predict() R function caret package errors: "newdata" rows different, "type" not accepted

1
votes

  • I am running a logistic regression analysis using the caret package.

  • Data is input as a 18x6 matrix

  • everything is fine so far except the predict() function.

  • R is telling me the type parameter is supposed to be raw or prob but raw just spits out an exact copy of the last column (the values of the binomial variable). prob gives me the following error:

"Error in dimnames(out)[[2]] <- modelFit$obsLevels : length of 'dimnames' [2] not equal to array extent In addition: Warning message: 'newdata' had 7 rows but variables found have 18 rows"

install.packages("pbkrtest")
install.packages("caret")
install.packages('e1071', dependencies=TRUE)
#install.packages('caret', dependencies = TRUE)
require(caret)
library(caret)

A=matrix(
  c(
    64830,18213,4677,24761,9845,17504,22137,12531,5842,28827,51840,4079,1000,2069,969,9173,11646,946,66161,18852,5581,27219,10159,17527,23402,11409,8115,31425,55993,0,0,1890,1430,7873,12779,627,68426,18274,5513,25687,10971,14104,19604,13438,6011,30055,57242,0,0,2190,1509,8434,10492,755,69716,18366,5735,26556,11733,16605,20644,15516,5750,31116,64330,0,0,1850,1679,9233,12000,500,73128,18906,5759,28555,11951,19810,22086,17425,6152,28469,72020,0,0,1400,1750,8599,12000,500,1,1,1,0,1,0,0,0,0,1,0,1,1,1,1,1,1,1
  ),

  nrow = 18,
  ncol = 6,
  byrow = FALSE)  #"bycol" does NOT exist
################### data set as vectors
a<-c(64830,18213,4677,24761,9845,17504,22137,12531,5842,28827,51840,4079,1000,2069,969,9173,11646,946)
b<-c(66161,18852,5581,27219,10159,17527,23402,11409,8115,31425,55993,0,0,1890,1430,7873,12779,627)
c<-c(68426,18274,5513,25687,10971,14104,19604,13438,6011,30055,57242,0,0,2190,1509,8434,10492,755)
d<-c(69716,18366,5735,26556,11733,16605,20644,15516,5750,31116,64330,0,0,1850,1679,9233,12000,500)
e<-c(73128,18906,5759,28555,11951,19810,22086,17425,6152,28469,72020,0,0,1400,1750,8599,12000,500)
f<-c(1,1,1,0,1,0,0,0,0,1,0,1,1,1,1,1,1,1)
######################

n<-nrow(A);
K<-ncol(A)-1;

Train <- createDataPartition(f, p=0.6, list=FALSE)  #60% of data set is used as training.
training <- A[ Train, ]
testing <- A[ -Train, ]
nrow(training)

#this is the logistic formula:
#estimates from logistic regression characterize the relationship between the predictor and response variable on a log-odds scale
mod_fit <- train(f ~ a + b + c + d +e,  data=training, method="glm", family="binomial")
mod_fit

#this isthe exponential function to calculate the odds ratios for each preditor:
exp(coef(mod_fit$finalModel))

predict(mod_fit, newdata=training)
predict(mod_fit, newdata=testing, type="prob")
3
rmachine-learninglogistic-regressionr-carettraining-data
Just curious, why do you need to define both A data frame and the a~f vectors? - zyurnaidi
@zyurnaidi I did that because I was comparing the binomial variable to the other columns (f vs a-e). I didn't know of any other way to run the logistic formula. Is there a better way? - Gabriel
Sure, we only need to set up the data as data frame, then access each column by its name. See the answers. - zyurnaidi

3 Answers

0
votes

I'm not very sure to understand, but A is a matrix of (a,b,c,d,e,f). So you don't need to create two objects.

install.packages("pbkrtest")
install.packages("caret")
install.packages('e1071', dependencies=TRUE)
#install.packages('caret', dependencies = TRUE)
require(caret)
library(caret)

A=matrix(
  c(
        64830,18213,4677,24761,9845,17504,22137,12531,5842,28827,51840,4079,1000,2069,969,9173,11646,946,66161,18852,5581,27219,10159,17527,23402,11409,8115,31425,55993,0,0,1890,1430,7873,12779,627,68426,18274,5513,25687,10971,14104,19604,13438,6011,30055,57242,0,0,2190,1509,8434,10492,755,69716,18366,5735,26556,11733,16605,20644,15516,5750,31116,64330,0,0,1850,1679,9233,12000,500,73128,18906,5759,28555,11951,19810,22086,17425,6152,28469,72020,0,0,1400,1750,8599,12000,500,1,1,1,0,1,0,0,0,0,1,0,1,1,1,1,1,1,1
  ),
  nrow = 18,
  ncol = 6,
  byrow = FALSE)  #"bycol" does NOT exist

A <- data.frame(A)
colnames(A) <- c('a','b','c','d','e','f')
A$f <- as.factor(A$f)

Train <- createDataPartition(A$f, p=0.6, list=FALSE)  #60% of data set is used as training.
training <- A[ Train, ]
testing <- A[ -Train, ]
nrow(training)

And to predict a variable you must enter the explanatory variables and not the variable to predict

mod_fit <- train(f ~ a + b + c + d +e,  data=training, method="glm", family="binomial")
mod_fit

#this isthe exponential function to calculate the odds ratios for each preditor:
exp(coef(mod_fit$finalModel))

predict(mod_fit, newdata=training[,-which(colnames(training)=="f")])
predict(mod_fit, newdata=testing[,-which(colnames(testing)=="f")])
0
votes

Short answer, you should not include the explained variable, which is f in your predict equation. So you should do:

predict(mod_fit, newdata=training[, -ncol(training])
predict(mod_fit, newdata=testing[, -ncol(testing])

The issue with the warning message 'newdata' had 11 rows but variables found have 18 rows is because you run the regression using the whole data set (18 observations), but predict using just part of it (either 11 or 7).

EDIT: To simplify the data creation and glm processes we can do:

library(caret)
A <- data.frame(a = c(64830,18213,4677,24761,9845,17504,22137,12531,5842,28827,51840,4079,1000,2069,969,9173,11646,946),
                b = c(66161,18852,5581,27219,10159,17527,23402,11409,8115,31425,55993,0,0,1890,1430,7873,12779,627),
                c = c(68426,18274,5513,25687,10971,14104,19604,13438,6011,30055,57242,0,0,2190,1509,8434,10492,755),
                d = c(69716,18366,5735,26556,11733,16605,20644,15516,5750,31116,64330,0,0,1850,1679,9233,12000,500),
                e = c(73128,18906,5759,28555,11951,19810,22086,17425,6152,28469,72020,0,0,1400,1750,8599,12000,500),
                f = c(1,1,1,0,1,0,0,0,0,1,0,1,1,1,1,1,1,1))

Train <- createDataPartition(f, p=0.6, list=FALSE)  #60% of data set is used as training.
training <- A[ Train, ]
testing <- A[ -Train, ]

mod_fit <- train(f ~ a + b + c + d + e,  data=training, method="glm", family="binomial")
0
votes

I try to run logistic regression model. I wrote this code:

install.packages('caret')
library(caret)
setwd('C:\\Users\\BAHOZ\\Documents\\')
D<-read.csv(file = "D.csv",header = T)
D<-read.csv(file = 'DataSet.csv',header=T)
names(D)
set.seed(111134)
Train<-createDataPartition(D$X, p=0.7,list = FALSE)
training<-D[Train,]
length(training$age)
testing<-D[-Train,]
length(testing$age)
mod_fit<-train(X~age + gender  + total.Bilirubin + direct.Bilirubin + total.proteins + albumin + A.G.ratio+SGPT + SGOT + Alkphos,data=training,method="glm", family="binomial")
summary(mod_fit)
exp(coef(mod_fit$finalModel))

And I recived this message for last command:

     (Intercept)              age           gender  total.Bilirubin direct.Bilirubin   total.proteins          albumin        A.G.ratio 
  0.01475027       1.01596886       1.03857883       1.00022899       1.78188072       1.00065332       1.01380334       1.00115742 
        SGPT             SGOT          Alkphos 
  3.93498241       0.05616662      38.29760014

By running this command I could predict my data,

predict(mod_fit , newdata=testing)

But if I set type="prob" or type="raw"

predict(mod_fit , newdata=testing, type = "prob")

it falls in error:

Error in dimnames(out) <- *vtmp* :

length of 'dimnames' [2] not equal to array extent