Identifying positions of the last TRUEs in a sequence of TRUEs and FALSEs

Question

I have a vector of TRUEs and FALSEs:

x <- c(F,F,F,T,T,T,F,F,F,T,T,T,F,T,T)

I'd like to elegantly (and in base) identify the position of the last TRUE before it changes to FALSE.

The following works, though, it seems like it could be simplified:

c((x[-1] != x[-length(x)]),T) & x
> FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE  TRUE

Input and output:

Simoniac

Answer 1

We may look where x is greater than shifted x with 0 appended.

x>c(x[-1],0)
# [1] FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE  TRUE

Misfile

Answer 2

Taking advantage of diff with an appended FALSE to catch the implied TRUE-to-FALSE at the end.

diff(c(x,FALSE)) == -1
# [1] FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE FALSE FALSE  TRUE
#[13] FALSE FALSE  TRUE

Greenebaum

More golfy diff(c(x,0))<0. – Misfile

Answer 3

Check rle

rlex = rle(x)
end = cumsum(rlex$lengths)
x&(seq(length(x)) %in% end)
[1] FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE  TRUE

Another layout suggested by Frank

seq_along(x) %in% with(rle(x), cumsum(lengths)[values])
[1] FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE  TRUE

Conduct

Answer 4

Non-base solution for identifying the last TRUE before a FALSE.

 library(dplyr)

 y <- data.frame(x = c(FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE,FALSE,
 FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE))

 y %>% 
   mutate(lasttrue = case_when(x == TRUE & lead(x) == FALSE ~ TRUE,
                               TRUE ~ FALSE))

Edit:

y %>% 
  mutate(lasttrue = case_when(x > lead(x) ~ T,
                              T ~ F)) 

O

Returns the wrong result with the last two elements are FALSE, TRUE – Simoniac

Answer 5

Another version with rle

x[setdiff(seq_along(x), with(rle(x), cumsum(lengths) * values))] <- FALSE
x
#[1] FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE  TRUE

Dictum

Answer 6

An option with duplicated

library(data.table)
!duplicated(rleid(x), fromLast = TRUE) & x
#[1] FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE  TRUE

Wyn

Answer 7

benchmarks

Thanks for all the solutions. If anyone is interested in benchmarks:

library(dplyr)
library(data.table)

set.seed(1)
x <- sample(c(TRUE, FALSE), 1000000, replace = T)
y <- data.frame(x = x) # For M. Viking's solution
x_dt <- x # For Ronak Shah's solution

microbenchmark::microbenchmark(Khaynes = {Khaynes <- c((x[-1] != x[-length(x)]),T) & x},
                                jay.sf = {jay.sf <- x>c(x[-1],0)},
                                jay.sf_2 = {jay.sf_2 <- diff(c(x,0))<0},
                                thelatemail = {thelatemail <- diff(c(x,FALSE)) == -1}, 
                                WeNYoBen = {rlex = rle(x); end = cumsum(rlex$lengths); WeNYoBen <- x&(seq(length(x)) %in% end)}, 
                                M._Viking = {M._Viking <- y %>% mutate(lasttrue = case_when(x > lead(x) ~ T, T ~ F))}, 
                                akrun = {akrun <- !duplicated(rleid(x), fromLast = TRUE) & x},
                                frank = {frank <- seq_along(x) %in% with(rle(x), cumsum(lengths)[values])}, 
                                Ronak_Shah = {x_dt[setdiff(seq_along(x_dt), with(rle(x_dt), cumsum(lengths) * values))] <- FALSE},
                                times = 50)
# Output:
    # Unit: milliseconds
    #         expr      min       lq      mean    median       uq      max neval
    #      Khaynes  23.0283  26.5010  31.76180  31.71290  37.1449  46.3824    50
    #       jay.sf  13.0630  13.5373  17.84056  13.77135  20.5462  73.5926    50
    #     jay.sf_2  26.1960  27.7653  35.25296  36.39615  39.3686  61.8858    50
    #  thelatemail  24.8204  26.7178  32.51675  33.50165  36.6328  41.9279    50
    #     WeNYoBen  83.9070  98.4700 107.79965 101.88475 107.1933 170.2940    50
    #    M._Viking  73.5963  83.4467  93.99603  86.58535  94.0915 151.7075    50
    #        akrun  42.5265  43.2879  48.42697  44.98085  51.1533 105.2836    50
    #        frank  81.9115  90.1559  95.40261  93.97015  98.2921 129.6162    50
    #   Ronak_Shah 109.0678 121.8230 133.10690 125.63930 133.7222 231.5350    50

all.equal(Khaynes, jay.sf)
all.equal(Khaynes, jay.sf_2)
all.equal(Khaynes, thelatemail)
all.equal(Khaynes, WeNYoBen)
all.equal(Khaynes, M._Viking$lasttrue) # When the last element is TRUE it will return false.
all.equal(Khaynes, akrun)
all.equal(Khaynes, frank)
all.equal(Khaynes, x_dt) # Ronak Shah solution.

Simoniac

Nice follow up. @Misfile 's x>c(x[-1],0) is the winner. – O