R not printing the right thing in loop - r

Question

I am trying to print the "result" of using table function, but when I tried to use the code here, I got something very strange:
for (i in 1:4){
print (table(paste("group",i,"$", "BMI_obese",sep=""), paste("group",i,"$","A1.1", sep="")))
}
This is the result in R output:
group1$A1.1
group1$BMI_obese 1
group2$A1.1
group2$BMI_obese 1
group3$A1.1
group3$BMI_obese 1
group4$A1.1
group4$BMI_obese 1
But when I type out the statement without typing inside the loop:
table(group2$BMI_obese, group2$A1.1)
I got what I want:
1 2 3 4 5
0 51 20 9 8 0
1 37 20 15 6 4
Does anyone know which part of my for loop code is not correct or can be modified to fit my purpose of printing the loop table result?
Hi, all but now I have another problem. I am trying to add an inner loop which will take the column name as an argument, because I would like to loop through mulitiple column for each of the group data (i.e. for group1, I would like to have table of BMI_obese vs A1.1, BMI_obese vs A1.2 ... BMI_obese vs A1.15. This is my code, but somehow it is not working, I think it is because it is not recognizing the A1.1, A1.2,... as an column taking from the data group1, group2, group3, group4. But instead it is treated as a string I think. I am not sure how to fix it:
for (i in 2:4) {
for (j in c("A1.1","A1.2"))
{
print(with(get(paste0("group", i)),table(BMI_obese,j)))
}
}
I keep getting this error message:
Error in table(BMI_obese, j) : all arguments must have the same length

Answer 1

Okay, you are trying to construct a variable name using paste and then do a table. You are simply passing the name of the variable to table, not the variable object itself. For this sort of approach you want to use get()
for (i in 1:4) {
with(get(paste0("group", i), table(BMI_obese, A1.1))
}
#example saving as a list (using lapply rather than for loop)
group1 <- data.frame(x=LETTERS[1:10], y=(1:10)[sample(10, replace=TRUE)])
group2 <- data.frame(x=LETTERS[1:10], y=(1:10)[sample(10, replace=TRUE)])
result <- lapply(1:2, function(i) with(get(paste0("group", i)), table(x, y)))
#look at first six rows of each:
head(result[[1]])
head(result[[2]])
#example illustrating fetching objects from a string name
data(mtcars)
head(with(get("mtcars"), table(disp, cyl)))
head(with(get("mtcars"), table(disp, "cyl")))
#Error in table(disp, "cyl") : all arguments must have the same length
head(with(get("mtcars"), table(disp, get("cyl"))))

Answer 2

You could also use a combination of eval and parse like this:
x1 <- c(sample(10, 100, replace = TRUE))
y1 <- c(sample(10, 100, replace = TRUE))
table(eval(parse(text = paste0("x", 1))),
eval(parse(text = paste0("y", 1))))
But I'd also say it is not the nicest practice to access variables that way...

Answer 3

Your types are used wrong. See the difference:
table(group2$BMI_obese, group2$A1.1)
and
table(paste(...),paste(...))
So what type does paste return? Certainly some string.
EDIT:
paste(...) was not meant to be syntactically correct but an abbreviation for paste("group",i,"$", "BMI_obese",sep=""), or whatever you paste together.
paste(...) is returning some string. If you put that result into a table, you get a table of strings (the unexpected result that you got). What you want to do is acessing variables or fields with the name which is returned by your paste(...). Just an an eval to your paste like Daniel said and do it like this.
for (i in 1:4){
print (table(eval(paste("group",i,"$", "BMI_obese",sep="")),eval(paste("group",i,"$","A1.1", sep=""))))
}