Permutations and combinations of all the columns in R - r

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

Related

How to capture the most important variables in Bootstrapped models in R?

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

How to use lapply with get.confusion_matrix() in R?

I am performing a PLS-DA analysis in R using the mixOmics package. I have one binary Y variable (presence or absence of wetland) and 21 continuous predictor variables (X) with values ranging from 1 to 100.
I have made the model with the data_training dataset and want to predict new outcomes with the data_validation dataset. These datasets have exactly the same structure.
My code looks like:
library(mixOmics)
model.plsda<-plsda(X,Y, ncomp = 10)
myPredictions <- predict(model.plsda, newdata = data_validation[,-1], dist = "max.dist")
I want to predict the outcome based on 10, 9, 8, ... to 2 principal components. By using the get.confusion_matrix function, I want to estimate the error rate for every number of principal components.
prediction <- myPredictions$class$max.dist[,10] #prediction based on 10 components
confusion.mat = get.confusion_matrix(truth = data_validatie[,1], predicted = prediction)
get.BER(confusion.mat)
I can do this seperately for 10 times, but I want do that a little faster. Therefore I was thinking of making a list with the results of prediction for every number of components...
library(BBmisc)
prediction_test <- myPredictions$class$max.dist
predictions_components <- convertColsToList(prediction_test, name.list = T, name.vector = T, factors.as.char = T)
...and then using lapply with the get.confusion_matrix and get.BER function. But then I don't know how to do that. I have searched on the internet, but I can't find a solution that works. How can I do this?
Many thanks for your help!
Without reproducible there is no way to test this but you need to convert the code you want to run each time into a function. Something like this:
confmat <- function(x) {
prediction <- myPredictions$class$max.dist[,x] #prediction based on 10 components
confusion.mat = get.confusion_matrix(truth = data_validatie[,1], predicted = prediction)
get.BER(confusion.mat)
}
Now lapply:
results <- lapply(10:2, confmat)
That will return a list with the get.BER results for each number of PCs so results[[1]] will be the results for 10 PCs. You will not get values for prediction or confusionmat unless they are included in the results returned by get.BER. If you want all of that, you need to replace the last line to the function with return(list(prediction, confusionmat, get.BER(confusion.mat)). This will produce a list of the lists so that results[[1]][[1]] will be the results of prediction for 10 PCs and results[[1]][[2]] and results[[1]][[3]] will be confusionmat and get.BER(confusion.mat) respectively.

Output multiple vectors from for loop in R

As someone relatively new to R I'm having an issue with creating a for loop.
I have a very large data set with 9000 observations and 25 categorical variables, which I've transformed into binary data and preformed hierarchical clustering. Now I want to try K-Modes clustering to produce an Elbow Plot using the "within-cluster simple-matching distance for each cluster", which is outputted from kmodes$withindiff. I can sum this for each of the k in 1:8 clusters to get the Elbow Plot.
library(klaR)
for(k in 1:8)
{
WCSM[k] <- sum(kmodes(data,k,iter.max=100)$withindiff)
}
plot(1:8,WCSM,type="b", xlab="Number of Clusters",ylab="Within-Cluster
Simple-Matching Distance Summed", main="K-modes Elbow Plot")
My issue is that I want further output from k-modes. For each k in 1:8 I would like to get the vector of integers indicating the cluster to which each object is allocated to given by kmodes$cluster. I need to create a for loop that loops through each k in 1:8 and saves each of the outputs into 8 separate vectors. But I don't know how to do such a for loop. I could just run the 8 lines of code separately but they each take 15mins to run with iter.max=10 so increasing this to iter.max=100 will need to be left running overnight so a loop would be useful.
cl.kmodes2=kmodes(data, 2,iter.max=100)
cl.kmodes3=kmodes(data, 3,iter.max=100)
cl.kmodes4=kmodes(data, 4,iter.max=100)
cl.kmodes5=kmodes(data, 5,iter.max=100)
cl.kmodes6=kmodes(data, 6,iter.max=100)
cl.kmodes7=kmodes(data, 7,iter.max=100)
cl.kmodes8=kmodes(data, 8,iter.max=100)
Ultimately I want to compare the results from the hierarchical binary clustering to the k-modes clustering by getting the Adjusted Rand Index. For example, cutting the tree at k=4 for the hierarchical cluster and comparing this to a 4 cluster solution from k-modes:
dist.binary = dist(data, method="binary")
cl.binary = hclust(dist.binary, method="complete")
hcl.4 = cutree(cl.binary, k = 4)
tab = table(hcl.4, cl.kmodes4$cluster)
library(e1071)
classAgreement(tab)
I agree with Imo, using a list is the best solution.
If you don't want to do that, you could also use assign() to create a new vector in every iteration:
library(klaR)
for(k in 1:8) {
assign(paste("cl.kmodes", k, sep = ""), kmodes(data, k, iter.max = 100))
}
The best method is to put the output from your clusters into a named list:
library(klaR)
myClusterList <- list()
for(k in 1:8) {
myClusterList[[paste0("k.", i)]] <- kmodes(data, i,iter.max=100)
}
You can then pull out the any of the contents easily:
sum(myClusterList[["k.1"]]$withindiff)
or
sum(myClusterList[[1]]$withindiff)
You can also save the list to use in future R sessions, see ?save.

Using a for loop for performing several regressions

I am currently performing a style analysis using the following method: http://www.r-bloggers.com/style-analysis/ . It is a constrained regression of one asset on a number of benchmarks, over a rolling 36 month window.
My problem is that I need to perform this regression for a fairly large number of assets and doing it one by one would take a huge amount of time. To be more precise: Is there a way to tell R to regress columns 1-100 one by one on colums 101-116. Of course this also means printing 100 different plots, one for each asset. I am new to R and have been stuck for several days now.
I hope it doesn't matter that the following excerpt isn't reproducible, since the code works as originally intended.
# Style Regression over Window, constrained
#--------------------------------------------------------------------------
# setup
load.packages('quadprog')
style.weights[] = NA
style.r.squared[] = NA
# Setup constraints
# 0 <= x.i <= 1
constraints = new.constraints(n, lb = 0, ub = 1)
# SUM x.i = 1
constraints = add.constraints(rep(1, n), 1, type = '=', constraints)
# main loop
for( i in window.len:ndates ) {
window.index = (i - window.len + 1) : i
fit = lm.constraint( hist.returns[window.index, -1], hist.returns[window.index, 1], constraints )
style.weights[i,] = fit$coefficients
style.r.squared[i,] = fit$r.squared
}
# plot
aa.style.summary.plot('Style Constrained', style.weights, style.r.squared, window.len)
Thank you very much for any tips!
"Is there a way to tell R to regress columns 1-100 one by one on colums 101-116."
Yes! You can use a for loop, but you there's also a whole family of 'apply' functions which are appropriate. Here's a generalized solution with a random / toy dataset and using lm(), but you can sub in whatever regression function you want
# data frame of 116 cols of 20 rows
set.seed(123)
dat <- as.data.frame(matrix(rnorm(116*20), ncol=116))
# with a for loop
models <- list() # empty list to store models
for (i in 1:100) {
models[[i]] <-
lm(formula=x~., data=data.frame(x=dat[, i], dat[, 101:116]))
}
# with lapply
models2 <-
lapply(1:100,
function(i) lm(formula=x~.,
data=data.frame(x=dat[, i], dat[, 101:116])))
# compare. they give the same results!
all.equal(models, models2)
# to access a single model, use [[#]]
models2[[1]]

How to find significant correlations in a large dataset

I'm using R.
My dataset has about 40 different Variables/Vektors and each has about 80 entries. I'm trying to find significant correlations, that means I want to pick one variable and let R calculate all the correlations of that variable to the other 39 variables.
I tried to do this by using a linear modell with one explaining variable that means: Y=a*X+b.
Then the lm() command gives me an estimator for a and p-value of that estimator for a. I would then go on and use one of the other variables I have for X and try again until I find a p-value thats really small.
I'm sure this is a common problem, is there some sort of package or function that can try all these possibilities (Brute force),show them and then maybe even sorts them by p-value?
You can use the function rcorr from the package Hmisc.
Using the same demo data from Richie:
m <- 40
n <- 80
the_data <- as.data.frame(replicate(m, runif(n), simplify = FALSE))
colnames(the_data) <- c("y", paste0("x", seq_len(m - 1)))
Then:
library(Hmisc)
correlations <- rcorr(as.matrix(the_data))
To access the p-values:
correlations$P
To visualize you can use the package corrgram
library(corrgram)
corrgram(the_data)
Which will produce:
In order to print a list of the significant correlations (p < 0.05), you can use the following.
Using the same demo data from #Richie:
m <- 40
n <- 80
the_data <- as.data.frame(replicate(m, runif(n), simplify = FALSE))
colnames(the_data) <- c("y", paste0("x", seq_len(m - 1)))
Install Hmisc
install.packages("Hmisc")
Import library and find the correlations (#Carlos)
library(Hmisc)
correlations <- rcorr(as.matrix(the_data))
Loop over the values printing the significant correlations
for (i in 1:m){
for (j in 1:m){
if ( !is.na(correlations$P[i,j])){
if ( correlations$P[i,j] < 0.05 ) {
print(paste(rownames(correlations$P)[i], "-" , colnames(correlations$P)[j], ": ", correlations$P[i,j]))
}
}
}
}
Warning
You should not use this for drawing any serious conclusion; only useful for some exploratory analysis and formulate hypothesis. If you run enough tests, you increase the probability of finding some significant p-values by random chance: https://www.xkcd.com/882/. There are statistical methods that are more suitable for this and that do do some adjustments to compensate for running multiple tests, e.g. https://en.wikipedia.org/wiki/Bonferroni_correction.
Here's some sample data for reproducibility.
m <- 40
n <- 80
the_data <- as.data.frame(replicate(m, runif(n), simplify = FALSE))
colnames(the_data) <- c("y", paste0("x", seq_len(m - 1)))
You can calculate the correlation between two columns using cor. This code loops over all columns except the first one (which contains our response), and calculates the correlation between that column and the first column.
correlations <- vapply(
the_data[, -1],
function(x)
{
cor(the_data[, 1], x)
},
numeric(1)
)
You can then find the column with the largest magnitude of correlation with y using:
correlations[which.max(abs(correlations))]
So knowing which variables are correlated which which other variables can be interesting, but please don't draw any big conclusions from this knowledge. You need to have a proper think about what you are trying to understand, and which techniques you need to use. The folks over at Cross Validated can help.
If you are trying to predict y using only one variable than you have to take the one that is mainly correlated with y.
To do this just use the command which.max(abs(cor(x,y))). If you want to use more than one variable in your model then you have to consider something like the lasso estimator
One option is to run a correlation matrix:
cor_result=cor(data)
write.csv(cor_result, file="cor_result.csv")
This correlates all the variables in the file against each other and outputs a matrix.

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