I am just trying my hands on lambda functions of purrr family.
Suppose I have to do some operation iteratively over result of previous iteration in vector through accumulate I can do it through .x and .y, where .x is result of application on previous element and .y is current element. Also assume that the function/iteration is 2x+3y i.e. double the previous result and add three times of current element, it can be done through like.
accumulate(1:10, ~2*.x + 3*.y)
[1] 1 8 25 62 139 296 613 1250 2527 5084
#OR
accumulate(1:10, ~2*.x + 3*.y, .init = 2.5)
[1] 2.5 8.0 22.0 53.0 118.0 251.0 520.0 1061.0 2146.0 4319.0 8668.0
However, I am not able to do so these type of iterations in accumulate2
accumulate2(1:5, 1:5, ~2*.x + 3*.y +.z)
Error in reduce2_impl(.x, .y, .f, ..., .init = .init, .acc = TRUE) :
`.y` does not have length 4
OR
accumulate2(1:5, 1:4, ~2*.x + 3*.y +.z)
Error in .f(out, .x[[x_i]], .y[[y_i]], ...) : object '.z' not found
What actually I understood from purrr::accumulate was that, it is a two argument function where first argument is result of previous iteration and second argument is vector passed into it directly. On similar lines accumulate2 is a three argument function where first argument is result of previous computation and other two arguments are passed directly.
Can someone tell where I am erring or am wrong in understanding the logic or what should be the correct way of doing it.
The .x and .y are only specified when there are two arguments. With more than 2, we can use ..1, ..2, ..3 in their order of occurrence
library(purrr)
accumulate2(1:5, 1:4, ~2*..1 + 3*..2 +..3)
#[[1]]
#[1] 1
#[[2]]
#[1] 9
#[[3]]
#[1] 29
#[[4]]
#[1] 73
#[[5]]
#[1] 165