# Sum the leading X rows of a column?

992 views
3

I have a code where I run a check on the leading rows of a value, in order to calculate a coverage value. I have a sample of the code below:

``````library(data.table)

df <- data.frame(
dept = c(rep('FIREDEPT', 5), rep('WATERDEPT', 5)),
month = 201808:201812,
initial_stock = sample(75884:85347, 10),
variable_predicted = sample(50000:100000, 10))

df <- mutate(df, calculation = ifelse(initial_stock <= (shift(variable_predicted, type="lead", fill = 0)
+ shift(variable_predicted, type="lead", fill = 0, n = 2)
+ shift(variable_predicted, type="lead", fill = 0, n = 3)),
(3 + (initial_stock
- shift(variable_predicted, type="lead", fill = 0)
- shift(variable_predicted, type="lead", fill = 0, n = 2)
) / shift(variable_predicted, type="lead", fill = 0, n = 3)) * 30,
0))
``````

the thing is, the leading value is much larger and I have to do this in a series of ifelses for multiple formulas. Is there a way for me to get the result of this part:

``````(shift(variable_predicted, type="lead", fill = 0)
+ shift(variable_predicted, type="lead", fill = 0, n = 2)
+ shift(variable_predicted, type="lead", fill = 0, n = 3)),
``````

without using a series of shifts?

8

`shift` is vectorized, so you can give it a vector for `n`, and then add the results together with `Reduce('+', ...)`

``````a <-
with(df,
+ shift(variable_predicted, type="lead", fill = 0, n = 2)
+ shift(variable_predicted, type="lead", fill = 0, n = 3)))

b <-
with(df,
Reduce(`+`, shift(variable_predicted, n = 1:3, fill = 0, type = 'lead')))

identical(a, b)
# [1] TRUE
``````

posted this