So my dataframe is structured like so:

for each row in s, I would like to take the lag(s), add it to column x, then set it to the value of s.

my output data would therefore look like:

I tried the following function, but after fiddling I was only able to get all NA's or all 0's:

mydata$s = lag(mydata$s) mydata$x  

Note - if it helps, I can remove the first row.

CodePudding user response：

It works for me. Set up:

mydata <- data.frame(x = c(NA, 13, -3, 2, -4), s = c(0, 13, 10, 12, 8) )
mydata$s <- lag(mydata$s) mydata$x 

Gives:

mydata
   x  s
1 NA NA
2 13 13
3 -3 10
4  2 12
5 -4  8

The difference is my first s is NA. That should be expected as the first x is NA.

CodePudding user response：

You can use cumsum() to perform the job, and also replace NA with 0 during the calculation (without changing your original dataset).

library(tidyverse)

df %>% mutate(s = cumsum(ifelse(is.na(x), 0, x)))

   x  s
1 NA  0
2 13 13
3 -3 10
4  2 12
5 -4  8

CodePudding user response：

Base R solution:

mydata$s <- c(mydata$x[1], cumsum(mydata$x[-1]))

Data:

mydata <- data.frame(x = c(NA, 13, -3, 2, -4))