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.
Related
For a paper I'm writing I have subsetted a larger dataset into 3 groups, because I thought the strength of correlations between 2 variables in those groups would differ (they did). I want to see if subsetting my data into random groupings would also significantly affect the strength of correlations (i.e., whether what I'm seeing is just an effect of subsetting, or if those groupings are actually significant).
To this end, I am trying to generate n new data frames by randomly sampling 150 rows from an existing dataset, and then want to calculate correlation coefficients for two variables in those n new data frames, saving the correlation coefficient and significance in a new file.
But, HOW?
I can do it manually, e.g., with dplyr, something like
newdata <- sample_n(Random_sample_data, 150)
output <- cor.test(newdata$x, newdata$y, method="kendall")
I'd obviously like to not type this out 1000 or 100000 times, and have been trying things with loops and lapply (see below) but they've not worked (undoubtedly due to something really obvious that I'm missing!).
Here I have tried to assign each row to a different group, with 10 groups in total, and then to do correlations between x and y by those groups:
Random_sample_data<-select(Range_corrected, x, y)
cat <- sample(1:10, 1229, replace=TRUE)
Random_sample_cats<-cbind(Random_sample_data,cat)
correlation <- function(c) {
c <- cor.test(x,y, method="kendall")
return(c)
}
b<- daply(Random_sample_cats, .(cat), correlation)
Error message:
Error in cor.test(x, y, method = "kendall") :
object 'x' not found
Once you have the code for what you want to do once, you can put it in replicate to do it n times. Here's a reproducible example on built-in data
result = replicate(n = 10, expr = {
newdata <- sample_n(mtcars, 10)
output <- cor.test(newdata$wt, newdata$qsec, method="kendall")
})
replicate will save the result of the last line of what you did (output <- ...) for each replication. It will attempt to simplify the result, in this case cor.test returns a list of length 8, so replicate will simplify the results to a matrix with 8 rows and 10 columns (1 column per replication).
You may want to clean up the results a little bit so that, e.g., you only save the p-value. Here, we store only the p-value, so the result is a vector with one p-value per replication, not a matrix:
result = replicate(n = 10, expr = {
newdata <- sample_n(mtcars, 10)
cor.test(newdata$wt, newdata$qsec, method="kendall")$p.value
})
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
I ran a rfe Model with around 400 variables and got the result that the optimal model uses 40 variables. However, plotting the standard deviations of the error based on cross validation shows that the 40 variable model performs only slightly better than a model with only 10 variables. That's why I'd like to go for the model with 10 variables. I would like to use the 10 variables which perform best for a ten- variable-model and train the model again.
How can I get the 10 variables which lead to the model performance shown in the rfe object?
Since I use rerank=TRUE, I cannot just pick the 10 highest ranked variables from varImp(rfeModel$fit) right? (Would this work if I was not using "rerank" ?)
I'm also struggling with the differences between the output from varImp(rfeModel$fit), varImp(rfeModel), pickVars(rfeModel$variables,40).
What is the right way to get the best performing variables with regard to the size of interest?
The following example can be used:
data(BloodBrain)
x <- scale(bbbDescr[,-nearZeroVar(bbbDescr)])
x <- x[, -findCorrelation(cor(x), .8)]
x <- as.data.frame(x)
set.seed(1)
rfProfile <- rfe(x, logBBB,
sizes = c(2, 5, 10, 20),
method="nnet",
maxit=10,
rfeControl(functions = caretFuncs,
returnResamp="all",
method="cv",
rerank=TRUE))
print(rfProfile), varImp(rfProfile$fit), varImp(rfProfile), pickVars(rfProfile$variables, rfProfile$optsize)
The simplest thing to do is to use the update function:
new_profile <- update(rfProfile, x = x, y = logBBB, size = 10)
Internally, it uses this code:
selectedVars <- rfProfile$variables
bestVar <- rfProfile$control$functions$selectVar(selectedVars, 10)
Max
I want to measure the features importance with the cforest function from the party library.
My output variable has something like 2000 samples in class 0 and 100 samples in class 1.
I think a good way to avoid bias due to class unbalance is to train each tree of the forest using a subsample such that the number of elements of class 1 is the same of the number of element in class 0.
Is there anyway to do that? I am thinking to an option like n_samples = c(20, 20)
EDIT:
An example of code
> iris.cf <- cforest(Species ~ ., data = iris,
+ control = cforest_unbiased(mtry = 2)) #<--- Here I would like to train the forest using a balanced subsample of the data
> varimp(object = iris.cf)
Sepal.Length Sepal.Width Petal.Length Petal.Width
0.048981818 0.002254545 0.305818182 0.271163636
>
EDIT:
Maybe my question is not clear enough.
Random forest is a set of decision trees. In general the decision trees are constructed using only a random subsample of the data. I would like that the used subsample has the same numbers of element in the class 1 and in the class 0.
EDIT:
The function that I am looking for is for sure available in the randomForest package
sampsize
Size(s) of sample to draw. For classification, if sampsize is a vector of the length the number of strata, then sampling is stratified by strata, and the elements of sampsize indicate the numbers to be drawn from the strata.
I need the same for the party package. Is there any way to get it?
I will assume you know what you want to accomplish, but don't know enough R to do that.
Not sure if the function provides balancing of data as an argument, but you can do it manually. Below is the code I quickly threw together. More elegant solution might exist.
# just in case
myData <- iris
# replicate everything *10* times. Replicate is just a "loop 10 times".
replicate(10,
{
# split dataset by class and add separate classes to list
splitList <- split(myData, myData$Species)
# sample *20* random rows from each matrix in a list
sampledList <- lapply(splitList, function(dat) { dat[sample(20),] })
# combine sampled rows to a data.frame
sampledData <- do.call(rbind, sampledList)
# your code below
res.cf <- cforest(Species ~ ., data = sampledData,
control = cforest_unbiased(mtry = 2)
)
varimp(object = res.cf)
}
)
Hope you can take it from here.
users
I am trying to develop a local model (PLSR) which is predicting a query sample by a model built on the 10 most similar samples using the code below (not the full model yet, just a part of it). I got stuck when trying to predict the query sample (second to last line). The model is actually predicting something, ("prd") but not the query sample!
Here is my code:
require("pls")
set.seed(10000) # generate some sample data
mat <- replicate(100, rnorm(100))
y <- as.matrix(mat[,1], drop=F)
x <- mat[,2:100]
eD <- dist(x, method="euclidean") # create a distance matrix
eDm <- as.matrix(eD)
Looping over all 100 samples and extracting their 10 most similar samples for subsequent model building and prediction of query sample:
for (i in 1:nrow(eDm)) {
kni <- head(order(eDm[,i]),11)[-1] # add 10 most similar samples to kni
pls1 <- plsr(y[kni,] ~ x[kni,], ncomp=5, validation="CV") # run plsr on sel. samples
prd <- predict(pls1, ncomp=5, newdata=x[[i]]) # predict query sample ==> I suspect there is something wrong with this expression: newdata=x[[i]]
}
I can't figure out how to address the query sample properly - many thanks i.a. for any help!
Best regards,
Chega
You are going to run into all sorts of pain building models with formulae like that. Also the x[[i]] isn't doing what you think it is - you need to supply a data frame usually to these modelling functions. In this case a matrix seems fine too.
I get all your code working OK if I use:
prd <- predict(pls1, ncomp=5, newdata=x[i, ,drop = FALSE])
giving
> predict(pls1, ncomp=5, newdata=x[i,,drop = FALSE])
, , 5 comps
y[kni, ]
[1,] 0.6409897
What you were seeing with your code are the fitted values for the training data.
> fitted(pls1)[, , 5, drop = FALSE]
, , 5 comps
y[kni, ]
1 0.1443274
2 0.2706769
3 1.1407780
4 -0.2345429
5 -1.0468221
6 2.1353091
7 0.8267103
8 3.3242296
9 -0.5016016
10 0.6781804
This is convention in R when you either don't supply newdata or the object you are supplying makes no sense and doesn't contain the covariates required to generate predictions.
I would have fitted the model as follows:
pls1 <- plsr(y ~ x, ncomp=5, validation="CV", subset = kni)
where I use the subset argument for its intended purpose; to select the rows of the input data to fit the model with. You get nicer output from the models; the labels use y instead of y[kni, ] etc, plus this general convention will serve you well in other modelling tools, where R will expect newdata to be a data frame with names exactly the same as those mentioned in the model formula. In your case, with your code, that would mean creating a data frame with names like x[kni, ] which are not easy to do, for good reason!