Updating a vector outside the loop in map(), using R

Question

I have the following simple vector:

a = c(1,0,0,1,0,0,0,0)

and I would like to obtain a vector (b) such that for each indicator x in a, if a[x] is 1, we let it as is, and if it is 0, we compute a[x-1] + 1, until the next 1:

b = c(1,2,3,1,2,3,4,5)

I tried using map():

map(
  .x = seq(1,(length(a))),
  .f = function(x) {
    a[x] = ifelse(a[x]==1, a[x], a[x-1]+1)
    a})

Obviously this does not work because map does not update the a vector. How can I do this using map(). Is it even possible to update a something outside map() ?

My solution to your previous question, uses purrr::reduce, which may correspond to what you are looking for. — PaulS
– PaulS, Commented May 2, 2022 at 11:15

Aron Strandberg · Accepted Answer · 2022-05-02 11:21:16Z

2

If you just change it to use the superassignment operator <<-, the way you attempted it does in fact work.

a = c(1,0,0,1,0,0,0,0)

map(
  .x = seq(1,(length(a))),
  .f = function(x) {
    a[x] <<- ifelse(a[x]==1, a[x], a[x-1]+1)
    a})
a

#> [1] 1 2 3 1 2 3 4 5

edited May 2, 2022 at 11:21

answered May 2, 2022 at 11:14

Aron Strandberg

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2 Comments

Rtist Over a year ago

Could you explain a little bit more what the superassignment does?

Maël Over a year ago

It's actually called 'scoping assignment' if you wanna make research yourself

Maël · Accepted Answer · 2022-05-02 11:12:31Z

1

Maybe a solution close to what you're looking (i.e. that would mimic a for loop) for is purrr::accumulate.

accumulate(1:8, .f = ~ ifelse(a[.y] == 1, 1, .x + 1))
#[1] 1 2 3 1 2 3 4 5

edited May 2, 2022 at 11:12

answered May 2, 2022 at 11:07

Maël

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3 Comments

Rtist Over a year ago

I did not know the accumulate function, very nice. Yet I think the conceptual question remains: how can I update an outside object in map?

Rtist Over a year ago

Can you explaine what is the .y argument? from ?accumulate i see only .x

Maël Over a year ago

This is indeed not well documented: .y is the value in the list, .x is the previous value. See here.

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