I am trying to run a simple naive bayes model (trying to redo what I have seen the datacamp course).
I am using the R naivebayes package.
The training dataset is where9am which looks like this:
My first problem is the following... when I have several predictions in a dataframe thursday9am...
... and I use the following code:
locmodel <- naive_bayes(location ~ daytype, data = where9am)
my_pred <- predict(locmodel, thursday9am)
I get a series of <NA> while it works well with the correct prediction if the thursday9am dataframe only contains a single observation.
The second problem is the following: when I use the following code to get the associated probabilities...
locmodel <- naive_bayes(location ~ daytype, data = where9am, type = c("class", "prob"))
predict(locmodel, thursday9am , type = "prob")
... even if I have only one observation in thursday9am, I get a series of <NaN>.
I am not sure what I am doing wrong.
Related
I'm trying to get a summary plot using fastshap explain function as in the code below.
p_function_G<- function(object, newdata)
caret::predict.train(object,
newdata =
newdata,
type = "prob")[,"AntiSocial"] # select G class
# Calculate the Shapley values
#
# boostFit: is a caret model using catboost algorithm
# trainset: is the dataset used for bulding the caret model.
# The dataset contains 4 categories W,G,R,GM
# corresponding to 4 diferent animal behaviors
library(caret)
shap_values_G <- fastshap::explain(xgb_fit,
X = game_train,
pred_wrapper =
p_function_G,
nsim = 50,
newdata= game_train[which(game_test=="AntiSocial"),])
)
However I'm getting error
Error in 'stop_vctrs()':
can't combine latitude and gender <factor<919a3>>
What's the way out?
I see that you are adapting code from Julia Silge's Predict ratings for board games Tutorial. The original code used SHAPforxgboost for generating SHAP values, but you're using the fastshap package.
Because Shapley explanations are only recently starting to gain traction, there aren't very many standard data formats. fastshap does not like tidyverse tibbles, it only takes matrices or matrix-likes.
The error occurs because, by default, fastshap attempts to convert the tibble to a matrix. But this fails, because matrices can only have one type (f.x. either double or factor, not both).
I also ran into a similar issue and found that you can solve this by passing the X parameter as a data.frame. I don't have access to your full code but you could you try replacing the shap_values_G code-block as so:
shap_values_G <- fastshap::explain(xgb_fit,
X = game_train,
pred_wrapper =
p_function_G,
nsim = 50,
newdata= as.data.frame(game_train[which(game_test=="AntiSocial"),]))
)
Wrap newdata with as.data.frame. This converts the tibble to a dataframe and so shouldn't upset fastshap.
I'm trying to build a histogram of residual values, however the first step I'm taking to do that is to run a linear model. R will not recognize the column name as an object.
The first three lines of code run fine. The second two give me an error saying the object area_ha cant be found, however, it is one of eight column titles in my data. Any advice on creating a linear model and a histogram to graph residuals would also be very helpful.
dat<-read.csv("/Users/sara/Desktop/birdsinforest.csv", header=TRUE)
linearmodel=lm(abundance ~ area_ha, data = dat)
summary(linearmodel)
area_ha$abundance_predicted = predict(linearmodel)
area_ha$residual = area_ha$abundance - area_ha$abundance_predicted
This is the error I get after running the last two lines of code:
Error in area_ha$abundance_predicted = predict(linearmodel) :
object 'area_ha' not found
Your code:
dat<-read.csv("/Users/sara/Desktop/birdsinforest.csv", header=TRUE)
linearmodel=lm(abundance ~ area_ha, data = dat)
summary(linearmodel)
area_ha$abundance_predicted = predict(linearmodel)
area_ha$residual = area_ha$abundance - area_ha$abundance_predicted
In the above code, area_ha seems like a variable (column name) and not data.frame since you're using it to fit a linear model. You should try the last two lines of code as below:
dat$abundance_predicted <- predict(linearmodel)
dat$residual <- dat$abundance - dat$abundance_predicted
Problem
I am using R 3.3.3 on Windows 10 (x64 bit). I get the following prediction output from the glmmPQL prediction function as follows:
library(MASS)
library(nlme)
library(dplyr)
model<-glmmPQL(a ~ b + c + d, data = trainingDataSet, family = binomial, random = list( ~ 1 | e), correlation = corAR1())
The prediction values are given as follows:
p <- predict(model, newdata=testingDataSet, type="response",level=0) (1.0)
The output it gives is as follows:
I then try to measure the performance of this output using the following code:
pr <- prediction(p, testingDataSet$a)(1.1)
It gives us the following error as follows:
Error in prediction(p, testingDataSet$a) :
Format of predictions is invalid. (1.2)
I have successfully been able to use the prediction method in R using other functions (glm,svm,nn) when the data looks something like as follows:
model<-glm(a ~ b + c + e, family = binomial(link = 'logit'), data = trainingDataSet)
p <- predict(model, newdata=testingDataSet, type="response") (1.3)
Attempts
I believe the fix to the above problem is to get it into the format shown below (1.3). I have tried the following things using R and have been failing.
I have tried casting p in 1.0 using as.numeric() and as.list() and other things. I want to get look like the p R object in 1.3. In other words, I believe the format is reason why things not working for me?
No matter what mutate or casting I try, I can't seem to get it into the form in 1.3 and image shown as desired. Especially with the index as columns features.
I'm coming up empty handed on stackoverflow and the R help files. When I use the command class(p) both tell me they are numeric.
Question
Give the above, can someone tell me how I can use R to get the output from glmmPQL in a format that the prediction function can use as shown above please?
In other words, how can I make sure the output in 1.0 can made to match the output in 1.3 in R? My attempts have failed and I would deeply appreciate someone more skilled in R to point out where I am failing?
If you use as.numeric(p) then you'll get the values you want - then the only difference is that the GLM output has names. You can add these in with something like:
p <- as.numeric(p)
names(p) <- 1:length(p)
If this doesn't work, you can use str(p) to examine the structure of the object in more depth.
I'm fairly new to using the caret library and it's causing me some problems. Any
help/advice would be appreciated. My situations are as follows:
I'm trying to run a general linear model on some data and, when I run it
through the confusionMatrix, I get 'the data and reference factors must have
the same number of levels'. I know what this error means (I've run into it before), but I've double and triple checked my data manipulation and it all looks correct (I'm using the right variables in the right places), so I'm not sure why the two values in the confusionMatrix are disagreeing. I've run almost the exact same code for a different variable and it works fine.
I went through every variable and everything was balanced until I got to the
confusionMatrix predict. I discovered this by doing the following:
a <- table(testing2$hold1yes0no)
a[1]+a[2]
1543
b <- table(predict(modelFit,trainTR2))
dim(b)
[1] 1538
Those two values shouldn't disagree. Where are the missing 5 rows?
My code is below:
set.seed(2382)
inTrain2 <- createDataPartition(y=HOLD$hold1yes0no, p = 0.6, list = FALSE)
training2 <- HOLD[inTrain2,]
testing2 <- HOLD[-inTrain2,]
preProc2 <- preProcess(training2[-c(1,2,3,4,5,6,7,8,9)], method="BoxCox")
trainPC2 <- predict(preProc2, training2[-c(1,2,3,4,5,6,7,8,9)])
trainTR2 <- predict(preProc2, testing2[-c(1,2,3,4,5,6,7,8,9)])
modelFit <- train(training2$hold1yes0no ~ ., method ="glm", data = trainPC2)
confusionMatrix(testing2$hold1yes0no, predict(modelFit,trainTR2))
I'm not sure as I don't know your data structure, but I wonder if this is due to the way you set up your modelFit, using the formula method. In this case, you are specifying y = training2$hold1yes0no and x = everything else. Perhaps you should try:
modelFit <- train(trainPC2, training2$hold1yes0no, method="glm")
Which specifies y = training2$hold1yes0no and x = trainPC2.
I am doing just a regular logistic regression using the caret package in R. I have a binomial response variable coded 1 or 0 that is called a SALES_FLAG and 140 numeric response variables that I used dummyVars function in R to transform to dummy variables.
data <- dummyVars(~., data = data_2, fullRank=TRUE,sep="_",levelsOnly = FALSE )
dummies<-(predict(data, data_2))
model_data<- as.data.frame(dummies)
This gives me a data frame to work with. All of the variables are numeric. Next I split into training and testing:
trainIndex <- createDataPartition(model_data$SALE_FLAG, p = .80,list = FALSE)
train <- model_data[ trainIndex,]
test <- model_data[-trainIndex,]
Time to train my model using the train function:
model <- train(SALE_FLAG~. data=train,method = "glm")
Everything runs nice and I get a model. But when I run the predict function it does not give me what I need:
predict(model, newdata =test,type="prob")
and I get an ERROR:
Error in dimnames(out)[[2]] <- modelFit$obsLevels :
length of 'dimnames' [2] not equal to array extent
On the other hand when I replace "prob" with "raw" for type inside of the predict function I get prediction but I need probabilities so I can code them into binary variable given my threshold.
Not sure why this happens. I did the same thing without using the caret package and it worked how it should:
model2 <- glm(SALE_FLAG ~ ., family = binomial(logit), data = train)
predict(model2, newdata =test, type="response")
I spend some time looking at this but not sure what is going on and it seems very weird to me. I have tried many variations of the train function meaning I didn't use the formula and used X and Y. I used method = 'bayesglm' as well to check and id gave me the same error. I hope someone can help me out. I don't need to use it since the train function to get what I need but caret package is a good package with lots of tools and I would like to be able to figure this out.
Show us str(train) and str(test). I suspect the outcome variable is numeric, which makes train think that you are doing regression. That should also be apparent from printing model. Make it a factor if you want to do classification.
Max