Get indexes of a vector of numbers in another vector

Question

Let's suppose we have the following vector:

v <- c(2,2,3,5,8,0,32,1,3,12,5,2,3,5,8,33,1)

Given a sequence of numbers, for instance c(2,3,5,8), I am trying to find what the position of this sequence of numbers is in the vector v. The result I expect is something like:

FALSE TRUE TRUE TRUE TRUE FALSE FALSE FALSE FALSE FALSE FALSE TRUE TRUE TRUE TRUE FALSE FALSE

I am trying to use which(v == c(2,3,5,8)), but it doesn't give me what I am looking for.

Keep in mind this will **not** be possible with floats due to the usual binary representation limits. You might be able to modify any of the given solutions, replacing `==` with `all.equal` or `cgwtools::approxeq` (tooting my own horn there) — Carl Witthoft, Feb 07 '18 at 14:34
Seems like you just a string search algorithm https://en.wikipedia.org/wiki/String_searching_algorithm — Alexander, Feb 08 '18 at 04:20
@Alexander A string search algorithm isn't the most efficient solution in this case. See [the benchmarks](https://stackoverflow.com/a/48708439/2204410) for examples. — Jaap, Feb 10 '18 at 07:08
@Jaap Which of those is a string search algorithm? Which string search algorithm is it, in particular? To be clear: I'm not suggesting to convert this data into a string type just to use a pre-made substring search algorithm on it. The data can be counted as a string itself, with an alphabet consistent of numbers rather than characters — Alexander, Feb 11 '18 at 19:09
link these 2 qns https://stackoverflow.com/questions/16244006/matching-a-sequence-in-a-larger-vector — chinsoon12, Jul 24 '21 at 06:18

talat · Accepted Answer · 2018-02-07T10:08:59.370

27

Using base R you could do the following:

v <- c(2,2,3,5,8,0,32,1,3,12,5,2,3,5,8,33,1)
x <- c(2,3,5,8)

idx <- which(v == x[1])
idx[sapply(idx, function(i) all(v[i:(i+(length(x)-1))] == x))]
# [1]  2 12

This tells you that the exact sequence appears twice, starting at positions 2 and 12 of your vector v.

It first checks the possible starting positions, i.e. where v equals the first value of x and then loops through these positions to check if the values after these positions also equal the other values of x.

edited Feb 07 '18 at 10:08

answered Feb 07 '18 at 10:05

talat

68,970
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3

Worth noting, `idx <- which(v == x[1])` is an important step. While other answers are going through all 1:4 shift variations 14 times, this answer does it in 3 steps. – zx8754 Feb 07 '18 at 10:51

Jaap · Answer 2 · 2018-02-09T15:02:38.717

17

Two other approaches using the shift-function trom data.table:

library(data.table)

# option 1
which(rowSums(mapply('==',
                     shift(v, type = 'lead', n = 0:(length(x) - 1)),
                     x)
              ) == length(x))

# option 2
which(Reduce("+", Map('==',
                      shift(v, type = 'lead', n = 0:(length(x) - 1)),
                      x)
             ) == length(x))

both give:

[1]  2 12

To get a full vector of the matching positions:

l <- length(x)
w <- which(Reduce("+", Map('==',
                           shift(v, type = 'lead', n = 0:(l - 1)),
                           x)
                  ) == l)
rep(w, each = l) + 0:(l-1)

which gives:

[1]  2  3  4  5 12 13 14 15

The benchmark which was included earlier in this answer has been moved to a separate community wiki answer.

Used data:

v <- c(2,2,3,5,8,0,32,1,3,12,5,2,3,5,8,33,1)
x <- c(2,3,5,8)

edited Feb 09 '18 at 15:02

answered Feb 07 '18 at 10:24

Jaap

81,064
34
182
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2

Many of these solutions don't give the desired output, the extra step is not cost free – moodymudskipper Feb 08 '18 at 11:29
1

Not sure whose idea it was, but getting a full vector of matching positions concatenated together seems like a bad idea. If I test with `x = c(1,1,1)` then I may find positions appearing multiple times. Besides, it's redundant -- the informational content of the first position is enough... Anyway, not a big deal, just my two cents ... noticed it all over the benchmarks. – Frank Feb 09 '18 at 22:11
1

@Frank I don't think it is a bad idea necessarily. It depends on what you want to do with it. I included it in the benchmarks to make sure every solution returns the same and thus get a fair comparison. – Jaap Feb 10 '18 at 07:04

jogo · Answer 3 · 2018-02-10T09:08:52.077

15

You can use rollapply() from zoo

v <- c(2,2,3,5,8,0,32,1,3,12,5,2,3,5,8,33,1)
x <- c(2,3,5,8)

library("zoo")
searchX <- function(x, X) all(x==X)
rollapply(v, FUN=searchX, X=x, width=length(x))

The result TRUEshows you the beginning of the sequence.
The code could be simplified to rollapply(v, length(x), identical, x) (thanks to G. Grothendieck):

set.seed(2)
vl <- as.numeric(sample(1:10, 1e6, TRUE))
# vm <- vl[1:1e5]
# vs <- vl[1:1e4]
x <- c(2,3,5)

library("zoo")
searchX <- function(x, X) all(x==X)
i1 <- rollapply(vl, FUN=searchX, X=x, width=length(x))
i2 <- rollapply(vl, width=length(x), identical, y=x)

identical(i1, i2)

For using identical() both arguments must be of the same type (num and int are not the same).
If needed == coerces int to num; identical() does not any coercion.

edited Feb 10 '18 at 09:08

answered Feb 07 '18 at 10:03

jogo

12,469
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42

Could you check your 2nd solution? As you can see in [the benchmark answer](https://stackoverflow.com/a/48708439/2204410) it doesn't return the same output as the other answers. – Jaap Feb 09 '18 at 15:04
1

I tried (also unsuccesfully) to repair it as well. I will remove it from the benchmarks. – Jaap Feb 09 '18 at 16:30
1

The code could be simplified to `rollapply(v, length(x), identical, x)` where `v` and `x` must be of the same type, e.g. both integer or both double, since for example `identical(5L, 5)` is `FALSE`. – G. Grothendieck Feb 10 '18 at 03:18
1

@G.Grothendieck Thx, that was indeed the issue. When both are of the same type, the solution with `identical` works. – Jaap Feb 10 '18 at 07:16

Frank · Answer 4 · 2018-02-07T21:38:19.483

I feel like looping should be efficient:

w = seq_along(v)
for (i in seq_along(x)) w = w[v[w+i-1L] == x[i]]

w 
# [1]  2 12

This should be writable in C++ following @SymbolixAU approach for extra speed.

A basic comparison:

# create functions for selected approaches
redjaap <- function(v,x)
  which(Reduce("+", Map('==', shift(v, type = 'lead', n = 0:(length(x) - 1)), x)) == length(x))
loop <- function(v,x){
  w = seq_along(v)
  for (i in seq_along(x)) w = w[v[w+i-1L] == x[i]]
  w
}

# check consistency
identical(redjaap(v,x), loop(v,x))
# [1] TRUE

# check speed
library(microbenchmark)
vv <- rep(v, 1e4)
microbenchmark(redjaap(vv,x), loop(vv,x), times = 100)
# Unit: milliseconds
#            expr      min       lq      mean   median       uq       max neval cld
#  redjaap(vv, x) 5.883809 8.058230 17.225899 9.080246 9.907514  96.35226   100   b
#     loop(vv, x) 3.629213 5.080816  9.475016 5.578508 6.495105 112.61242   100  a 

# check consistency again
identical(redjaap(vv,x), loop(vv,x))
# [1] TRUE

SymbolixAU · Answer 5 · 2018-07-16T22:23:19.907

Here are two Rcpp solutions. The first one returns the location of v that is the starting position of the sequence.

library(Rcpp)

v <- c(2,2,3,5,8,0,32,1,3,12,5,2,3,5,8,33,1)
x <- c(2,3,5,8)

cppFunction('NumericVector SeqInVec(NumericVector myVector, NumericVector mySequence) {

    int vecSize = myVector.size();
    int seqSize = mySequence.size();
    NumericVector comparison(seqSize);
    NumericVector res(vecSize);

    for (int i = 0; i < vecSize; i++ ) {

        for (int j = 0; j < seqSize; j++ ) {
                comparison[j] = mySequence[j] == myVector[i + j];
        }

        if (sum(comparison) == seqSize) {
            res[i] = 1;
        }else{
            res[i] = 0;
        }
    }

    return res;

    }')

SeqInVec(v, x)
#[1] 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0

This second one returns the index values (as per the other answers) of every matched entry in the sequence.

cppFunction('NumericVector SeqInVec(NumericVector myVector, NumericVector mySequence) {

  int vecSize = myVector.size();
  int seqSize = mySequence.size();
  NumericVector comparison(seqSize);
  NumericVector res(vecSize);
  int foundCounter = 0;

  for (int i = 0; i < vecSize; i++ ) {

    for (int j = 0; j < seqSize; j++ ) {
      comparison[j] = mySequence[j] == myVector[i + j];
    }

    if (sum(comparison) == seqSize) {
      for (int j = 0; j < seqSize; j++ ) {
        res[foundCounter] = i + j + 1;
        foundCounter++;
      }
    }
  }

  IntegerVector idx = seq(0, (foundCounter-1));
  return res[idx];
}')

SeqInVec(v, x)
# [1]  2  3  4  5 12 13 14 15

Optimising

As @MichaelChirico points out in their comment, further optimisations can be made. For example, if we know the first entry in the sequence doesn't match a value in the vector, we don't need to do the rest of the comparison

cppFunction('NumericVector SeqInVecOpt(NumericVector myVector, NumericVector mySequence) {

  int vecSize = myVector.size();
  int seqSize = mySequence.size();
  NumericVector comparison(seqSize);
  NumericVector res(vecSize);
  int foundCounter = 0;

  for (int i = 0; i < vecSize; i++ ) {

    if (myVector[i] == mySequence[0]) {
        for (int j = 0; j < seqSize; j++ ) {
          comparison[j] = mySequence[j] == myVector[i + j];
        }

        if (sum(comparison) == seqSize) {
          for (int j = 0; j < seqSize; j++ ) {
            res[foundCounter] = i + j + 1;
            foundCounter++;
          }
        }
    }
  }

  IntegerVector idx = seq(0, (foundCounter-1));
  return res[idx];
}')

The answer with benchmarks shows the performance of these approaches

since you're examining subsequent elements, shouldn't there be a way to optimize by skipping elements we already know don't start the sequence? e.g. in OPs example when checking at the second `2` we already know the 3rd element is not `2` so we can skip checking the elements after `3` — MichaelChirico, Feb 11 '18 at 11:18
@MichaelChirico - yes we can - thanks for the pointer. I've edited it with a new solution, but welcome any other suggestions — SymbolixAU, Feb 11 '18 at 21:02
2-3x speed-up, nice! I guess the improvement depends on the length of the "search" string and its density (% of `TRUE` values). — MichaelChirico, Feb 12 '18 at 00:09
@MichaelChirico - yes that will likely be a factor. I've also tested a variation where it will increment `i` by the size of the search string, rather than one each time. In this example I didn't see any improvement, however. — SymbolixAU, Feb 12 '18 at 00:13
This solution has a problem when searched sequence is at the end of vector. For example, if you change v to (appending the searched sequence) v <- c(2,2,3,5,8,0,32,1,3,12,5,2,3,5,8,33,1,2,3,5,8) x <- c(2,3,5,8) the last occurrence of x in v will not be found. — Samo, Jul 16 '18 at 18:15
@Samo - yes I see what you're saying. I think the `i` loop will need to iterate to `i < vecSize`, rather than to `i < (vecSize - seqSize)`. Let me double check. — SymbolixAU, Jul 16 '18 at 21:53

score 9 · Answer 6 · edited Jul 16 '18 at 22:22

A benchmark on the posted answers:

Load the needed packages:

library(data.table)
library(microbenchmark)
library(Rcpp)
library(zoo)

Creating vector with which the benchmarks will be run:

set.seed(2)
vl <- sample(1:10, 1e6, TRUE)
vm <- vl[1:1e5]
vs <- vl[1:1e4]
x <- c(2,3,5)

Testing whether all solution give the same outcome on the small vector vs:

> all.equal(jaap1(vs,x), jaap2(vs,x))
[1] TRUE
> all.equal(jaap1(vs,x), docendo(vs,x))
[1] TRUE
> all.equal(jaap1(vs,x), a5c1(vs,x))
[1] TRUE
> all.equal(jaap1(vs,x), jogo1(vs,x))
[1] TRUE
> all.equal(jaap1(vs,x), moody(vs,x))
[1] "Numeric: lengths (24, 873) differ"
> all.equal(jaap1(vs,x), cata1(vs,x))
[1] "Numeric: lengths (24, 0) differ"
> all.equal(jaap1(vs,x), u989(vs,x))
[1] TRUE
> all.equal(jaap1(vs,x), frank(vs,x))
[1] TRUE
> all.equal(jaap1(vs,x), symb(vs,x))
[1] TRUE
> all.equal(jaap1(vs, x), symbOpt(vs, x))
[1] TRUE

Further inspection of the cata1 and moody solutions learns that they don't give the desired output. They are therefore not included in the benchmarks.

The benchmark for the smallest vector vs:

mbs <- microbenchmark(jaap1(vs,x), jaap2(vs,x), docendo(vs,x), a5c1(vs,x),
                      jogo1(vs,x), u989(vs,x), frank(vs,x), symb(vs,x), symbOpt(vs, x),
                      times = 100)

gives:

 print(mbs, order = "median")

 Unit: microseconds
          expr       min         lq        mean     median         uq        max neval
  symbOpt(vs, x)    40.658    47.0565    78.47119    51.5220    56.2765   2170.708   100
     symb(vs, x)   106.208   112.7885   151.76398   117.0655   123.7450   1976.360   100
    frank(vs, x)   121.303   129.0515   203.13616   132.1115   137.9370   6193.837   100
    jaap2(vs, x)   187.973   218.7805   322.98300   235.0535   255.2275   6287.548   100
    jaap1(vs, x)   306.944   341.4055   452.32426   358.2600   387.7105   6376.805   100
     a5c1(vs, x)   463.721   500.9465   628.13475   516.2845   553.2765   6179.304   100
  docendo(vs, x)  1139.689  1244.0555  1399.88150  1313.6295  1363.3480   9516.529   100
     u989(vs, x)  8048.969  8244.9570  8735.97523  8627.8335  8858.7075  18732.750   100
    jogo1(vs, x) 40022.406 42208.4870 44927.58872 43733.8935 45008.0360 124496.190   100

The benchmark for the medium vector vm:

mbm <- microbenchmark(jaap1(vm,x), jaap2(vm,x), docendo(vm,x), a5c1(vm,x),
                      jogo1(vm,x), u989(vm,x), frank(vm,x), symb(vm,x), symbOpt(vm, x),
                      times = 100)

gives:

print(mbm, order = "median")

Unit: microseconds
           expr        min          lq        mean      median         uq        max neval
 symbOpt(vm, x)    357.452    405.0415    974.9058    763.0205   1067.803   7444.126   100
    symb(vm, x)   1032.915   1117.7585   1923.4040   1422.1930   1753.044  17498.132   100
   frank(vm, x)   1158.744   1470.8170   1829.8024   1826.1330   1935.641   6423.966   100
   jaap2(vm, x)   1622.183   2872.7725   3798.6536   3147.7895   3680.954  14886.765   100
   jaap1(vm, x)   3053.024   4729.6115   7325.3753   5607.8395   6682.814  87151.774   100
    a5c1(vm, x)   5487.547   7458.2025   9612.5545   8137.1255   9420.684  88798.914   100
 docendo(vm, x)  10780.920  11357.7440  13313.6269  12029.1720  13411.026  21984.294   100
    u989(vm, x)  83518.898  84999.6890  88537.9931  87675.3260  90636.674 105681.313   100
   jogo1(vm, x) 471753.735 512979.3840 537232.7003 534780.8050 556866.124 646810.092   100

The benchmark for the largest vector vl:

mbl <- microbenchmark(jaap1(vl,x), jaap2(vl,x), docendo(vl,x), a5c1(vl,x),
                      jogo1(vl,x), u989(vl,x), frank(vl,x), symb(vl,x), symbOpt(vl, x),
                      times = 100)

gives:

  print(mbl, order = "median")

Unit: milliseconds
           expr         min          lq       mean     median         uq       max neval
 symbOpt(vl, x)    4.679646    5.768531   12.30079    6.67608   11.67082  118.3467   100
    symb(vl, x)   11.356392   12.656124   21.27423   13.74856   18.66955  149.9840   100
   frank(vl, x)   13.523963   14.929656   22.70959   17.53589   22.04182  132.6248   100
   jaap2(vl, x)   18.754847   24.968511   37.89915   29.78309   36.47700  145.3471   100
   jaap1(vl, x)   37.047549   52.500684   95.28392   72.89496  138.55008  234.8694   100
    a5c1(vl, x)   54.563389   76.704769  116.89269   89.53974  167.19679  248.9265   100
 docendo(vl, x)  109.824281  124.631557  156.60513  129.64958  145.47547  296.0214   100
    u989(vl, x) 1380.886338 1413.878029 1454.50502 1436.18430 1479.18934 1632.3281   100
   jogo1(vl, x) 4067.106897 4339.005951 4472.46318 4454.89297 4563.08310 5114.4626   100

The used functions of each solution:

jaap1 <- function(v,x) {
  l <- length(x);
  w <- which(rowSums(mapply('==', shift(v, type = 'lead', n = 0:(length(x) - 1)), x) ) == length(x));
  rep(w, each = l) + 0:(l-1)
}

jaap2 <- function(v,x) {
  l <- length(x);
  w <- which(Reduce("+", Map('==', shift(v, type = 'lead', n = 0:(length(x) - 1)), x)) == length(x));
  rep(w, each = l) + 0:(l-1)
}

docendo <- function(v,x) {
  l <- length(x);
  idx <- which(v == x[1]);
  w <- idx[sapply(idx, function(i) all(v[i:(i+(length(x)-1))] == x))];
  rep(w, each = l) + 0:(l-1)
}

a5c1 <- function(v,x) {
  l <- length(x);
  w <- which(colSums(t(embed(v, l)[, l:1]) == x) == l);
  rep(w, each = l) + 0:(l-1)
}

jogo1 <- function(v,x) {
  l <- length(x);
  searchX <- function(x, X) all(x==X);
  w <- which(rollapply(v, FUN=searchX, X=x, width=l));
  rep(w, each = l) + 0:(l-1)
}

moody <- function(v,x) {
  l <- length(x);
  v2 <- as.numeric(factor(c(v,NA),levels = x));
  v2[is.na(v2)] <- l+1;
  which(diff(v2) == 1)
}

cata1 <- function(v,x) {
  l <- length(x);
  w <- which(sapply(lapply(seq(length(v)-l)-1, function(i) v[seq(x)+i]), identical, x));
  rep(w, each = l) + 0:(l-1)
}

u989 <- function(v,x) {
  l <- length(x);
  s <- paste(v, collapse = '-');
  p <- paste0('\\b', paste(x, collapse = '-'), '\\b');
  i <- c(1, unlist(gregexpr(p, s)));
  m <- substring(s, head(i,-1), tail(i,-1));
  ln <- lengths(strsplit(m, '-'));
  w <- cumsum(c(ln[1], ln[-1]-1));
  rep(w, each = l) + 0:(l-1)
}

frank <- function(v,x) {
  l <- length(x);
  w = seq_along(v);
  for (i in seq_along(x)) w = w[v[w+i-1L] == x[i]];
  rep(w, each = l) + 0:(l-1)
}

cppFunction('NumericVector SeqInVec(NumericVector myVector, NumericVector mySequence) {

            int vecSize = myVector.size();
            int seqSize = mySequence.size();
            NumericVector comparison(seqSize);
            NumericVector res(vecSize);
            int foundCounter = 0;

            for (int i = 0; i < vecSize; i++ ) {

            for (int j = 0; j < seqSize; j++ ) {
            comparison[j] = mySequence[j] == myVector[i + j];
            }

            if (sum(comparison) == seqSize) {
            for (int j = 0; j < seqSize; j++ ) {
            res[foundCounter] = i + j + 1;
            foundCounter++;
            }
            }
            }

            IntegerVector idx = seq(0, (foundCounter-1));
            return res[idx];
            }')

symb <- function(v,x) {SeqInVec(v, x)}

cppFunction('NumericVector SeqInVecOpt(NumericVector myVector, NumericVector mySequence) {

  int vecSize = myVector.size();
  int seqSize = mySequence.size();
  NumericVector comparison(seqSize);
  NumericVector res(vecSize);
  int foundCounter = 0;

  for (int i = 0; i < vecSize; i++ ) {

        if (myVector[i] == mySequence[0]) {
        for (int j = 0; j < seqSize; j++ ) {
          comparison[j] = mySequence[j] == myVector[i + j];
        }

        if (sum(comparison) == seqSize) {
          for (int j = 0; j < seqSize; j++ ) {
            res[foundCounter] = i + j + 1;
            foundCounter++;
          }
        }
        }
  }

  IntegerVector idx = seq(0, (foundCounter-1));
  return res[idx];
}')

symbOpt <- function(v,x) {SeqInVecOpt(v,x)}

Since this is a cw-answer I'll add my own benchmark of some of the answers.

library(data.table)
library(microbenchmark)

set.seed(2); v <- sample(1:100, 5e7, TRUE); x <- c(2,3,5)

jaap1 <- function(v, x) { 
  which(rowSums(mapply('==',shift(v, type = 'lead', n = 0:(length(x) - 1)),
                       x)) == length(x)) 
}

jaap2 <- function(v, x) {
  which(Reduce("+", Map('==',shift(v, type = 'lead', n = 0:(length(x) - 1)),
                        x)) == length(x))
}

dd1 <- function(v, x) {
  idx <- which(v == x[1])
  idx[sapply(idx, function(i) all(v[i:(i+(length(x)-1))] == x))]
}

dd2 <- function(v, x) {
  idx <- which(v == x[1L])
  xl <- length(x) - 1L
  idx[sapply(idx, function(i) all(v[i:(i+xl)] == x))]
}

frank <- function(v, x) {
  w = seq_along(v)
  for (i in seq_along(x)) w = w[v[w+i-1L] == x[i]]
  w 
}

all.equal(jaap1(v, x), dd1(v, x))
all.equal(jaap2(v, x), dd1(v, x))
all.equal(dd2(v, x), dd1(v, x))
all.equal(frank(v, x), dd1(v, x))

bm <- microbenchmark(jaap1(v, x), jaap2(v, x), dd1(v, x), dd2(v, x), frank(v, x), 
                     unit = "relative", times = 25)

plot(bm)

Imgur

bm
Unit: relative
        expr      min       lq     mean   median       uq       max neval
 jaap1(v, x) 4.487360 4.591961 4.724153 4.870226 4.660023 3.9361093    25
 jaap2(v, x) 2.026052 2.159902 2.116204 2.282644 2.138106 2.1133068    25
   dd1(v, x) 1.078059 1.151530 1.119067 1.257337 1.201762 0.8646835    25
   dd2(v, x) 1.000000 1.000000 1.000000 1.000000 1.000000 1.0000000    25
 frank(v, x) 1.400735 1.376405 1.442887 1.427433 1.611672 1.3440097    25

Bottom line: without knowing the real data, all these benchmarks don't tell the whole story.

score 4 · Answer 7 · edited May 12 '22 at 11:06

Here's a solution that leverages binary search on secondary indices in data.table. (Great vignette here)

This method has quite a bit of overhead so it's not particularly competitive on the 1e4 length vector in the benchmark, but it hangs near the top of the pack as the size increases.

matt <- function(v,x){
  l <- length(x);
  SL <- seq_len(l-1);
  DT <- data.table(Seq_0 = v);
  for (i in SL) set(DT, j = eval(paste0("Seq_",i)), value = shift(DT[["Seq_0"]],n = i, type = "lead"));
  w <- DT[as.list(x),on = paste0("Seq_",c(0L,SL)), which = TRUE];
  rep(w, each = l) + 0:(l-1)
}

Benchmarking

library(data.table)
library(microbenchmark)
library(Rcpp)
library(zoo)

set.seed(2)
vl <- sample(1:10, 1e6, TRUE)
vm <- vl[1:1e5]
vs <- vl[1:1e4]
x <- c(2,3,5)

Vector Length 1e4

Unit: microseconds
           expr       min        lq       mean     median        uq       max neval
    symb(vs, x)   138.342   143.048   161.6681   153.1545   159.269   259.999    10
   frank(vs, x)   176.634   184.129   198.8060   193.2850   200.701   257.050    10
   jaap2(vs, x)   282.231   299.025   342.5323   316.5185   337.760   524.212    10
   jaap1(vs, x)   490.013   528.123   568.6168   538.7595   547.268   731.340    10
    a5c1(vs, x)   706.450   742.270   751.3092   756.2075   758.859   793.446    10
     dd2(vs, x)  1319.098  1348.082  2061.5579  1363.2265  1497.960  7913.383    10
 docendo(vs, x)  1427.768  1459.484  1536.6439  1546.2135  1595.858  1696.070    10
     dd1(vs, x)  1377.502  1406.272  2217.2382  1552.5030  1706.131  8084.474    10
    matt(vs, x)  1928.418  2041.597  2390.6227  2087.6335  2430.470  4762.909    10
    u989(vs, x)  8720.330  8821.987  8935.7188  8882.0190  9106.705  9163.967    10
   jogo1(vs, x) 47123.615 47536.700 49158.2600 48449.2390 50957.035 52496.981    10

Vector Length 1e5

Unit: milliseconds
           expr        min         lq       mean     median         uq        max neval
    symb(vm, x)   1.319921   1.378801   1.464972   1.423782   1.577006   1.682156    10
   frank(vm, x)   1.671155   1.739507   1.806548   1.760738   1.844893   2.097404    10
   jaap2(vm, x)   2.298449   2.380281   2.683813   2.432373   2.566581   4.310258    10
    matt(vm, x)   3.195048   3.495247   3.577080   3.607060   3.687222   3.844508    10
   jaap1(vm, x)   4.079117   4.179975   4.776989   4.496603   5.206452   6.295954    10
    a5c1(vm, x)   6.488621   6.617709   7.366226   6.720107   6.877529  12.500510    10
     dd2(vm, x)  12.595699  12.812876  14.990739  14.058098  16.758380  20.743506    10
 docendo(vm, x)  13.635357  13.999721  15.296075  14.729947  16.151790  18.541582    10
     dd1(vm, x)  13.474589  14.177410  15.676348  15.446635  17.150199  19.085379    10
    u989(vm, x)  94.844298  95.026733  96.309658  95.134400  97.460869 100.536654    10
   jogo1(vm, x) 575.230741 581.654544 621.824297 616.474265 628.267155 723.010738    10

Vector Length 1e6

Unit: milliseconds
           expr        min         lq       mean     median         uq        max neval
    symb(vl, x)   13.34294   13.55564   14.01556   13.61847   14.78210   15.26076    10
   frank(vl, x)   17.35628   17.45602   18.62781   17.56914   17.88896   25.38812    10
    matt(vl, x)   20.79867   21.07157   22.41467   21.23878   22.56063   27.12909    10
   jaap2(vl, x)   22.81464   22.92414   22.96956   22.99085   23.02558   23.10124    10
   jaap1(vl, x)   40.00971   40.46594   43.01407   41.03370   42.81724   55.90530    10
    a5c1(vl, x)   65.39460   65.97406   69.27288   66.28000   66.72847   83.77490    10
     dd2(vl, x)  127.47617  132.99154  161.85129  134.63168  157.40028  342.37526    10
     dd1(vl, x)  140.06140  145.45085  154.88780  154.23280  161.90710  171.60294    10
 docendo(vl, x)  147.07644  151.58861  162.20522  162.49216  165.49513  183.64135    10
    u989(vl, x) 2022.64476 2041.55442 2055.86929 2054.92627 2066.26187 2088.71411    10
   jogo1(vl, x) 5563.31171 5632.17506 5863.56265 5872.61793 6016.62838 6244.63205    10

score 3 · Answer 8 · edited May 11 '22 at 23:19

Here is a string-based approach in base R:

str <- paste(v, collapse = '-')
# "2-2-3-5-8-0-32-1-3-12-5-2-3-5-8-33-1"

pattern <- paste0('\\b', paste(x, collapse = '-'), '\\b')
# "\\b2-3-5-8\\b"

inds <- unlist(gregexpr(pattern, str)) # (1)
# 3 25
sapply(inds, function(i) lengths(strsplit(substr(str, 1, i),'-'))) # (2)

# [1]  2 12

\\b is used for exact matching.
(1) Finds the positions at which pattern is seen within str.
(2) Getting back the respective indices within the original vector v.

As for running-time efficiency, here is a much faster solution than my first solution:

str <- paste(v, collapse = '-')
pattern <- paste0('\\b', paste(x, collapse = '-'), '\\b')

inds <- c(1, unlist(gregexpr(pattern, str)))

m <- substring(str, head(inds,-1), tail(inds,-1))
ln <- lengths(strsplit(m, '-'))
cumsum(c(ln[1], ln[-1]-1))

score 1 · Answer 9 · edited May 11 '22 at 23:16

1

Note: This doesn't always give the desired output

We can convert v to factors and keep only consecutive values in our transformed vector:

v2 <- as.numeric(factor(c(v,NA),levels = x)) # [1]  1  1  2  3  4 NA NA NA ...
v2[is.na(v2)] <- length(x)+1                 # [1]  1  1  2  3  4  5  5  5 ...
output <- diff(v2) ==1
# [1] FALSE  TRUE  TRUE  TRUE  TRUE FALSE FALSE FALSE FALSE FALSE FALSE  TRUE  TRUE  TRUE  TRUE FALSE FALSE

Data

v <- c(2,2,3,5,8,0,32,1,3,12,5,2,3,5,8,33,1)
x <- c(2,3,5,8)

edited May 11 '22 at 23:16

Peter Mortensen

30,738
21
105
131

answered Feb 07 '18 at 13:03

moodymudskipper

46,417
11
121
167

1

that's pretty computationally intensive. – Carl Witthoft Feb 07 '18 at 14:30
is it ? I don't know, it's the only fully vectorized solution so far, too many copies ? – moodymudskipper Feb 07 '18 at 14:33
I plead guilty to not having run `microbenchmark` on the various answers here. It's just a gut feeling because of the number of class coercions going on there. – Carl Witthoft Feb 07 '18 at 14:40
@CarlWitthoft, I guess that the answers by catastrophic-failure, which both utilise nested loops, will be much slower. But I too haven't tested any. – talat Feb 07 '18 at 14:51

score 0 · Answer 10 · edited May 12 '22 at 11:07

I have modified talat's solution as I found this didn't work in all scenarios.

Firstly, if this step idx[sapply(idx, function(i) all(v[i:(i+(length(x)-1))] == x))] will produce NAs if v[i:(i+(length(x)-1))] == x)) contains NAs and no FALSE. Secondly, in order to match the desired outcome I have used the indices to create the final logical vector as desired.

seq_detect <- function(v, x) {
  #If the integer is not detected then return early a vector of all falses
  if(!any(v == x[1])){
    return(vector(length = length(v)))
  }
  #Create an index of v where the first value in x appears
  idx <- which(v == x[1])

  #See if each of those indices do indeed match the whole pattern
  index_seq_start_raw <- idx[sapply(idx, function(i) all(v[i:(i+(length(x)-1))] == x))]

  #These may return NAs if above index outside range of 1:length(v)
  if(all(is.na(index_seq_start_raw))){
    return(vector(length = length(v)))
  }

  #If some NAs then remove these
  (index_seq_start <- index_seq_start_raw[!is.na(index_seq_start_raw)])

  #Create template of FALSES for output
  output <- vector(length = length(v))

  #Loop over index_seq_start and replace any matches with TRUEs
  for(i in seq_along(1:length(index_seq_start))){
    output[(index_seq_start[i]):(index_seq_start[i]+3)] <- TRUE
  }
  output
}

#This works on both the following pairs of vectors, where as due to indexing
#issues @talat's solution causes an error with v1 and x1.
v <- c(2, 2, 3, 5, 8, 0, 32, 1, 3, 12, 5, 2, 3, 5, 8, 33, 1)
x <- c(2, 3, 5, 8)

[1] FALSE  TRUE  TRUE  TRUE  TRUE FALSE FALSE FALSE FALSE FALSE FALSE  TRUE  TRUE  TRUE  TRUE FALSE FALSE

v1 <- c(1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1)
x1 <- c(1, 2, 2, 1)

[1] FALSE FALSE FALSE FALSE FALSE  TRUE  TRUE  TRUE  TRUE FALSE FALSE FALSE FALSE

Get indexes of a vector of numbers in another vector

10 Answers10

Optimising

A benchmark on the posted answers:

Benchmarking

Vector Length 1e4

Vector Length 1e5

Vector Length 1e6

Linked

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