for loop & if function in R - r

Question

I was writing a loop with if function in R. The table is like below:
ID category
1 a
1 b
1 c
2 a
2 b
3 a
3 b
4 a
5 a
I want to use the for loop with if function to add another column to count each grouped ID, like below count column:
ID category Count
1 a 1
1 b 2
1 c 3
2 a 1
2 b 2
3 a 1
3 b 2
4 a 1
5 a 1
My code is (output is the table name):
for (i in 2:nrow(output1)){
if(output1[i,1] == output[i-1,1]){
output1[i,"rn"]<- output1[i-1,"rn"]+1
}
else{
output1[i,"rn"]<-1
}
}
But the result returns as all count column values are all "1".
ID category Count
1 a 1
1 b 1
1 c 1
2 a 1
2 b 1
3 a 1
3 b 1
4 a 1
5 a 1
Please help me out... Thanks

Answer 1

There are packages and vectorized ways to do this task, but if you are practicing with loops try:
output1$rn <- 1
for (i in 2:nrow(output1)){
if(output1[i,1] == output1[i-1,1]){
output1[i,"rn"]<- output1[i-1,"rn"]+1
}
else{
output1[i,"rn"]<-1
}
}
With your original code, when you made this call output1[i-1,"rn"]+1 in the third line of your loop, you were referencing a row that didn't exist on the first pass. By first creating the row and filling it with the value 1, you give the loop something explicit to refer to.
output1
# ID category rn
# 1 1 a 1
# 2 1 b 2
# 3 1 c 3
# 4 2 a 1
# 5 2 b 2
# 6 3 a 1
# 7 3 b 2
# 8 4 a 1
# 9 5 a 1
With the package dplyr you can accomplish it quickly with:
library(dplyr)
output1 %>% group_by(ID) %>% mutate(rn = 1:n())
Or with data.table:
library(data.table)
setDT(output1)[,rn := 1:.N, by=ID]
With base R you can also use:
output1$rn <- with(output1, ave(as.character(category), ID, FUN=seq))
There are vignettes and tutorials on the two packages mentioned, and by searching ?ave in the R console for the last approach.

Answer 2

looping solution will be painfully slow for bigger data. Here is one line solution using data.table:
require(data.table)
a<-data.table(ID=c(1,1,1,2,2,3,3,4,5),category=c('a','b','c','a','b','a','b','a','a'))
a[,':='(category_count = 1:.N),by=.(ID)]

Answer 3

what you want is actually a column of factor level. do this
df$count=as.numeric(df$category)
this will give out put as
ID category count
1 1 a 1
2 1 b 2
3 1 c 3
4 2 a 1
5 2 b 2
6 3 a 1
7 3 b 2
8 4 a 1
9 5 a 1
provided your category is already a factor. if not first convert to factor
df$category=as.factor(df$category)
df$count=as.numeric(df$category)