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 is the position of this sequence of numbers 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.
Thanks beforehand.
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.
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)
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.
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
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
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
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)
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.
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.
Hats off to everyone else posting solutions, learning a lot from this question.
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)
}
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)
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
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
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
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.pattern is seen within str.v.UPDATE
As for the discussion of 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))
EDIT: some have noted that my answer doesn't always give the desired output, I might fix it later, caution meanwhile!
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)
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