Identify and replace duplicate from a column vector
r, xts
Solution
You were on the right track with `diff`, but I think it's simpler than you're making it:
a[c(F, diff(a) == 0)] <- 0
a
[1] 1 0 0 2 3 2 0 0 0 1 0 0 0 0 2 3 4 0 1 0
Edit
To make this solution another column, then you can just assign it by:
data$b <- ifelse(c(F, diff(data$a)==0), 0, data$a)
Or by making a copy:
data$b <- data$a
data$b[c(F, diff(data$b) ==0)] <- 0
Problem
I've got two columns which are as under: ``` a<- c(1,1,1,2,3,2,2,2,2,1,0,0,0,0,2,3,4,4,1,1) date<- Sys.Date()-20:1 data<- xts(a,date) colnames(data)<- "a" data ``` Here we can see that there are lot of duplicate elements, ie. they are repeated ones. I want a code which can replace all the elements which are consecutive and duplicate by 0 except for the first element. The result which i require is ``` a<- c(1,0,0,2,3,2,0,0,0,1,0,0,0,0,2,3,4,0,1,0) ``` I've tried what i've learnt from my earlier post ``` ifelse(data$a == c(data$a[1]-1,data$a[(1:length(data$a)-1)]) , 0 , data$a) ``` and also i've tried ``` data$a<- replace(data$a, duplicated(c(0, cumsum(abs(diff(data$a))))), 0) ``` but both the codes are not working in xts. Though both the above mentioned code is working for normal vector.