Vectorizing loop over vector elements

Question

I find it hard to come up with a fast solution to the following problem:

I have a vector of observations, which indicates the time of observation of certain phenomena. <

Answer 1

I suspect that your 0 values are actually NA values. Here I make them NA and than use na.locf (Last Observation Carried Forward) from package zoo:

example <- c(0,0,0,1,0,1,1,0,0,0,-1,0,0,-1,-1,0,0,1,0,0)
res <- example
#res[res==0] <- NA
#the same but faster
res <- res/res*res
library(zoo)
res <- na.locf(res,  na.rm = FALSE)
res[is.na(res)] <- 0
cbind(example, res)
#       example res
#  [1,]       0   0
#  [2,]       0   0
#  [3,]       0   0
#  [4,]       1   1
#  [5,]       0   1
#  [6,]       1   1
#  [7,]       1   1
#  [8,]       0   1
#  [9,]       0   1
# [10,]       0   1
# [11,]      -1  -1
# [12,]       0  -1
# [13,]       0  -1
# [14,]      -1  -1
# [15,]      -1  -1
# [16,]       0  -1
# [17,]       0  -1
# [18,]       1   1
# [19,]       0   1
# [20,]       0   1

Answer 2

I am quite sure somebody will approach a better pure-R solution, but my first try is to use only 1 loop as follows:

x <- c(0,0,0,1,0,1,1,0,0,0,-1,0,0,-1,-1,0,0,1,0,0)

last <- x[1]
for (i in seq_along(x)) {
   if (x[i] == 0) x[i] <- last
   else last <- x[i] 
}

x
## [1]  0  0  0  1  1  1  1  1  1  1 -1 -1 -1 -1 -1 -1 -1  1  1  1

The above easily translates to an effective C++ code:

Rcpp::cppFunction('
NumericVector elimzeros(NumericVector x) {
   int n = x.size();
   NumericVector y(n);
   double last = x[0];
   for (int i=0; i<n; ++i) {
      if (x[i] == 0)
         y[i] = last;
      else
         y[i] = last = x[i];
   }
   return y;
}
')

elimzeros(x)
## [1]  0  0  0  1  1  1  1  1  1  1 -1 -1 -1 -1 -1 -1 -1  1  1  1

Some benchmarks:

set.seed(123L)
x <- sample(c(-1,0,1), replace=TRUE, 100000)
# ...
microbenchmark::microbenchmark(
   gagolews(x),
   gagolews_Rcpp(x),
   Roland(x),
   AndreyShabalin_match(x),
   AndreyShabalin_findInterval(x),
   AndreyShabalin_cumsum(x),
   unit="relative"
)
## Unit: relative
##                            expr        min         lq     median         uq        max neval
##                     gagolews(x) 167.264538 163.172532 162.703810 171.186482 110.604258   100
##                gagolews_Rcpp(x)   1.000000   1.000000   1.000000   1.000000   1.000000   100
##                       Roland(x)  33.817744  34.374521  34.544877  35.633136  52.825091   100
##         AndreyShabalin_match(x)  45.217805  43.819050  44.105279  44.800612  58.375625   100
##  AndreyShabalin_findInterval(x)  45.191419  43.832256  44.283284  45.094304  23.819259   100
##        AndreyShabalin_cumsum(x)   8.701682   8.367212   8.413992   9.938748   5.676467   100

Answer 3

I'll try to be the one to offer a pure R solution:

example <- c(0,0,0,1,0,1,1,0,0,0,-1,0,0,-1,-1,0,0,1,0,0);

cs = cumsum(example!=0);
mch = match(cs, cs);
desired.output = example[mch];

print(cbind(example,desired.output))

UPD: It may be faster to calculate mch above with

mch = findInterval(cs-1,cs)+1

UPD2: I like the answer by @Roland. It can be shortened to two lines:

NN = (example != 0);
desired.output = c(example[1], example[NN])[cumsum(NN) + 1L];