Problems following a code example. InformationValue::Woe - r

Question

I'm learning new feature selection methods with this entry of a blog:
https://www.machinelearningplus.com/machine-learning/feature-selection/
Point 9. And I stumbled upon some problems. First is the CV, which I have solved.
library(InformationValue)
adult <- read.csv('https://archive.ics.uci.edu/ml/machine-learning-databases/adult/adult.data',
sep = ',', fill = F, strip.white = T,stringsAsFactors = FALSE)
colnames(adult) <- c('age', 'WORKCLASS', 'fnlwgt', 'EDUCATION',
'educatoin_num', 'MARITALSTATUS', 'OCCUPATION', 'RELATIONSHIP', 'RACE', 'SEX',
'capital_gain', 'capital_loss', 'hours_per_week', 'NATIVECOUNTRY', 'ABOVE50K')
inputData <- adult
print(head(inputData))
But then I can't solve the next chunk
# Choose Categorical Variables to compute Info Value.
cat_vars <- c ("WORKCLASS", "EDUCATION", "MARITALSTATUS", "OCCUPATION", "RELATIONSHIP", "RACE", "SEX", "NATIVECOUNTRY") # get all categorical variables
# Init Output
df_iv <- data.frame(VARS=cat_vars, IV=numeric(length(cat_vars)), STRENGTH=character(length(cat_vars)), stringsAsFactors = F) # init output dataframe
# Get Information Value for each variable
for (factor_var in factor_vars){
df_iv[df_iv$VARS == factor_var, "IV"] <- InformationValue::IV(X=inputData[, factor_var], Y=inputData$ABOVE50K)
df_iv[df_iv$VARS == factor_var, "STRENGTH"] <- attr(InformationValue::IV(X=inputData[, factor_var], Y=inputData$ABOVE50K), "howgood")
}
# Sort
df_iv <- df_iv[order(-df_iv$IV), ]
df_iv
And I keep getting 0 values in IV and, of course, Not predictive in the column of the dataframe.
I've tried to do a
factor_vars=cat_vars
But it doesn't seems to work and quite frankly I can't figure out why this doesn't work.

Answer 1

Just solved it. In first instance the argument of stringsAsFactors = FALSE its unnecesary, since we need factors.
Then, consulting the IV function and looking at the summary of the dataset, i noticed that despise its a factor the function requieres a numeric input, the function cannot extract its "value" (level). So we must work arround it.
as.numeric(inputData$ABOVE50K)
"Solves it" Although maybe i should change the values since it gives 1-2 instead of the classic 0-1 response. Im working on it.
I think theres got to be an easiest solution, but:
levels(inputData$ABOVE50K)
inputData$ABOVE50K2 = as.numeric(inputData$ABOVE50K)
inputData$ABOVE50K3= ifelse(inputData$ABOVE50K2 ==1,0, ifelse(inputData$ABOVE50K2==2,1,NA))
inputData$ABOVE50K3 <- factor(inputData$ABOVE50K3)
And the output is the same. So there is no need to change the levels to 0-1.
# Choose Categorical Variables to compute Info Value.
cat_vars <- c ("WORKCLASS", "EDUCATION", "MARITALSTATUS", "OCCUPATION", "RELATIONSHIP", "RACE", "SEX", "NATIVECOUNTRY") # get all categorical variables
factor_vars= cat_vars
# Init Output
df_iv <- data.frame(VARS=cat_vars, IV=numeric(length(cat_vars)), STRENGTH=character(length(cat_vars)), stringsAsFactors = F) # init output dataframe
# Get Information Value for each variable
for (factor_var in factor_vars){
df_iv[df_iv$VARS == factor_var, "IV"] <- InformationValue::IV(X=inputData[, factor_var], Y=inputData$ABOVE50K3)
df_iv[df_iv$VARS == factor_var, "STRENGTH"] <- attr(InformationValue::IV(X=inputData[, factor_var], Y=inputData$ABOVE50K3), "howgood")
}
# Sort
df_iv <- df_iv[order(-df_iv$IV), ]
df_iv