CODEWITHSUNDEEP
r
dimitris_ps
dimitris_ps

Reputation: 5951

Conditionally replace elements of a vector based on an index

It's best explained with an example.

I have a vector, or column from data.frame named vec:

vec <- c(NA, NA, 1, NA, NA, NA, NA, 1, NA, NA, NA, NA, NA, NA)

I would like a vectorized process (not a for loop) to change the three trailing NA when a 1 is observed.

The end vector would be:

c(NA, NA, 1, 1, 1, 1, NA, 1, 1, 1, 1, NA, NA, NA)

If we had:

vec <- c(NA, NA, 1, NA, 1, NA, NA, 1, NA, NA, NA, NA, NA, NA)

The end vector would look like:

c(NA, NA, 1, 1, 1, 1, 1, 1, 1, 1, 1, NA, NA, NA)

A very badly written solution is:

vec2 <- vec
for(i in index(v)){
  if(!is.na(v[i])) vec2[i] <- 1
  if(i>3){
    if(!is.na(vec[i-1])) vec2[i] <- 1
    if(!is.na(vec[i-2])) vec2[i] <- 1
    if(!is.na(vec[i-3])) vec2[i] <- 1
  }
  if(i==3){
    if(!is.na(vec[i-1])) vec2[i] <- 1
    if(!is.na(vec[i-2])) vec2[i] <- 1
  }
  if(i==2){
    if(!is.na(vec[i-1])) vec2[i] <- 1
  }
}

Upvotes: 13

Views: 2632

Answers (7)

nicola
nicola

Reputation: 24480

Another option:

`[<-`(vec,c(outer(which(vec==1),1:3,"+")),1)
# [1] NA NA  1  1  1  1 NA  1  1  1  1 NA NA NA

Although the above works with the examples, it stretches the length of vec if a 1 is found in the last positions. Better to make a simple check and wrap into a function:

threeNAs<-function(vec) {
    ind<-c(outer(which(vec==1),1:3,"+"))
    ind<-ind[ind<=length(vec)]
    `[<-`(vec,ind,1)
}

Upvotes: 19

Jaap
Jaap

Reputation: 83215

Another fast solution:

vec[rep(which(vec == 1), each = 3) + c(1:3)] <- 1

which gives:

> vec
 [1] NA NA  1  1  1  1 NA  1  1  1  1 NA NA NA

Benchmarking is only really useful when done on larger datasets. A benchmark with a 10k larger vector and the several posted solutions:

library(microbenchmark)

microbenchmark(ans.jaap = {vec <- rep(c(NA, NA, 1, NA, NA, NA, NA, 1, NA, NA, NA, NA, NA, NA),1e4); 
                           vec[rep(which(vec == 1), each = 3) + c(1:3)] <- 1},
               ans.989 = {vec <- rep(c(NA, NA, 1, NA, NA, NA, NA, 1, NA, NA, NA, NA, NA, NA),1e4);
                          r <- which(vec==1);
                          vec[c(mapply(seq, r, r+3))] <- 1},
               ans.sotos = {vec <- rep(c(NA, NA, 1, NA, NA, NA, NA, 1, NA, NA, NA, NA, NA, NA),1e4);
                            vec[unique(as.vector(t(sapply(which(vec == 1), function(i) seq(i+1, length.out = 3)))))] <- 1},
               ans.gregor = {vec <- rep(c(NA, NA, 1, NA, NA, NA, NA, 1, NA, NA, NA, NA, NA, NA),1e4);
                             vec[is.na(vec)] <- 0;
                             n <- length(vec);
                             vec <- vec + c(0, vec[1:(n-1)]) + c(0, 0, vec[1:(n-2)]) + c(0, 0, 0, vec[1:(n-3)]);
                             vec[vec == 0] <- NA},
               ans.moody = {vec <- rep(c(NA, NA, 1, NA, NA, NA, NA, 1, NA, NA, NA, NA, NA, NA),1e4);
                            output <- sapply(1:length(vec),function(i){any(!is.na(vec[max(0,i-3):i]))});
                            output[output] <- 1;
                            output[output==0] <- NA},
               ans.nicola = {vec <- rep(c(NA, NA, 1, NA, NA, NA, NA, 1, NA, NA, NA, NA, NA, NA),1e4);
                             `[<-`(vec,c(outer(which(vec==1),1:3,"+")),1)})

which gives the following benchmark:

Unit: microseconds
       expr        min         lq       mean     median         uq        max neval   cld
   ans.jaap   1778.905   1937.414   3064.686   2100.595   2257.695  86233.593   100 a    
    ans.989  87688.166  89638.133  96992.231  90986.269  93326.393 182431.366   100   c  
  ans.sotos 125344.157 127968.113 132386.664 130117.438 132951.380 214460.174   100    d 
 ans.gregor   4036.642   5824.474  10861.373   6533.791   7654.587  87806.955   100  b   
  ans.moody 173146.810 178369.220 183698.670 180318.799 184000.062 264892.878   100     e
 ans.nicola    966.927   1390.486   1723.395   1604.037   1904.695   3310.203   100 a

Upvotes: 14

Sotos
Sotos

Reputation: 51582

A non-Vectorized solution, but nevertheless, another option using base R,

vec[unique(as.vector(t(sapply(which(vec == 1), function(i) seq(i+1, length.out = 3)))))] <- 1
vec
#[1] NA NA  1  1  1  1 NA  1  1  1  1 NA NA NA

vec1[unique(as.vector(t(sapply(which(vec1 == 1), function(i) seq(i+1, length.out = 3)))))] <- 1
vec1
#[1] NA NA  1  1  1  1  1  1  1  1  1 NA NA NA

Upvotes: 2

moodymudskipper
moodymudskipper

Reputation: 47300

With sapply, any, and is.na:

output <- sapply(1:length(vec),function(i){any(!is.na(vec[max(0,i-3):i]))})
output[output] <- 1
output[output==0] <- NA

Upvotes: 0

989
989

Reputation: 12937

How about this:

r <- which(vec==1)
vec[c(mapply(seq, r, r+3))] <- 1

Examples:

vec <- c(NA, NA, 1, NA, NA, NA, NA, 1, NA, NA, NA, NA, NA, NA)
#[1] NA NA  1  1  1  1 NA  1  1  1  1 NA NA NA

vec <- c(NA, NA, 1, NA, 1, NA, NA, 1, NA, NA, NA, NA, NA, NA)
#[1] NA NA  1  1  1  1  1  1  1  1  1 NA NA NA

Upvotes: -1

SymbolixAU
SymbolixAU

Reputation: 26248

What really is 'vectorised', if not a loop written in a C-language?

Here's a C++ loop that benchmarks well. 

vec <- c(NA, NA, 1, NA, NA, NA, NA, 1, NA, NA, NA, NA, NA, NA)

library(Rcpp)

cppFunction('NumericVector fixVec(NumericVector myVec){

    int n = myVec.size();
    int foundCount = 0;

    for(int i = 0; i < n; i++){
      if(myVec[i] == 1) foundCount = 1; 

      if(ISNA(myVec[i])){
        if(foundCount >= 1 & foundCount <= 3){
          myVec[i] = 1;
          foundCount++;
        }
      }
    }
    return myVec;
    }')

fixVec(vec)
# [1] NA NA  1  1  1  1 NA  1  1  1  1 NA NA NA

Benchmarks

library(microbenchmark)

microbenchmark(
      ans.jaap = {
        vec <- rep(c(NA, NA, 1, NA, NA, NA, NA, 1, NA, NA, NA, NA, NA, NA),1e4); 
      vec[rep(which(vec == 1), each = 4) + c(0:3)] <- 1
},

    ans.nicola = {
        vec <- rep(c(NA, NA, 1, NA, NA, NA, NA, 1, NA, NA, NA, NA, NA, NA),1e4);
      `[<-`(vec,c(outer(which(vec==1),0:3,"+")),1)
        },

    ans.symbolix = {
        vec <- rep(c(NA, NA, 1, NA, NA, NA, NA, 1, NA, NA, NA, NA, NA, NA),1e4);
      vec <- fixVec(vec)
        }
)

# Unit: microseconds
# expr              min       lq      mean   median        uq       max neval
# ans.jaap     2017.789 2264.318 2905.2437 2579.315 3588.4850  4667.249   100
# ans.nicola   1242.002 1626.704 3839.4768 2095.311 3066.4795 81299.962   100
# ans.symbolix  504.577  533.426  838.5661  718.275  966.9245  2354.373   100


vec <- rep(c(NA, NA, 1, NA, NA, NA, NA, 1, NA, NA, NA, NA, NA, NA),1e4)
vec <- fixVec(vec)

vec2 <- rep(c(NA, NA, 1, NA, NA, NA, NA, 1, NA, NA, NA, NA, NA, NA),1e4)
vec2[rep(which(vec2 == 1), each = 4) + c(0:3)] <- 1

identical(vec, vec2)
# [1] TRUE

Upvotes: 5

Gregor de Cillia
Gregor de Cillia

Reputation: 7635

The following code does what you asked for. It involves "shifting" the vector and then adding the shifted versions

vec[is.na(vec)] <- 0                                 
n <- length(vec)                                     
vec <- vec + c(0, vec[1:(n-1)]) + c(0, 0, vec[1:(n-2)]) + c(0, 0, 0, vec[1:(n-3)])  
vec[vec == 0] <- NA                                    
vec[vec != 0] <- 1                                     

# vec                    |   0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0 ,0, 0
# c(0, vec[1:(n-1)])     | + 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0 ,0, 0
# c(0, 0, vec[1:(n-2)])  | + 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0 ,0
# c(0,0,0,vec[1:(n-3)])  | + 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0 
#                        |-------------------------------------------
#                        |   0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0

Upvotes: 3

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