I'm currently using RStudio for doing text mining on Support tickets, clustering them by their description (freetext). For this, I compare kmeans to EM algorithm. I prepared the data with the tm package, and now I try do apply clustering algorithms to the data matrix.
With the kmeans() function, I can use following Code snippet to Output the 5 most frequent Terms in text Clusters (kmeans21):
> for (i in 1:num_cluster) {
cat(paste("cluster ", i, ": ", sep = ""))
s <- sort(kmeans21$centers[i, ], decreasing = T)
cat(names(s)[1:5], "\n")
}
Until now, I couldnt find a function to do the same within the mclust package. My data has the following Format:
> bic21 <- MclustBIC(m1, G=21)
> emmodel21 <- summary(bic21, data = m1)
With the command
> emmodel21$classification
I can see the Cluster for each supportticket, but is there also the possibility to Output the most frequent Terms like in the first Code block for kmeans?
I think you can try
summary(mod1, parameters = TRUE)
Just tried the same example in the link
library(mclust)
data(diabetes)
X <- diabetes[,-1]
BIC <- mclustBIC(X)
mod1 <- Mclust(X, x = BIC)
summary(mod1, parameters = TRUE)
Slightly altering the first example in the vignette:
data(diabetes)
X <- diabetes[,-1]
mod <- mclust(X)
means <- mod$parameters$means
The means object is now a matrix containing the means of the clusters.
Related
I am trying to input matrix data into the brm() function to run a signal regression. brm is from the brms package, which provides an interface to fit Bayesian models using Stan. Signal regression is when you model one covariate using another within the bigger model, and you use the by parameter like this: model <- brm(response ~ s(matrix1, by = matrix2) + ..., data = Data). The problem is, I cannot input my matrices using the 'data' parameter because it only allows one data.frame object to be inputted.
Here are my code and the errors I obtained from trying to get around that constraint...
First off, my reproducible code leading up to the model-building:
library(brms)
#100 rows, 4 columns. Each cell contains a number between 1 and 10
Data <- data.frame(runif(100,1,10),runif(100,1,10),runif(100,1,10),runif(100,1,10))
#Assign names to the columns
names(Data) <- c("d0_10","d0_100","d0_1000","d0_10000")
Data$Density <- as.matrix(Data)%*%c(-1,10,5,1)
#the coefficients we are modelling
d <- c(-1,10,5,1)
#Made a matrix with 4 columns with values 10, 100, 1000, 10000 which are evaluation points. Rows are repeats of the same column numbers
Bins <- 10^matrix(rep(1:4,times = dim(Data)[1]),ncol = 4,byrow =T)
Bins
As mentioned above, since 'data' only allows one data.frame object to be inputted, I've tried other ways of inputting my matrix data. These methods include:
1) making the matrix within the brm() function using as.matrix()
signalregression.brms <- brm(Density ~ s(Bins,by=as.matrix(Data[,c(c("d0_10","d0_100","d0_1000","d0_10000"))])),data = Data)
#Error in is(sexpr, "try-error") :
argument "sexpr" is missing, with no default
2) making the matrix outside the formula, storing it in a variable, then calling that variable inside the brm() function
Donuts <- as.matrix(Data[,c(c("d0_10","d0_100","d0_1000","d0_10000"))])
signalregression.brms <- brm(Density ~ s(Bins,by=Donuts),data = Data)
#Error: The following variables can neither be found in 'data' nor in 'data2':
'Bins', 'Donuts'
3) inputting a list containing the matrix using the 'data2' parameter
signalregression.brms <- brm(Density ~ s(Bins,by=donuts),data = Data,data2=list(Bins = 10^matrix(rep(1:4,times = dim(Data)[1]),ncol = 4,byrow =T),donuts=as.matrix(Data[,c(c("d0_10","d0_100","d0_1000","d0_10000"))])))
#Error in names(dat) <- object$term :
'names' attribute [1] must be the same length as the vector [0]
None of the above worked; each had their own errors and it was difficult troubleshooting them because I couldn't find answers or examples online that were of a similar nature in the context of brms.
I was able to use the above techniques just fine for gam(), in the mgcv package - you don't have to define a data.frame using 'data', you can call on variables defined outside of the gam() formula, and you can make matrices inside the gam() function itself. See below:
library(mgcv)
signalregression2 <- gam(Data$Density ~ s(Bins,by = as.matrix(Data[,c("d0_10","d0_100","d0_1000","d0_10000")]),k=3))
#Works!
It seems like brms is less flexible... :(
My question: does anyone have any suggestions on how to make my brm() function run?
Thank you very much!
My understanding of signal regression is limited enough that I'm not convinced this is correct, but I think it's at least a step in the right direction. The problem seems to be that brm() expects everything in its formula to be a column in data. So we can get the model to compile by ensuring all the things we want are present in data:
library(tidyverse)
signalregression.brms = brm(Density ~
s(cbind(d0_10_bin, d0_100_bin, d0_1000_bin, d0_10000_bin),
by = cbind(d0_10, d0_100, d0_1000, d0_10000),
k = 3),
data = Data %>%
mutate(d0_10_bin = 10,
d0_100_bin = 100,
d0_1000_bin = 1000,
d0_10000_bin = 10000))
Writing out each column by hand is a little annoying; I'm sure there are more general solutions.
For reference, here are my installed package versions:
map_chr(unname(unlist(pacman::p_depends(brms)[c("Depends", "Imports")])), ~ paste(., ": ", pacman::p_version(.), sep = ""))
[1] "Rcpp: 1.0.6" "methods: 4.0.3" "rstan: 2.21.2" "ggplot2: 3.3.3"
[5] "loo: 2.4.1" "Matrix: 1.2.18" "mgcv: 1.8.33" "rstantools: 2.1.1"
[9] "bayesplot: 1.8.0" "shinystan: 2.5.0" "projpred: 2.0.2" "bridgesampling: 1.1.2"
[13] "glue: 1.4.2" "future: 1.21.0" "matrixStats: 0.58.0" "nleqslv: 3.3.2"
[17] "nlme: 3.1.149" "coda: 0.19.4" "abind: 1.4.5" "stats: 4.0.3"
[21] "utils: 4.0.3" "parallel: 4.0.3" "grDevices: 4.0.3" "backports: 1.2.1"
I have a dataset of BBC articles with two columns: 'category' and 'text'. I need to construct a Naive Bayes algorithm that predicts the category (i.e. business, entertainment) of an article based on type.
I'm attempting this with Quanteda and have the following code:
library(quanteda)
bbc_data <- read.csv('bbc_articles_labels_all.csv')
text <- textfile('bbc_articles_labels_all.csv', textField='text')
bbc_corpus <- corpus(text)
bbc_dfm <- dfm(bbc_corpus, ignoredFeatures = stopwords("english"), stem=TRUE)
# 80/20 split for training and test data
trainclass <- factor(c(bbc_data$category[1:1780], rep(NA, 445)))
testclass <- factor(c(bbc_data$category[1781:2225]))
bbcNb <- textmodel_NB(bbc_dfm, trainclass)
bbc_pred <- predict(bbcNb, testclass)
It seems to work smoothly until predict(), which gives:
Error in newdata %*% log.lik :
requires numeric/complex matrix/vector arguments
Can anyone provide insight on how to resolve this? I'm still getting the hang of text analysis and quanteda. Thank you!
Here is a link to the dataset.
As a stylistic note, you don't need to separately load the labels/classes/categories, the corpus will have them as one of its docvars:
library("quanteda")
text <- readtext::readtext('bbc_articles_labels_all.csv', text_field='text')
bbc_corpus <- corpus(text)
bbc_dfm <- dfm(bbc_corpus, remove = stopwords("english"), stem = TRUE)
all_classes <- docvars(bbc_corpus)$category
trainclass <- factor(replace(all_classes, 1780:length(all_classes), NA))
bbcNb <- textmodel_nb(bbc_dfm, trainclass)
You don't even need to specify a second argument to predict. If you don't, it will use the whole original dfm:
bbc_pred <- predict(bbcNb)
Finally, you may want to assess the predictive accuracy. This will give you a summary of the model's performance on the test set:
library(caret)
confusionMatrix(
bbc_pred$docs$predicted[1781:2225],
all_classes[1781:2225]
)
However, as #ken-benoit noted, there is a bug in quanteda which prevents prediction from working with more than two classes. Until that's fixed, you could binarize the classes with something like:
docvars(bbc_corpus)$category <- factor(
ifelse(docvars(bbc_corpus)$category=='sport', 'sport', 'other')
)
(note that this must be done before you extract all_classes from bbc_corpus above).
I am trying to reproduce the error which I get with my dataset using a general dataset. Please correct me if I am missing something.
After fitting a Classification tree using library(party), I am trying to get the split conditions of the tree on each node. I managed to write a code, which i believed was working fine, until I found a bug. Could anyone help me to solve it?
My code:
require(party)
iris$Petal.Width <- as.factor(iris$Petal.Width)#imp to convert to factorial
(ct <- ctree(Species ~ ., data = iris))
plot(ct)
#print(ct)
a <- ct #convert it to s4 object
t <- a#tree
#recursive function to traverse the tree and get the splitting conditions
recurse_tree <- function(tree,ret_list=list(),sub_list=list()){
if(!tree$terminal){
sub_list$assign <-list(tree$psplit$splitpoint,tree$psplit$variableName,class(tree$psplit))
names(sub_list)[which(names(sub_list)=="assign")] <- paste("node",tree$nodeID,sep="")
ret_list <- recurse_tree(tree$left, ret_list, sub_list)
ret_list <- recurse_tree(tree$right, ret_list, sub_list)
}
if(tree$terminal){
ret_list$assign <- c(sub_list, tree$prediction)
names(ret_list)[which(names(ret_list)=="assign")] <- paste("node",tree$nodeID,sep="")
return(ret_list)
}
return(ret_list)
}
result <- recurse_tree(t) #call to the functions
Now, the result gives me the list of of all nodes and split conditions and predictions (I assumed). But, when I check the split conditions for
expected output on Node5: {1.1, 1.2, 1.6, 1.7 } # from printing the tree print(ct), I get this
output I get on Node5 from my function: {"1" , "1.3" ,"1.4" ,"1.5" } which is basically the split condition of Node6, which is wrong. How did I get this?
z <- result[2] #I know node5 is second in the list
z <- unlist(z,recursive = F,use.names = T) #unlist
levels(z[[3]][[1]]) [which((z[[3]][[1]])==0)] #to find levels of corresponding values
What I doubt, my function(recurse_tree) is always giving me the split conditions of the right terminal node and not left node. Any help will be appreciated.
I have a list of lm models objects with possible repeated, so I'd like to find a way of checking if some of these lm objects are equal, if so them delete it. In words, I want to "deduplicate" my list.
I'd appreciate very much any help.
An example of the problem:
## Creates outcome and predictors
outcome <- c(names(mtcars)[1:3])
predictors <- c(names(mtcars)[4:11])
dataset <- mtcars
## Creates model list
model_list <- lapply(seq_along((predictors)), function(n) {
left_hand_side <- outcome[1]
right_hand_side <- apply(X = combn(predictors, n), MARGIN = 2, paste, collapse = " + ")
paste(left_hand_side, right_hand_side, sep = " ~ ")
})
## Convert model list into a verctor
model_vector <- unlist(model_list)
## Fit linear models to all itens from the vector of models
list_of_fit <- lapply(model_vector, function(x) {
formula <- as.formula(x)
fit <- step(lm(formula, data = dataset))
fit
})
# Exclude possible missing
list_of_fit <- Filter(Negate(function(x) is.null(unlist(x))), list_of_fit)
# These models are the same in my list
lm253 <- list_of_fit[[253]];lm253
lm254 <- list_of_fit[[254]];lm254
lm255 <- list_of_fit[[255]];lm255
I want to exclude duplicated entries in list_of_fit.
It seems wasteful to fit so many models and then throw away most of them. Your object names make your code hard to read for me, but it seems your models can be distinguished based on their formula. Maybe this helps:
lista_de_ajustes[!duplicated(vapply(lista_de_ajustes,
function(m) deparse(m$call),
FUN.VALUE = "a"))]
I made a simple correction in you code Roland, so it worked for me.
I changed from deparse(m$call) to deparse(formula(m)), due this I'm able to compare the complete formulas.
lista_de_ajustes[!duplicated(vapply(lista_de_ajustes, function(m) deparse(formula(m)), FUN.VALUE = "a"))]
Thank you very much!
I want to compare between two different classification methods, namely ctree and C5.0 in the libraries partyand c50 respectively, the comparison is to test their sensitivity to the initial start points. The test should be carried 30 times for each time the number of wrong classified items are calculated and stored in a vector then by using t-test I hope to see if they are really different or not.
library("foreign"); # for read.arff
library("party") # for ctree
library("C50") # for C5.0
trainTestSplit <- function(data, trainPercentage){
newData <- list();
all <- nrow(data);
splitPoint <- floor(all * trainPercentage);
newData$train <- data[1:splitPoint, ];
newData$test <- data[splitPoint:all, ];
return (newData);
}
ctreeErrorCount <- function(st,ss){
set.seed(ss);
model <- ctree(Class ~ ., data=st$train);
class <- st$test$Class;
st$test$Class <- NULL;
pre = predict(model, newdata=st$test, type="response");
errors <- length(which(class != pre)); # counting number of miss classified items
return(errors);
}
C50ErrorCount <- function(st,ss){
model <- C5.0(Class ~ ., data=st$train, seed=ss);
class <- st$test$Class;
pre = predict(model, newdata=st$test, type="class");
errors <- length(which(class != pre)); # counting number of miss classified items
return(errors);
}
compare <- function(n = 30){
data <- read.arff(file.choose());
set.seed(100);
errors = list(ctree = c(), c50 = c());
seeds <- floor(abs(rnorm(n) * 10000));
for(i in 1:n){
splitData <- trainTestSplit(data, 0.66);
errors$ctree[i] <- ctreeErrorCount(splitData, seeds[i]);
errors$c50[i] <- C50ErrorCount(splitData, seeds[i]);
}
cat("\n\n");
cat("============= ctree Vs C5.0 =================\n");
cat(paste(errors$ctree, " ", errors$c50, "\n"))
tt <- t.test(errors$ctree, errors$c50);
print(tt);
}
The program shown is supposedly doing the job of comparison, but because of the number of errors does not change in the vectors then the t.test function produces an error. I used iris inside R (but changing class to Class) and Winchester breast cancer data which can be downloaded here to test it but any data can be used as long as it has Class attribute
But I get in to the problem that the result of both methods remain constant and not changes while I am changing the random seed, theoretically ,as described in their documentation,both of the functions use random seeds, ctree uses set.seed(x) while C5.0 uses an argument called seed to set seed, unfortunatly I can not find the effect.
Could you please tell me how to control initials of these functions
ctrees does only depend on a random seed in the case where you configure it to use a random selection of input variables (ie that mtry > 0 within ctree_control). See http://cran.r-project.org/web/packages/party/party.pdf (p. 11)
In regards to C5.0-trees the seed is used this way:
ctrl = C5.0Control(sample=0.5, seed=ss);
model <- C5.0(Class ~ ., data=st$train, control = ctrl);
Notice that the seed is used to select a sample of the data, not within the algoritm itself. See http://cran.r-project.org/web/packages/C50/C50.pdf (p. 5)