stratified splitting the data - r

Question

I have a large data set and like to fit different logistic regression for each City, one of the column in my data. The following 70/30 split works without considering City group.
indexes <- sample(1:nrow(data), size = 0.7*nrow(data))
train <- data[indexes,]
test <- data[-indexes,]
But this does not guarantee the 70/30 split for each city.
lets say that I have City A and City B, where City A has 100 rows, and City B has 900 rows, totaling 1000 rows. Splitting the data with above code will give me 700 rows for train and 300 for test data, but it does not guarantee that i will have 70 rows for City A, and 630 rows for City B in the train data. How do i do that?
Once i have the training data split-ed to 70/30 fashion for each city,i will run logistic regression for each city ( I know how to do this once i have the train data)

Answer 1

Try createDataPartition from caret package. Its document states: By default, createDataPartition does a stratified random split of the data.
library(caret)
train.index <- createDataPartition(Data$Class, p = .7, list = FALSE)
train <- Data[ train.index,]
test <- Data[-train.index,]
it can also be used for stratified K-fold like:
ctrl <- trainControl(method = "repeatedcv",
repeats = 3,
...)
# when calling train, pass this train control
train(...,
trControl = ctrl,
...)
check out caret document for more details

Answer 2

The package splitstackshape has a nice function stratified which can do this as well, but this is a bit better than createDataPartition because it can use multiple columns to stratify at once. It can be used with one column like:
library(splitstackshape)
set.seed(42) # good idea to set the random seed for reproducibility
stratified(data, c('City'), 0.7)
Or with multiple columns:
stratified(data, c('City', 'column2'), 0.7)

Answer 3

The typical way is with split
lapply( split(dfrm, dfrm$City), function(dd){
indexes= sample(1:nrow(dd), size = 0.7*nrow(dd))
train= dd[indexes, ] # Notice that you may want all columns
test= dd[-indexes, ]
# analysis goes here
}
If you were to do it in steps as you attempted above it would be like this:
cities <- split(data,data$city)
idxs <- lapply(cities, function (d) {
indexes <- sample(1:nrow(d), size=0.7*nrow(d))
})
train <- data[ idxs[[1]], ] # for the first city
test <- data[ -idxs[[1]], ]
I happen to think the is the clumsy way to do it, but perhaps breaking it down into small steps will let you examine the intermediate values.

Answer 4

Your code works just fine as is, if City is a column, simply run training data as train[,2]. You can do this easily for each one with a lambda function
logReg<-function(ind) {
reg<-glm(train[,ind]~WHATEVER)
....
return(val) }
Then run sapply over the vector of city indexes.

Answer 5

Another possible way, similar to IRTFMs answer (e.g., using only base-r) is to use the following. Note that this answer returns a stratified index, which can be used like the index calculated in the question.
p <- 0.7
strats <- your_data$the_stratify_variable
rr <- split(1:length(strats), strats)
idx <- sort(as.numeric(unlist(sapply(rr, function(x) sample(x, length(x) * p)))))
train <- your_data[idx, ]
test <- your_data[-idx, ]
Example:
p <- 0.7
strats <- mtcars$cyl
rr <- split(1:length(strats), strats)
idx <- sort(as.numeric(unlist(sapply(rr, function(x) sample(x, length(x) * p)))))
train <- mtcars[idx, ]
test <- mtcars[-idx, ]
table(mtcars$cyl) / nrow(mtcars)
#> 4 6 8
#> 0.34375 0.21875 0.43750
table(train$cyl) / nrow(train)
#> 4 6 8
#> 0.35 0.20 0.45
table(test$cyl) / nrow(test)
#> 4 6 8
#> 0.3333333 0.2500000 0.4166667
We see that all datasets all (mtcars), train, and test have roughly the same class distributions!