How to use lag/lead in mutate with only one initial value? - r

Question

Sample df:
library(tidyverse)
iris <- iris[1:10,]
iris$testlag <- NA
iris[[1,"testlag"]] <- 5
Sepal.Length Sepal.Width Petal.Length Petal.Width Species testlag
1 5.1 3.5 1.4 0.2 setosa 5
2 4.9 3.0 1.4 0.2 setosa NA
3 4.7 3.2 1.3 0.2 setosa NA
4 4.6 3.1 1.5 0.2 setosa NA
5 5.0 3.6 1.4 0.2 setosa NA
6 5.4 3.9 1.7 0.4 setosa NA
7 4.6 3.4 1.4 0.3 setosa NA
8 5.0 3.4 1.5 0.2 setosa NA
9 4.4 2.9 1.4 0.2 setosa NA
10 4.9 3.1 1.5 0.1 setosa NA
In the testlag column, I'm interesting in using dplyr::lag() to retrieve the previous value and add some column, for example Petal.Length to it. As I have only one initial value, each subsequent calculation requires it to work iteratively, so I thought something like mutate would work.
I first tried doing something like this:
iris %>% mutate_at("testlag", ~ lag(.) + Petal.Length)
But this removed the first value, and only gave a valid value for the second row and NAs for the rest. Intuitively I know why it's removing the first value, but I thought the nature of mutate would allow it to work for the rest of the values, so I don't know how to fix that.
Of course using base R I could something like:
for (idx in 2:nrow(iris)) {
iris[[idx, "testlag"]] <-
lag(iris$testlag)[idx] + iris[[idx, "Petal.Length"]]
}
But I would prefer to implement this in tidyverse syntax.
Edit: Desired output (from my for loop)
Sepal.Length Sepal.Width Petal.Length Petal.Width Species testlag
1 5.1 3.5 1.4 0.2 setosa 5.0
2 4.9 3.0 1.4 0.2 setosa 6.4
3 4.7 3.2 1.3 0.2 setosa 7.7
4 4.6 3.1 1.5 0.2 setosa 9.2
5 5.0 3.6 1.4 0.2 setosa 10.6
6 5.4 3.9 1.7 0.4 setosa 12.3
7 4.6 3.4 1.4 0.3 setosa 13.7
8 5.0 3.4 1.5 0.2 setosa 15.2
9 4.4 2.9 1.4 0.2 setosa 16.6
10 4.9 3.1 1.5 0.1 setosa 18.1

Answer 1

Does this work for you?
library(tidyverse)
library("data.table")
iris <- iris[1:10,]
iris$testlag <- NA
iris[[1,"testlag"]] <- 5
iris %>% mutate (testlag = lag(first(testlag) + cumsum(Petal.Length)))
Result:
Sepal.Length Sepal.Width Petal.Length Petal.Width Species testlag
1 5.1 3.5 1.4 0.2 setosa NA
2 4.9 3.0 1.4 0.2 setosa 6.4
3 4.7 3.2 1.3 0.2 setosa 7.8
4 4.6 3.1 1.5 0.2 setosa 9.1
5 5.0 3.6 1.4 0.2 setosa 10.6
6 5.4 3.9 1.7 0.4 setosa 12.0
7 4.6 3.4 1.4 0.3 setosa 13.7
8 5.0 3.4 1.5 0.2 setosa 15.1
9 4.4 2.9 1.4 0.2 setosa 16.6
10 4.9 3.1 1.5 0.1 setosa 18.0
Since technically there is no N-1 Petal length when N = 1, I left the first value of testlag NA. Do you really need it to be initial value? If you need, this will work:
iris %>% mutate (testlag = lag(first(testlag) + cumsum(Petal.Length), default=first(testlag)))

Answer 2

The function you're looking for is tidyr::fill
library(tidyverse)
iris <- iris[1:10,]
iris$testlag <- NA
iris[[1,"testlag"]] <- 5
iris %>% fill(testlag, .direction = "down")
# Note the default is 'down', but I included here for completeness
This takes the specified column (testlag in this case), and copies any values in that column to the values below. This also works if you have a value in a subset of the rows: it copies the value down until it reaches a new value, then it picks up with that one.
For example:
library(tidyverse)
iris <- iris[1:10,]
iris$testlag <- NA
iris[[1,"testlag"]] <- 5
iris[[5,"testlag"]] <- 10
Sepal.Length Sepal.Width Petal.Length Petal.Width Species testlag
1 5.1 3.5 1.4 0.2 setosa 5
2 4.9 3.0 1.4 0.2 setosa NA
3 4.7 3.2 1.3 0.2 setosa NA
4 4.6 3.1 1.5 0.2 setosa NA
5 5.0 3.6 1.4 0.2 setosa 10
6 5.4 3.9 1.7 0.4 setosa NA
7 4.6 3.4 1.4 0.3 setosa NA
8 5.0 3.4 1.5 0.2 setosa NA
9 4.4 2.9 1.4 0.2 setosa NA
10 4.9 3.1 1.5 0.1 setosa NA
Applying this function...
iris %>% fill(testlag, .direction = "down")
Gives
Sepal.Length Sepal.Width Petal.Length Petal.Width Species testlag
1 5.1 3.5 1.4 0.2 setosa 5
2 4.9 3.0 1.4 0.2 setosa 5
3 4.7 3.2 1.3 0.2 setosa 5
4 4.6 3.1 1.5 0.2 setosa 5
5 5.0 3.6 1.4 0.2 setosa 10
6 5.4 3.9 1.7 0.4 setosa 10
7 4.6 3.4 1.4 0.3 setosa 10
8 5.0 3.4 1.5 0.2 setosa 10
9 4.4 2.9 1.4 0.2 setosa 10
10 4.9 3.1 1.5 0.1 setosa 10