CODEWITHSUNDEEP
rdataframe
barker
barker

Reputation: 1055

R - Find the sum for the lagging record and add another column to current value iteratively

So my dataframe is structured like so:

x s
NA 0
13 0
-3 0
2 0
-4 0

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:

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

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.

Upvotes: 1

Views: 252

Answers (3)

hello_friend
hello_friend

Reputation: 5798

Base R solution:

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

Data:

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

Upvotes: 1

benson23
benson23

Reputation: 19142

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

Upvotes: 1

Brian Syzdek
Brian Syzdek

Reputation: 948

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.

Upvotes: 1

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