Hey dear stackoverflowers,
I would like to perform (multiple) regression analysis on a large customer data set, trying to predict amount spent after initial purchase based on various independent variables, observed during the first purchase.
In this data set, for the same combination of input variable values (say gender=male, age=30, income=40k, first_purchase_value = 99,90), I can have multiple observartions with varying y values (i.e. multiple customers share the same independent variable attributes, but behave differently according to their observed y values).
Is this a problem for regression analysis, i.e. do I have to condense these observations by e.g. averaging? I am getting negative R2 values, that's why I'm asking (I know that a linear model might also just be the wrong assumption here) ...
Thank you for helping me. I tried using the search function, but was unable to find similar topics (probably because the question is silly?).
Cheers!
Edit: This is the code I'm using:
spl <- sample.split(data$spent, SplitRatio = 0.75)
data_train <- subset(data, spl == TRUE)
data_test <- subset(data, spl == FALSE)
model_lm_spent <- lm(spent ~ ., data = data_train)
summary(model_lm_spent)
model_lm_predictions_spent <- predict(model_lm_spent, newdata = data_test)
SSE_spent = sum((data_test$spent - model_lm_predictions_spent)^2)
SST_spent = sum((data_test$spent - mean(data$spent))^2)
1 - SSE_spent/SST_spent
Related
I am trying to perform a simple RDA using the vegan package to test the effects of depth, basin and sector on genetic population structure using the following data frame.
datafile.
The "ALL" variable is the genetic population assignment (structure).
In case the link to my data doesn't work well, I'll paste a snippet of my data frame here.
I read in the data this way:
RDAmorph_Oct6 <- read.csv("RDAmorph_Oct6.csv")
My problems are two-fold:
1) I can't seem to get my genetic variable to read correctly. I have tried three things to fix this.
gen=rda(ALL ~ Depth + Basin + Sector, data=RDAmorph_Oct6, na.action="na.exclude")
Error in eval(specdata, environment(formula), enclos = globalenv()) :
object 'ALL' not found
In addition: There were 12 warnings (use warnings() to see them)
so, I tried things like:
> gen=rda("ALL ~ Depth + Basin + Sector", data=RDAmorph_Oct6, na.action="na.exclude")
Error in colMeans(x, na.rm = TRUE) : 'x' must be numeric
so I specified numeric
> RDAmorph_Oct6$ALL = as.numeric(RDAmorph_Oct6$ALL)
> gen=rda("ALL ~ Depth + Basin + Sector", data=RDAmorph_Oct6, na.action="na.exclude")
Error in colMeans(x, na.rm = TRUE) : 'x' must be numeric
I am really baffled. I've also tried specifying each variable with dataset$variable, but this doesn't work either.
The strange thing is, I can get an rda to work if I look the effects of the environmental variables on a different, composite, variable
MC = RDAmorph_Oct6[,5:6]
H_morph_var=rda(MC ~ Depth + Basin + Sector, data=RDAmorph_Oct6, na.action="na.exclude")
Note that I did try to just extract the ALL column for the genetic rda above. This didn't work either.
Regardless, this leads to my second problem.
When I try to plot the rda I get a super weird plot. Note the five dots in three places. I have no idea where these come from.
I will have to graph the genetic rda, and I figure I'll come up with the same issue, so I thought I'd ask now.
I've been though several tutorials and tried many iterations of each issue. What I have provided here is I think the best summary. If anyone can give me some clues, I would much appreciate it.
The documentation, ?rda, says that the left-hand side of the formula specifying your model needs to be a data matrix. You can't pass it the name of a variable in the data object as the left-hand side (or at least if this was ever anticipated, doing so exposes bugs in how we parse the formula which is what leads to further errors).
What you want is a data frame containing a variable ALL for the left-hand side of the formula.
This works:
library('vegan')
df <- read.csv('~/Downloads/RDAmorph_Oct6.csv')
ALL <- df[, 'ALL', drop = FALSE]
Notice the drop = FALSE, which stops R from dropping the empty dimension (i.e. converting the single column data frame to a vector.
Then your original call works:
ord <- rda(ALL ~ Basin + Depth + Sector, data = df, na.action = 'na.exclude')
The problem is that rda expects a separate df for the first part of the formula (ALL in your code), and does not use the one in the data = argument.
As mentioned above, you can create a new df with the variable needed for analysis, but here's a oneline solution that should also work:
gen <- rda(RDAmorph_Oct6$ALL ~ Depth + Basin + Sector, data = RDAmorph_Oct6, na.action = na.exclude)
This is partly similar to Gavin simpson's answer. There is also a problem with the categorical vectors in your data frame. You can either use library(data.table) and the rowid function to set the categorical variables to unique integers. Most preferably, not use them. I also wanted to set the ID vector as site names, but I am too lazy now.
library(data.table)
RDAmorph_Oct6 <- read.csv("C:/........../RDAmorph_Oct6.csv")
#remove NAs before. I like looking at my dataframes before I analyze them.
RDAmorph_Oct6 <- na.omit(RDAmorph_Oct6)
#I removed one duplicate
RDAmorph_Oct6 <- RDAmorph_Oct6[!duplicated(RDAmorph_Oct6$ID),]
#Create vector with only ALL
ALL <- RDAmorph_Oct6$ALL
#Create data frame with only numeric vectors and remove ALL
dfn <- RDAmorph_Oct6[,-c(1,4,11,12)]
#Select all categorical vectors.
dfc <- RDAmorph_Oct6[,c(1,11,12)]
#Give the categorical vectors unique integers doesn't do this for ID (Why?).
dfc2 <- as.data.frame(apply(dfc, 2, function(x) rowid(x)))
#Bind back with numeric data frame
dfnc <- cbind.data.frame(dfn, dfc2)
#Select only what you need
df <- dfnc[c("Depth", "Basin", "Sector")]
#The rest you know
rda.out <- rda(ALL ~ ., data=df, scale=T)
plot(rda.out, scaling = 2, xlim=c(-3,2), ylim=c(-1,1))
#Also plot correlations
plot(cbind.data.frame(ALL, df))
Sector and depth have the highest variation. Almost logical, since there are only three vectors used. The assignment of integers to the categorical vector has probably no meaning at all. The function assigns from top to bottom unique integers to the following unique character string. I am also not really sure which question you want to answer. Based on this you can organize the data frame.
I have a simple Random Forest model I have created and tested in R. For now I have excluded an internal company ID from my training/testing data frames. Is there a way in R that I could include this column in my data and have the training/execution of my model ignore the field?
I obviously would not want the model to try and incorporate it as a variable, but upon an export of the data with a column added of the predicted outcome, I would need that internal id to tie back in other customer data so I know what customers have been categorized as
I am just using the out of the box random forest function from the randomForest library
#divide data into training and test sets
set.seed(3)
id<-sample(2,nrow(Churn_Model_Data_v2),prob=c(0.7,0.3),replace = TRUE)
churn_train<-Churn_Model_Data_v2[id==1,]
churn_test<-Churn_Model_Data_v2[id==2,]
#changes Churn data 1/2 to a factor for model
Churn_Model_Data_v2$`Churn` <- as.factor(Churn_Model_Data_v2$`Churn`)
churn_train$`Churn` <- as.factor(churn_train$`Churn`)
#churn_test$`Churn` <- as.factor(churn_test$`Churn`)
bestmtry <- tuneRF(churn_train,churn_train$`Churn`, stepFactor = 1.2,
improve =0.01, trace=T, plot=T )
#creates model based on training data, views model
churn_forest <- randomForest(`Churn`~. , data= churn_train )
churn_forest
#shows us what variables are most important
importance(churn_forest)
varImpPlot(churn_forest)
#predicts churn diagnosis on test data
predict_churn <- predict(churn_forest, newdata = churn_test, type="class")
predict_churn
A simple example of excluding a particular column or set of columns is as follows
library(MASS)
temp<-petrol
randomForest(No ~ .,data = temp[, !(colnames(temp) %in% c("SG"))]) # One Way
randomForest(No ~ .-SG,data = temp) #Another way with similar result
This method of exclusion is commonly valid across other fuctions/alogorithms in R too.
I am trying to implement a Naive Bayes model in R based on known information:
Age group, e.g. "18-24" and "25-34", etc.
Gender, "male" and "female"
Region, "London" and "Wales", etc.
Income, "£10,000 - £15,000", etc.
Job, "Full Time" and "Part Time", etc.
I am experiencing errors when implementing. My code is as per below:
library(readxl)
iphone <- read_excel("~/Documents/iPhone_1k.xlsx")
View(iphone)
summary(iphone)
iphone
library(caTools)
library(e1071)
set.seed(101)
sample = sample.split(iphone$Gender, SplitRatio = .7)
train = subset(iphone, sample == TRUE)
test = subset(iphone, sample == FALSE)
nB_model <- naiveBayes(Gender ~ Region + Retailer, data = train)
pred <- predict(nB_model, test, type="raw")
In the above scenario, I have an excel file called iPhone_1k (1,000 entries relating to people who have visited a website to buy an iPhone). Each row is a person visiting the website and the above demographics are known.
I have been trying to make the model work and have resorted to following the below link that uses only two variables (I would like to use a minimum of 4 but introduce more, if possible):
https://rpubs.com/dvorakt/144238
I want to be able to use these demographics to predict which retailer they will go to (also known for each instance in the iPhone_1k file). There are only 3 options. Can you please advise how to complete this?
P.S. Below is a screenshot of a simplified version of the data I have used to keep it simple in R. Once I get some code to work, I'll expand the number of variables and entries.
You are setting the problem incorrectly. It should be:
naiveBayes(Retailer ~ Gender + Region + AgeGroup, data = train)
or in short
naiveBayes(Retailer ~ ., data = train)
Also you might need to convert the columns into factors if they are characters. You can do it for all columns, right after reading from excel, by
iphone[] <- lapply(iphone, factor)
Note that if you add numeric variables in the future, you should not apply this step on them.
I'm trying to use the random forest model to predict Gender based on Height, Weight and Number of siblings. I've gotten the data from a much larger data set that contains dozens of variables, but I've cleaned it into this "clean" data.frame with omitted NA values and only the 4 variables I care about, the last column being Gender.
I've tried fiddling with the code and searching everywhere but I can't find a concrete fix.
Here's the code:
ind <- sample(nrow(clean),0.8*nrow(clean))
train <- clean[ind,]
test <- clean[-ind,]
rf <- randomForest(Gender ~ ., data = train[,1:4], ntree = 20)
pred <- predict(rf, newdata = test[,-c(length(test))])
cm <- table(test$Gender, pred)
cm
and here's the output:
Error in `[.default`(table(observed = y, predicted = out.class), levels(y), : subscript out of bounds
Traceback:
1. randomForest(Gender ~ ., data = train[, 1:4], ntree = 20)
2. randomForest.formula(Gender ~ ., data = train[, 1:4], ntree = 20)
3. randomForest.default(m, y, ...)
4. table(observed = y, predicted = out.class)[levels(y), levels(y)]
5. `[.table`(table(observed = y, predicted = out.class), levels(y),
. levels(y))
6. NextMethod()
The problem is likely that you have some kind of a variable level in your test data that was not reflected in your training data. So when it goes to assign the outcome, it has no basis to do so.
It is impossible to say for sure without sample data, but it is the most likely scenario. Try setting a seed set.seed=3 and then change the seed number set.seed=28 and so on, a few times to see if you end up finding a combination where you do not get the error.
Compare the conflicted data frame with the un-conflicted one to see what is missing.
EDIT:
Also, try running str(train) and str(test) to be sure the fields have remained the same. You can share that if you like by editing your post.
If any of the columns are factors with levels missing (meaning it has 10 levels but only 8 are represented in the train with 9 or 10 in the test) it might be a problem. They should be balanced if you are trying to create a predictor for all possible outcomes.
If nothing else works, you can set a seed and remove predictors one at a time until it runs correctly, then look to see how the train and test sets are different in that removed column.
In the past few days I have developed multiple PLS models in R for spectral data (wavebands as explanatory variables) and various vegetation parameters (as individual response variables). In total, the dataset comprises of 56. The first 28 (training set) have been used for model calibration, now all I want to do is to predict the response values for the remaining 28 observations in the tesset. For some reason, however, R keeps on the returning the fitted values of the calibration set for a given number of components rather than predictions for the independent test set. Here is what the model looks like in short.
# first simulate some data
set.seed(123)
bands=101
data <- data.frame(matrix(runif(56*bands),ncol=bands))
colnames(data) <- paste0(1:bands)
data$height <- rpois(56,10)
data$fbm <- rpois(56,10)
data$nitrogen <- rpois(56,10)
data$carbon <- rpois(56,10)
data$chl <- rpois(56,10)
data$ID <- 1:56
data <- as.data.frame(data)
caldata <- data[1:28,] # define model training set
valdata <- data[29:56,] # define model testing set
# define explanatory variables (x)
spectra <- caldata[,1:101]
# build PLS model using training data only
library(pls)
refl.pls <- plsr(height ~ spectra, data = caldata, ncomp = 10, validation =
"LOO", jackknife = TRUE)
It was then identified that a model comprising of 3 components yielded the best performance without over-fitting. Hence, the following command was used to predict the values of the 28 observations in the testing set using the above calibrated PLS model with 3 components:
predict(refl.pls, ncomp = 3, newdata = valdata)
Sensible as the output may seem, I soon discovered that all this piece of code generates are the fitted values of the PLS model for the calibration/training data, rather than predictions. I discovered this because the below code, in which newdata = is omitted, yields identical results.
predict(refl.pls, ncomp = 3)
Surely something must be going wrong, although I cannot seem to find out what specifically is. Is there someone out there who can, and is willing to help me move in the right direction?
I think the problem is with the nature of the input data. Looking at ?plsr and str(yarn) that goes with the example, plsr requires a very specific data frame that I find tricky to work with. The input data frame should have a matrix as one of its elements (in your case, the spectral data). I think the following works correctly (note I changed the size of the training set so that it wasn't half the original data, for troubleshooting):
library("pls")
set.seed(123)
bands=101
spectra = matrix(runif(56*bands),ncol=bands)
DF <- data.frame(spectra = I(spectra),
height = rpois(56,10),
fbm = rpois(56,10),
nitrogen = rpois(56,10),
carbon = rpois(56,10),
chl = rpois(56,10),
ID = 1:56)
class(DF$spectra) <- "matrix" # just to be certain, it was "AsIs"
str(DF)
DF$train <- rep(FALSE, 56)
DF$train[1:20] <- TRUE
refl.pls <- plsr(height ~ spectra, data = DF, ncomp = 10, validation =
"LOO", jackknife = TRUE, subset = train)
res <- predict(refl.pls, ncomp = 3, newdata = DF[!DF$train,])
Note that I got the spectral data into the data frame as a matrix by protecting it with I which equates to AsIs. There might be a more standard way to do this, but it works. As I said, to me a matrix inside of a data frame is not completely intuitive or easy to grok.
As to why your version didn't work quite right, I think the best explanation is that everything needs to be in the one data frame you pass to plsr for the data sources to be completely unambiguous.