CODEWITHSUNDEEP
rdplyrapply
Kou
Kou

Reputation: 175

How to use dplyr mutate to perform operation on a column when a lag variable and another column is involved

Suppose I have a data frame like this

> dat
  a          b         c
1 1  0.3321008 0.3321008
2 2 -0.2946729        NA
3 3 -0.1447266        NA
4 4 -0.9415429        NA
5 5 -1.0165080        NA

here is the dput

structure(list(a = 1:5, b = c(0.332100835317822, -0.294672931641969, 
-0.144726592564241, -0.941542877670977, -1.0165079846083), c = c(0.332100835317822, 
NA, NA, NA, NA)), .Names = c("a", "b", "c"), row.names = c(NA, 
-5L), class = "data.frame")

I would like to perform an operation on the c column such that c = lag(c)*b (except for the first element in c

I can do this using a simple for loop as below

for(i in (1:4)){
  dat$c[i+1] <- dat$c[i]*dat$b[i+1]
}

Output:

> dat
  a          b           c
1 1  0.3321008  0.33210084
2 2 -0.2946729 -0.09786113
3 3 -0.1447266  0.01416311
4 4 -0.9415429 -0.01333517
5 5 -1.0165080  0.01355531

How do I do this using dplyr mutate ? or using apply functions?

Upvotes: 0

Views: 884

Answers (1)

Frank
Frank

Reputation: 66819

By doing the math, we can avoid the iterative computation:

library(dplyr)
dat %>% mutate(c = cumprod(replace(b, 1, 1))*c[1])

Upvotes: 4

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