Related question - 1
I have a dataset like so:
> head(training_data)
year month channelGrouping visitStartTime visitNumber timeSinceLastVisit browser
1 2016 October Social 1477775021 1 0 Chrome
2 2016 September Social 1473037945 1 0 Safari
3 2017 July Organic Search 1500305542 1 0 Chrome
4 2017 July Organic Search 1500322111 2 16569 Chrome
5 2016 August Social 1471890172 1 0 Safari
6 2017 May Direct 1495146428 1 0 Chrome
operatingSystem isMobile continent subContinent country source medium
1 Windows 0 Americas South America Brazil youtube.com referral
2 Macintosh 0 Americas Northern America United States youtube.com referral
3 Windows 0 Americas Northern America Canada google organic
4 Windows 0 Americas Northern America Canada google organic
5 Macintosh 0 Africa Eastern Africa Zambia youtube.com referral
6 Android 1 Americas Northern America United States (direct)
isTrueDirect hits pageviews positiveTransaction
1 0 1 1 No
2 0 1 1 No
3 0 5 5 No
4 1 3 3 No
5 0 1 1 No
6 1 6 6 No
> str(training_data)
'data.frame': 1000 obs. of 18 variables:
$ year : int 2016 2016 2017 2017 2016 2017 2016 2017 2017 2016 ...
$ month : Factor w/ 12 levels "January","February",..: 10 9 7 7 8 5 10 3 3 12 ...
$ channelGrouping : chr "Social" "Social" "Organic Search" "Organic Search" ...
$ visitStartTime : int 1477775021 1473037945 1500305542 1500322111 1471890172 1495146428 1476003570 1488556031 1490323225 1480696262 ...
$ visitNumber : int 1 1 1 2 1 1 1 1 1 1 ...
$ timeSinceLastVisit : int 0 0 0 16569 0 0 0 0 0 0 ...
$ browser : chr "Chrome" "Safari" "Chrome" "Chrome" ...
$ operatingSystem : chr "Windows" "Macintosh" "Windows" "Windows" ...
$ isMobile : int 0 0 0 0 0 1 0 1 0 0 ...
$ continent : Factor w/ 5 levels "Africa","Americas",..: 2 2 2 2 1 2 3 3 2 4 ...
$ subContinent : chr "South America" "Northern America" "Northern America" "Northern America" ...
$ country : chr "Brazil" "United States" "Canada" "Canada" ...
$ source : chr "youtube.com" "youtube.com" "google" "google" ...
$ medium : chr "referral" "referral" "organic" "organic" ...
$ isTrueDirect : int 0 0 0 1 0 1 0 0 0 0 ...
$ hits : int 1 1 5 3 1 6 1 1 2 1 ...
$ pageviews : int 1 1 5 3 1 6 1 1 2 1 ...
$ positiveTransaction: Factor w/ 2 levels "No","Yes": 1 1 1 1 1 1 1 1 1 1 …
I then define my custom RMSLE function using Metrics package:
rmsleMetric <- function(data, lev = NULL, model = NULL){
out <- Metrics::rmsle(data$obs, data$pred)
names(out) <- c("rmsle")
return (out)
}
Then, I define the trainControl:
tc <- trainControl(method = "repeatedcv",
number = 5,
repeats = 5,
summaryFunction = rmsleMetric,
classProbs = TRUE)
My grid search:
tg <- expand.grid(alpha = 0, lambda = seq(0, 1, by = 0.1))
Finally, my model:
penalizedLogit_ridge <- train(positiveTransaction ~ .,
data = training_data,
metric="rmsle",
method = "glmnet",
family = "binomial",
trControl = tc,
tuneGrid = tg
)
When I try to run the command above, I get an error:
Something is wrong; all the rmsle metric values are missing:
rmsle
Min. : NA
1st Qu.: NA
Median : NA
Mean :NaN
3rd Qu.: NA
Max. : NA
NA's :11
Error: Stopping
In addition: There were 50 or more warnings (use warnings() to see the first 50)
Looking at warnings, I find:
1: In Ops.factor(1, actual) : ‘+’ not meaningful for factors
2: In Ops.factor(1, predicted) : ‘+’ not meaningful for factors
repeated 25 times
Since the same thing works fine if I change the metric to AUC using prSummary as my summary function, I don't believe that there are any issues with my data.
So, I believe that my function is wrong but I don't know how to figure out why it is wrong.
Any help is highly appreciated.
Your custom metric is not defined properly. If you use classProbs = TRUE and savePredictions = "final" with trainControl you will realize that there are two columns named according to your target classes which hold the predicted probabilities while the data$pred column holds the predicted class which can not be used to calculate the desired metric.
A proper way to define the function would be to get the possible levels and use them to extract the probabilities for one of the classes:
rmsleMetric <- function(data, lev = NULL, model = NULL){
lvls <- levels(data$obs)
out <- Metrics::rmsle(ifelse(data$obs == lev[2], 0, 1),
data[, lvls[1]])
names(out) <- c("rmsle")
return (out)
}
does it work:
library(caret)
library(mlbench)
data(Sonar)
tc <- trainControl(method = "repeatedcv",
number = 2,
repeats = 2,
summaryFunction = rmsleMetric,
classProbs = TRUE,
savePredictions = "final")
tg <- expand.grid(alpha = 0, lambda = seq(0, 1, by = 0.1))
penalizedLogit_ridge <- train(Class ~ .,
data = Sonar,
metric="rmsle",
method = "glmnet",
family = "binomial",
trControl = tc,
tuneGrid = tg)
#output
glmnet
208 samples
60 predictor
2 classes: 'M', 'R'
No pre-processing
Resampling: Cross-Validated (2 fold, repeated 2 times)
Summary of sample sizes: 105, 103, 104, 104
Resampling results across tuning parameters:
lambda rmsle
0.0 0.2835407
0.1 0.2753197
0.2 0.2768288
0.3 0.2797847
0.4 0.2827953
0.5 0.2856088
0.6 0.2881894
0.7 0.2905501
0.8 0.2927171
0.9 0.2947169
1.0 0.2965505
Tuning parameter 'alpha' was held constant at a value of 0
rmsle was used to select the optimal model using the largest value.
The final values used for the model were alpha = 0 and lambda = 1.
You can inspect caret::twoClassSummary - it is defined quite similarly.
Related
I am trying to build a simple Naive Bayes classifer for mushroom data. I want to use all of the variables as categorical predictors to predict if a mushroom is edible.
I am using caret package.
Here is my code in full:
##################################################################################
# Prepare R and R Studio environment
##################################################################################
# Clear the R studio console
cat("\014")
# Remove objects from environment
rm(list = ls())
# Install and load packages if necessary
if (!require(tidyverse)) {
install.packages("tidyverse")
library(tidyverse)
}
if (!require(caret)) {
install.packages("caret")
library(caret)
}
if (!require(klaR)) {
install.packages("klaR")
library(klaR)
}
#################################
mushrooms <- read.csv("agaricus-lepiota.data", stringsAsFactors = TRUE, header = FALSE)
na.omit(mushrooms)
names(mushrooms) <- c("edibility", "capShape", "capSurface", "cap-color", "bruises", "odor", "gill-attachment", "gill-spacing", "gill-size", "gill-color", "stalk-shape", "stalk-root", "stalk-surface-above-ring", "stalk-surface-below-ring", "stalk-color-above-ring", "stalk-color-below-ring", "veil-type", "veil-color", "ring-number", "ring-type", "spore-print-color", "population", "habitat")
# convert bruises to a logical variable
mushrooms$bruises <- mushrooms$bruises == 't'
set.seed(1234)
split <- createDataPartition(mushrooms$edibility, p = 0.8, list = FALSE)
train <- mushrooms[split, ]
test <- mushrooms[-split, ]
predictors <- names(train)[2:20] #Create response and predictor data
x <- train[,predictors] #predictors
y <- train$edibility #response
train_control <- trainControl(method = "cv", number = 1) # Set up 1 fold cross validation
edibility_mod1 <- train( #train the model
x = x,
y = y,
method = "nb",
trControl = train_control
)
When executing the train() function I get the following output:
Something is wrong; all the Accuracy metric values are missing:
Accuracy Kappa
Min. : NA Min. : NA
1st Qu.: NA 1st Qu.: NA
Median : NA Median : NA
Mean :NaN Mean :NaN
3rd Qu.: NA 3rd Qu.: NA
Max. : NA Max. : NA
NA's :2 NA's :2
Error: Stopping
In addition: Warning messages:
1: predictions failed for Fold1: usekernel= TRUE, fL=0, adjust=1 Error in predict.NaiveBayes(modelFit, newdata) :
Not all variable names used in object found in newdata
2: model fit failed for Fold1: usekernel=FALSE, fL=0, adjust=1 Error in x[, 2] : subscript out of bounds
3: In nominalTrainWorkflow(x = x, y = y, wts = weights, info = trainInfo, :
There were missing values in resampled performance measures.
x and y after script run:
> str(x)
'data.frame': 6500 obs. of 19 variables:
$ capShape : Factor w/ 6 levels "b","c","f","k",..: 6 6 1 6 6 6 1 1 6 1 ...
$ capSurface : Factor w/ 4 levels "f","g","s","y": 3 3 3 4 3 4 3 4 4 3 ...
$ cap-color : Factor w/ 10 levels "b","c","e","g",..: 5 10 9 9 4 10 9 9 9 10 ...
$ bruises : logi TRUE TRUE TRUE TRUE FALSE TRUE ...
$ odor : Factor w/ 9 levels "a","c","f","l",..: 7 1 4 7 6 1 1 4 7 1 ...
$ gill-attachment : Factor w/ 2 levels "a","f": 2 2 2 2 2 2 2 2 2 2 ...
$ gill-spacing : Factor w/ 2 levels "c","w": 1 1 1 1 2 1 1 1 1 1 ...
$ gill-size : Factor w/ 2 levels "b","n": 2 1 1 2 1 1 1 1 2 1 ...
$ gill-color : Factor w/ 12 levels "b","e","g","h",..: 5 5 6 6 5 6 3 6 8 3 ...
$ stalk-shape : Factor w/ 2 levels "e","t": 1 1 1 1 2 1 1 1 1 1 ...
$ stalk-root : Factor w/ 5 levels "?","b","c","e",..: 4 3 3 4 4 3 3 3 4 3 ...
$ stalk-surface-above-ring: Factor w/ 4 levels "f","k","s","y": 3 3 3 3 3 3 3 3 3 3 ...
$ stalk-surface-below-ring: Factor w/ 4 levels "f","k","s","y": 3 3 3 3 3 3 3 3 3 3 ...
$ stalk-color-above-ring : Factor w/ 9 levels "b","c","e","g",..: 8 8 8 8 8 8 8 8 8 8 ...
$ stalk-color-below-ring : Factor w/ 9 levels "b","c","e","g",..: 8 8 8 8 8 8 8 8 8 8 ...
$ veil-type : Factor w/ 1 level "p": 1 1 1 1 1 1 1 1 1 1 ...
$ veil-color : Factor w/ 4 levels "n","o","w","y": 3 3 3 3 3 3 3 3 3 3 ...
$ ring-number : Factor w/ 3 levels "n","o","t": 2 2 2 2 2 2 2 2 2 2 ...
$ ring-type : Factor w/ 5 levels "e","f","l","n",..: 5 5 5 5 1 5 5 5 5 5 ...
> str(y)
Factor w/ 2 levels "e","p": 2 1 1 2 1 1 1 1 2 1 ...
My environment is:
> R.version
_
platform x86_64-apple-darwin17.0
arch x86_64
os darwin17.0
system x86_64, darwin17.0
status
major 4
minor 0.3
year 2020
month 10
day 10
svn rev 79318
language R
version.string R version 4.0.3 (2020-10-10)
nickname Bunny-Wunnies Freak Out
> RStudio.Version()
$citation
To cite RStudio in publications use:
RStudio Team (2020). RStudio: Integrated Development Environment for R. RStudio, PBC, Boston, MA URL http://www.rstudio.com/.
A BibTeX entry for LaTeX users is
#Manual{,
title = {RStudio: Integrated Development Environment for R},
author = {{RStudio Team}},
organization = {RStudio, PBC},
address = {Boston, MA},
year = {2020},
url = {http://www.rstudio.com/},
}
$mode
[1] "desktop"
$version
[1] ‘1.3.1093’
$release_name
[1] "Apricot Nasturtium"
What you are trying to do is a bit tricky, most naive bayes implementation or at least the one you are using (from kLAR which is derived from e1071) uses a normal distribution. You can see under the details of naiveBayes help page from e1071:
The standard naive Bayes classifier (at least this implementation)
assumes independence of the predictor variables, and Gaussian
distribution (given the target class) of metric predictors. For
attributes with missing values, the corresponding table entries are
omitted for prediction.
And your predictors are categorical so this might be problematic. You can try to set kernel=TRUE and adjust=1 to force it towards normal, and avoid kernel=FALSE which will throw the error.
Before that we remove columns with only 1 level and sort out the column names, also in this case it's easier to use the formula and avoid the making dummy variables :
df = train
levels(df[["veil-type"]])
[1] "p"
df[["veil-type"]]=NULL
colnames(df) = gsub("-","_",colnames(df))
Grid = expand.grid(usekernel=TRUE,adjust=1,fL=c(0.2,0.5,0.8))
mod1 <- train(edibility~.,data=df,
method = "nb", trControl = trainControl(method="cv",number=5),
tuneGrid=Grid
)
mod1
Naive Bayes
6500 samples
21 predictor
2 classes: 'e', 'p'
No pre-processing
Resampling: Cross-Validated (5 fold)
Summary of sample sizes: 5200, 5200, 5200, 5200, 5200
Resampling results across tuning parameters:
fL Accuracy Kappa
0.2 0.9243077 0.8478624
0.5 0.9243077 0.8478624
0.8 0.9243077 0.8478624
Tuning parameter 'usekernel' was held constant at a value of TRUE
Tuning parameter 'adjust' was held constant at a value of 1
Accuracy was used to select the optimal model using the largest value.
The final values used for the model were fL = 0.2, usekernel = TRUE and
adjust = 1.
i'm trying to train a model with the caret Package (Random Forest), after running the "train" code, i get: Error: Can't find column RH_train in .data. Then, i tried converting the dependent variable (Rendimiento) to a factor but i get: Error: At least one of the class levels is not a valid R variable name; This will cause errors when class probabilities are generated because the variables names will be converted to X0, X0.5, X0.6, X0.65, X0.7, X0.75, X0.79, X0.8, X0.81, X0.82, X0.83, X0.85, X0.86, X0.87, X0.88, X0.9, X1 . Please use factor levels that can be used as valid R variable names (see ?make.names for help).
library(rpart)
library(rpart.plot)
library(RWekajars)
library(randomForest)
library(party)
library(caret)
library(e1071)
library(dplyr)
####Cargar base de datos####
setwd("C:/Users/Frankenstein/Downloads")
RH <- read_excel("RH.xlsx")
RH$`Año Ingreso`=NULL
RH$`Mes ingreso`=NULL
RH$`Status empleado para Gestión t`=NULL
RH$`Horario trabajo`=NULL
RH$Nacional=NULL
RH$Jefe=NULL
RH$`N Personal`=NULL
colnames(RH)
names(RH)[names(RH) == "Grado de distancia"] <- "Distancia"
names(RH)[names(RH) == "Clave para el estado civil"] <- "EstadoCivil"
names(RH)[names(RH) == "Clave de sexo"] <- "Sexo"
####Analizar la estructura del los datos#
str(RH)
Classes ‘tbl_df’, ‘tbl’ and 'data.frame': 325 obs. of 10 variables:
$ Rendimiento: num 0.6 0.8 0.85 0.86 0.85 0.8 1 0.86 1 0.9 ...
$ Edad : num 36 37 21 26 25 28 32 32 29 36 ...
$ Posición : chr "ANA" "ANA" "AUX" "AUX" ...
$ Sexo : num 1 1 0 1 1 1 1 1 1 1 ...
$ Distancia : num 5 3 1 1 3 2 2 4 5 4 ...
$ Estrato : num 2 5 3 3 3 3 5 2 5 6 ...
$ EstadoCivil: num 1 2 1 1 1 1 2 1 1 2 ...
$ Hijos : num 1 1 0 0 0 0 0 1 0 0 ...
$ Formación : chr "PREGRADO" "PREGRADO" "PREGRADO" "PREGRADO" ...
$ Educación : num 3 3 3 3 3 3 4 3 3 3 ...
hist(RH$Rendimiento)
summary(RH)
####Dividir datos en entrenamiento y testeo####
glimpse(RH_train)
glimpse(RH_test)
RH_train <- RH[1:243, ]
RH_test <- RH[244:325, ]
# Define the control
trControl <- trainControl(method = "cv",
number = 10,
search = "grid",
classProb=TRUE)
set.seed(1234)
RH_train$Rendimiento=factor(RH_train$Rendimiento)
RendimientoFactor=factor(RH_train$Rendimiento)
# Run the model
rf_default <- train(RH_train$Rendimiento ~ RH_train$Edad + RH_train$Sexo + RH_train$Distancia + RH_train$Estrato + RH_train$EstadoCivil + RH_train$Hijos + RH_train$Educación,
data=RH_train,
method = "rf",
metric = "Accuracy",
trControl = trControl)
I am running caret's neural network on the Bike Sharing dataset and I get the following error message:
In nominalTrainWorkflow(x = x, y = y, wts = weights, info = trainInfo,
: There were missing values in resampled performance measures.
I am not sure what the problem is. Can anyone help please?
The dataset is from:
https://archive.ics.uci.edu/ml/datasets/bike+sharing+dataset
Here is the coding:
library(caret)
library(bestNormalize)
data_hour = read.csv("hour.csv")
# Split dataset
set.seed(3)
split = createDataPartition(data_hour$casual, p=0.80, list=FALSE)
validation = data_hour[-split,]
dataset = data_hour[split,]
dataset = dataset[,c(-1,-2,-4)]
# View strucutre of data
str(dataset)
# 'data.frame': 13905 obs. of 14 variables:
# $ season : int 1 1 1 1 1 1 1 1 1 1 ...
# $ mnth : int 1 1 1 1 1 1 1 1 1 1 ...
# $ hr : int 1 2 3 5 8 10 11 12 14 15 ...
# $ holiday : int 0 0 0 0 0 0 0 0 0 0 ...
# $ weekday : int 6 6 6 6 6 6 6 6 6 6 ...
# $ workingday: int 0 0 0 0 0 0 0 0 0 0 ...
# $ weathersit: int 1 1 1 2 1 1 1 1 2 2 ...
# $ temp : num 0.22 0.22 0.24 0.24 0.24 0.38 0.36 0.42 0.46 0.44 ...
# $ atemp : num 0.273 0.273 0.288 0.258 0.288 ...
# $ hum : num 0.8 0.8 0.75 0.75 0.75 0.76 0.81 0.77 0.72 0.77 ...
# $ windspeed : num 0 0 0 0.0896 0 ...
# $ casual : int 8 5 3 0 1 12 26 29 35 40 ...
# $ registered: int 32 27 10 1 7 24 30 55 71 70 ...
# $ cnt : int 40 32 13 1 8 36 56 84 106 110 ...
## transform numeric data to Guassian
dataset_selected = dataset[,c(-13,-14)]
for (i in 8:12) { dataset_selected[,i] = predict(boxcox(dataset_selected[,i] +0.1))}
# View transformed dataset
str(dataset_selected)
#'data.frame': 13905 obs. of 12 variables:
#' $ season : int 1 1 1 1 1 1 1 1 1 1 ...
#' $ mnth : int 1 1 1 1 1 1 1 1 1 1 ...
#' $ hr : int 1 2 3 5 8 10 11 12 14 15 ...
#' $ holiday : int 0 0 0 0 0 0 0 0 0 0 ...
#' $ weekday : int 6 6 6 6 6 6 6 6 6 6 ...
#' $ workingday: int 0 0 0 0 0 0 0 0 0 0 ...
#' $ weathersit: int 1 1 1 2 1 1 1 1 2 2 ...
#' $ temp : num -1.47 -1.47 -1.35 -1.35 -1.35 ...
#' $ atemp : num -1.18 -1.18 -1.09 -1.27 -1.09 ...
#' $ hum : num 0.899 0.899 0.637 0.637 0.637 ...
#' $ windspeed : num -1.8 -1.8 -1.8 -0.787 -1.8 ...
#' $ casual : num -0.361 -0.588 -0.81 -1.867 -1.208 ...
# Train data with Neural Network model from caret
control = trainControl(method = 'repeatedcv', number = 10, repeats =3)
metric = 'RMSE'
set.seed(3)
fit = train(casual ~., data = dataset_selected, method = 'nnet', metric = metric, trControl = control, trace = FALSE)
Thanks for your help!
phivers comment is spot on, however I would still like to provide a more verbose answer on this concrete example.
In order to investigate what is going on in more detail one should add the argument savePredictions = "all" to trainControl:
control = trainControl(method = 'repeatedcv',
number = 10,
repeats = 3,
returnResamp = "all",
savePredictions = "all")
metric = 'RMSE'
set.seed(3)
fit = train(casual ~.,
data = dataset_selected,
method = 'nnet',
metric = metric,
trControl = control,
trace = FALSE,
form = "traditional")
now when running:
fit$results
#output
size decay RMSE Rsquared MAE RMSESD RsquaredSD MAESD
1 1 0e+00 0.9999205 NaN 0.8213177 0.009655872 NA 0.007919575
2 1 1e-04 0.9479487 0.1850270 0.7657225 0.074211541 0.20380571 0.079640883
3 1 1e-01 0.8801701 0.3516646 0.6937938 0.074484860 0.20787440 0.077960642
4 3 0e+00 0.9999205 NaN 0.8213177 0.009655872 NA 0.007919575
5 3 1e-04 0.9272942 0.2482794 0.7434689 0.091409600 0.24363651 0.098854133
6 3 1e-01 0.7943899 0.6193242 0.5944279 0.011560524 0.03299137 0.013002708
7 5 0e+00 0.9999205 NaN 0.8213177 0.009655872 NA 0.007919575
8 5 1e-04 0.8811411 0.3621494 0.6941335 0.092169810 0.22980560 0.098987058
9 5 1e-01 0.7896507 0.6431808 0.5870894 0.009947324 0.01063359 0.009121535
we notice the problem occurs when decay = 0.
lets filter the observations and predictions for decay = 0
library(tidyverse)
fit$pred %>%
filter(decay == 0) -> for_r2
var(for_r2$pred)
#output
0
we can observe that all of the predictions when decay == 0 are the same (have zero variance). The model exclusively predicts 0:
unique(for_r2$pred)
#output
0
So when the summary function tries to predict R squared:
caret::R2(for_r2$obs, for_r2$pred)
#output
[1] NA
Warning message:
In cor(obs, pred, use = ifelse(na.rm, "complete.obs", "everything")) :
the standard deviation is zero
Answer by #topepo (Caret package main developer). See detailed Github thread here.
It looks like it happens when you have one hidden unit and almost no
regularization. What is happening is that the model is predicting a
value very close to a constant (so that the RMSE is a little worse
than the basic st deviation of the outcome):
> ANN_cooling_fit$resample %>% dplyr::filter(is.na(Rsquared))
RMSE Rsquared MAE size decay Resample
1 8.414010 NA 6.704311 1 0e+00 Fold04.Rep01
2 8.421244 NA 6.844363 1 0e+00 Fold01.Rep03
3 7.855925 NA 6.372947 1 1e-04 Fold10.Rep07
4 7.963816 NA 6.428947 1 0e+00 Fold07.Rep09
5 8.492898 NA 6.901842 1 0e+00 Fold09.Rep09
6 7.892527 NA 6.479474 1 0e+00 Fold10.Rep10
> sd(mydata$V7)
[1] 7.962888
So it's nothing to really worry about; just some parameters that do very poorly.
The answer by #missuse is already very insightful to understand why this error happens.
So I just want to add some straightforward ways how to get rid of this error.
If in some cross-validation folds the predictions get zero variance, the model didn't converge. In such cases, you can try the neuralnet package which offers two parameters you can tune:
threshold : default value = 0.01. Set it to 0.3 and then try lower values 0.2, 0.1, 0.05.
stepmax : default value = 1e+05. Set it to 1e+08 and then try lower values 1e+07, 1e+06.
In most cases, it is sufficient to change the threshold parameter like this:
model.nn <- caret::train(formula1,
method = "neuralnet",
data = training.set[,],
# apply preProcess within cross-validation folds
preProcess = c("center", "scale"),
trControl = trainControl(method = "repeatedcv",
number = 10,
repeats = 3),
threshold = 0.3
)
I tried to use glm for estimate soccer teams strengths.
# data is dataframe (structure on bottom).
model <- glm(Goals ~ Home + Team + Opponent, family=poisson(link=log), data=data)
but get the error:
Error in if (any(y < 0)) stop("negative values not allowed for the 'Poisson' family") :
missing value where TRUE/FALSE needed
In addition: Warning message:
In Ops.factor(y, 0) : ‘<’ not meaningful for factors
data:
> data
Team Opponent Goals Home
1 5a51f2589d39c31899cce9d9 5a51f2579d39c31899cce9ce 3 1
2 5a51f2579d39c31899cce9ce 5a51f2589d39c31899cce9d9 0 0
3 5a51f2589d39c31899cce9da 5a51f2579d39c31899cce9cd 3 1
4 5a51f2579d39c31899cce9cd 5a51f2589d39c31899cce9da 0 0
> is.factor(data$Goals)
[1] TRUE
From the "details" section of documentation for glm() function:
A typical predictor has the form response ~ terms where response is the (numeric) response vector and terms is a series of terms which specifies a linear predictor for response.
So you want to make sure your Goals column is numeric:
df <- data.frame( Team= c("5a51f2589d39c31899cce9d9", "5a51f2579d39c31899cce9ce", "5a51f2589d39c31899cce9da", "5a51f2579d39c31899cce9cd"),
Opponent=c("5a51f2579d39c31899cce9ce", "5a51f2589d39c31899cce9d9", "5a51f2579d39c31899cce9cd", "5a51f2589d39c31899cce9da "),
Goals=c(3,0,3,0),
Home=c(1,0,1,0))
str(df)
#'data.frame': 4 obs. of 4 variables:
# $ Team : Factor w/ 4 levels "5a51f2579d39c31899cce9cd",..: 3 2 4 1
# $ Opponent: Factor w/ 4 levels "5a51f2579d39c31899cce9cd",..: 2 3 1 4
# $ Goals : num 3 0 3 0
# $ Home : num 1 0 1 0
model <- glm(Goals ~ Home + Team + Opponent, family=poisson(link=log), data=df)
Then here is the output:
> model
Call: glm(formula = Goals ~ Home + Team + Opponent, family = poisson(link = log),
data = df)
Coefficients:
(Intercept) Home Team5a51f2579d39c31899cce9ce
-2.330e+01 2.440e+01 -3.089e-14
Team5a51f2589d39c31899cce9d9 Team5a51f2589d39c31899cce9da Opponent5a51f2579d39c31899cce9ce
-6.725e-15 NA NA
Opponent5a51f2589d39c31899cce9d9 Opponent5a51f2589d39c31899cce9da
NA NA
Degrees of Freedom: 3 Total (i.e. Null); 0 Residual
Null Deviance: 8.318
Residual Deviance: 3.033e-10 AIC: 13.98
When working on a hierarchical/multilevel/panel dataset, it may be very useful to adopt a package which returns the within- and between-group standard deviations of the available variables.
This is something that with the following data in Stata can be easily done through the command
xtsum, i(momid)
I made a research, but I cannot find any R package which can do that..
edit:
Just to fix ideas, an example of hierarchical dataset could be this:
son_id mom_id hispanic mom_smoke son_birthweigth
1 1 1 1 3950
2 1 1 0 3890
3 1 1 0 3990
1 2 0 1 4200
2 2 0 1 4120
1 3 0 0 2975
2 3 0 1 2980
The "multilevel" structure is given by the fact that each mother (higher level) has two or more sons (lower level). Hence, each mother defines a group of observations.
Accordingly, each dataset variable can vary either between and within mothers or only between mothers. birtweigth varies among mothers, but also within the same mother. Instead, hispanic is fixed for the same mother.
For example, the within-mother variance of son_birthweigth is:
# mom1 means
bwt_mean1 <- (3950+3890+3990)/3
bwt_mean2 <- (4200+4120)/2
bwt_mean3 <- (2975+2980)/2
# Within-mother variance for birthweigth
((3950-bwt_mean1)^2 + (3890-bwt_mean1)^2 + (3990-bwt_mean1)^2 +
(4200-bwt_mean2)^2 + (4120-bwt_mean2)^2 +
(2975-bwt_mean3)^2 + (2980-bwt_mean3)^2)/(7-1)
While the between-mother variance is:
# overall mean of birthweigth:
# mean <- sum(data$son_birthweigth)/length(data$son_birthweigth)
mean <- (3950+3890+3990+4200+4120+2975+2980)/7
# within variance:
((bwt_mean1-mean)^2 + (bwt_mean2-mean)^2 + (bwt_mean3-mean)^2)/(3-1)
I don't know what your stata command should reproduce, but to answer the second part of question about
hierarchical structure , it is easy to do this with list.
For example, you define a structure like this:
tree = list(
"var1" = list(
"panel" = list(type ='p',mean = 1,sd=0)
,"cluster" = list(type = 'c',value = c(5,8,10)))
,"var2" = list(
"panel" = list(type ='p',mean = 2,sd=0.5)
,"cluster" = list(type="c",value =c(1,2)))
)
To create this lapply is convinent to work with list
tree <- lapply(list('var1','var2'),function(x){
ll <- list(panel= list(type ='p',mean = rnorm(1),sd=0), ## I use symbol here not name
cluster= list(type = 'c',value = rnorm(3))) ## R prefer symbols
})
names(tree) <-c('var1','var2')
You can view he structure with str
str(tree)
List of 2
$ var1:List of 2
..$ panel :List of 3
.. ..$ type: chr "p"
.. ..$ mean: num 0.284
.. ..$ sd : num 0
..$ cluster:List of 2
.. ..$ type : chr "c"
.. ..$ value: num [1:3] 0.0722 -0.9413 0.6649
$ var2:List of 2
..$ panel :List of 3
.. ..$ type: chr "p"
.. ..$ mean: num -0.144
.. ..$ sd : num 0
..$ cluster:List of 2
.. ..$ type : chr "c"
.. ..$ value: num [1:3] -0.595 -1.795 -0.439
Edit after OP clarification
I think that package reshape2 is what you want. I will demonstrate this here.
The idea here is in order to do the multilevel analysis we need to reshape the data.
First to divide the variables into two groups :identifier and measured variables.
library(reshape2)
dat.m <- melt(dat,id.vars=c('son_id','mom_id')) ## other columns are measured
str(dat.m)
'data.frame': 21 obs. of 4 variables:
$ son_id : Factor w/ 3 levels "1","2","3": 1 2 3 1 2 1 2 1 2 3 ...
$ mom_id : Factor w/ 3 levels "1","2","3": 1 1 1 2 2 3 3 1 1 1 ...
$ variable: Factor w/ 3 levels "hispanic","mom_smoke",..: 1 1 1 1 1 1 1 2 2 2 ...
$ value : num 1 1 1 0 0 0 0 1 0 0 ..
Once your have data in "moten" form , you can "cast" to rearrange it in the shape that you want:
# mom1 means for all variable
acast(dat.m,variable~mom_id,mean)
1 2 3
hispanic 1.0000000 0 0.0
mom_smoke 0.3333333 1 0.5
son_birthweigth 3943.3333333 4160 2977.5
# Within-mother variance for birthweigth
acast(dat.m,variable~mom_id,function(x) sum((x-mean(x))^2))
1 2 3
hispanic 0.0000000 0 0.0
mom_smoke 0.6666667 0 0.5
son_birthweigth 5066.6666667 3200 12.5
## overall mean of each variable
acast(dat.m,variable~.,mean)
[,1]
hispanic 0.4285714
mom_smoke 0.5714286
son_birthweigth 3729.2857143
I know this question is four years old, but recently I wanted to do the same in R and came up with the following function. It depends on dplyr and tibble. Where: df is the dataframe, columns is a numerical vector to subset the dataframe and individuals is the column with the individuals.
xtsumR<-function(df,columns,individuals){
df<-dplyr::arrange_(df,individuals)
panel<-tibble::tibble()
for (i in columns){
v<-df %>% dplyr::group_by_() %>%
dplyr::summarize_(
mean=mean(df[[i]]),
sd=sd(df[[i]]),
min=min(df[[i]]),
max=max(df[[i]])
)
v<-tibble::add_column(v,variacao="overal",.before=-1)
v2<-aggregate(df[[i]],list(df[[individuals]]),"mean")[[2]]
sdB<-sd(v2)
varW<-df[[i]]-rep(v2,each=12) #
varW<-varW+mean(df[[i]])
sdW<-sd(varW)
minB<-min(v2)
maxB<-max(v2)
minW<-min(varW)
maxW<-max(varW)
v<-rbind(v,c("between",NA,sdB,minB,maxB),c("within",NA,sdW,minW,maxW))
panel<-rbind(panel,v)
}
var<-rep(names(df)[columns])
n1<-rep(NA,length(columns))
n2<-rep(NA,length(columns))
var<-c(rbind(var,n1,n1))
panel$var<-var
panel<-panel[c(6,1:5)]
names(panel)<-c("variable","variation","mean","standard.deviation","min","max")
panel[3:6]<-as.numeric(unlist(panel[3:6]))
panel[3:6]<-round(unlist(panel[3:6]),2)
return(panel)
}