then, your for loop would have to run from 1:length(my.indices).
But alas, I have been spoiled by Mathematica, and thus cannot decipher the incredibly complicated world of R data types.
Solved it, below is the code with explanatory comments:
# find out the best model for this data
number_of_columns_to_model <- ncol(diabetes_training)-1
best_model <- c()
best_model_accuracy = 0
for(i in 2:2^number_of_columns_to_model-1){
# ignoring the first case i.e. i=1, as it doesn't represent any model
# convert the value of i to binary, e.g. i=5 will give combination = 0 0 0 0 0 1 0 1
combination = as.binary(i, n=number_of_columns_to_model) # from the binaryLogic package
model <- c()
for(i in 1:length(combination)){
# choose which columns to consider depending on the combination
if(combination[i])
model <- c(model, i)
}
for(x in k){
# for the columns decides by model, find out the accuracies of model for k=1:27
diabetes_prediction <- knn(train = diabetes_training[, model, with = FALSE], test = diabetes_test[, model, with = FALSE], cl = diabetes_train_labels, k = x)
accuracy[x] <- 100 * sum(diabetes_test_labels == diabetes_prediction)/length(diabetes_test_labels)
if( best_model_accuracy < accuracy[x] ){
best_model_accuracy = accuracy[x]
best_model = model
print(model)
}
}
}
I trained with Pima.tr and tested with Pima.te. KNN Accuracy for pre-processed predictors was 78% and 80% without pre-processing (and this because of the large influence of some variables).
The 80% performance is at par with a Logistic Regression model. You don't need to preprocess variables in Logistic Regression.
RandomForest, and Logistic Regression provide a hint on which variables to drop, so you don't need to go and perform all possible combinations.
Another way is to look at a matrix Scatter plot
You get a sense that there is difference between type 0 and type 1 when it comes to npreg, glu, bmi, age
You also notice the highly skewed ped and age, and you notice that there may be an anomaly data point between skin and and and other variables (you may need to remove that observation before going further)
Skin Vs Type box plot shows that for type Yes, an extreme outlier exist (try removing it)
You also notice that most of the boxes for Yes type are higher than No type=> the variables may add prediction to the model (you can confirm this through a Wilcoxon Rank Sum Test)
The high correlation between Skin and bmi means that you can use one or the other or an interact of both.
Another approach to reducing the number of predictors is to use PCA
Suppose I have a data set. There are some categorical variables and some numerical variables. I want to estimate the parameters of a model e^{X'b} for every categories and others. I am trying to do it in R code. I need to do it by creating design matrix for categorical variables like age==2 and age==3, where considering age==1 as reference category. But this program is not running and giving errors. What is the problem?
sex <- c("F","M","F","M","F","M")
age <- c(1,3,2,3,1,2) # categorical variable with three age categories
age <- as.factor(age)
dat <- data.frame(sex,age)
myfun <- function(par, data){
xx <- data
func <- exp(par[1]*(xx$age==2)+par[2]*(xx$age==3)+par[3]*factor(xx$sex))
return(-func)
}
optim(myfun, par=c(0.1,0.4,0.7), data=dat)
Your function myfun returns a vector of length 6 (because you multiply with xx$sex). It needs to be of length 1. Also, the optim function takes the par first and the function as second parameter.
EDIT: You need to rewrite your function to return a single value. -exp(X'b) is a vector of length of your observations. Maybe this goes in your direction:
myfun1 <- function(par, data) {
xx <- matrix(c(as.numeric(dat$sex), age, rep(1, nrow(dat))), ncol=3)
sum(-exp(xx %*% par))
}
optim(c(0.1,0.4,0.7), myfun1, data=dat)
Note that it would be more efficient to pass xx to optim since the calculation of xx is independent of the iteration.