Create a counter in a for loop in R - r

Question

I'm an unexperienced user of R and I need to create quite a complicated stuff.
My dataset looks like this :
dataset
a,b,c,d,e are different individuals.
I want to complete the D column as follows :
At the last line for each individual in the col A, D = sum(C)/(B-1).
Expected results should look like :
results
D4=sum(C2:C4)/(B4-1)=0.5
D6=sum(C5:C6)/(B6-1)=1, etc.
I attempted to deal with it with something like :
for(i in 2:NROW(dataset)){
dataset[i,4]<-ifelse(
(dataset[i,1]==data1[i-1,1]),sum(dataset[i,3])/(dataset[i,2]-1),NA
)
}
But it is obviously not sufficient, as it computes the D value for all the rows and not only the last for each individual, and it does not calculate the sum of C values for this individual.
And I really don't know how to figure it out. Do you guys have any advice ?
Many thanks.

Answer 1

If I understood your question correctly, then this is one approach to get to the desired result:
df <- data.frame(
A=c("a","a","a","b","b","c","c","c","d","e","e"),
B=c(3,3,3,2,2,3,3,3,1,2,2),
C=c(NA,1,0,NA,1,NA,0,1,NA,NA,0),
stringsAsFactors = FALSE)
for(i in 2:NROW(df)){
df[i,4]<-ifelse(
(df[i,1]!=df[i+1,1] | i == nrow(df)),sum(df[df$A == df[i,1],]$C, na.rm=TRUE)/(df[i,2]-1),NA
)
}
This code results in the following table:
A B C V4
1 a 3 NA NA
2 a 3 1 NA
3 a 3 0 0.5
4 b 2 NA NA
5 b 2 1 1.0
6 c 3 NA NA
7 c 3 0 NA
8 c 3 1 0.5
9 d 1 NA NaN
10 e 2 NA NA
11 e 2 0 0.0
The ifelse first tests if the individual of the current row of column A is different than the individual in the next row OR if it's the last row.
If it is the last row with this individual it takes the sum of column C (ignoring the NAs) of the rows with the individual present in column A divided by the value in column B minus one.
Otherwise it puts an NA in the fourth column.

Answer 2

Using dplyr you can try generating D for all rows and then remove where not required:
dftest %>%
group_by(A,B) %>%
dplyr::mutate(D = sum(C, na.rm=TRUE)/(B-1)) %>%
dplyr::mutate(D = if_else(row_number()== n(), D, as.double(NA)))
which gives:
Source: local data frame [11 x 4]
Groups: A, B [5]
A B C D
<chr> <dbl> <dbl> <dbl>
1 a 3 NA NA
2 a 3 1 NA
3 a 3 0 0.5
4 b 2 NA NA
5 b 2 1 1.0
6 c 3 NA NA
7 c 3 0 NA
8 c 3 1 0.5
9 d 1 NA NaN
10 e 2 NA NA
11 e 2 0 0.0