Question: Efficiently match all values of a vector in another vector

I'm looking to find an efficient method of matching all values of vector x in vector y rather than just the first position, as is returned by match().  What I'm after essentially is the default behavior of pmatch() but without partial matching:
x <- c(3L, 1L, 2L, 3L, 3L, 2L)
y <- c(3L, 3L, 3L, 3L, 1L, 3L)

Expected output:
pmatch(x, y)  
[1]  1  5 NA  2  3 NA

One way is to use ave() however this becomes slow and very memory inefficient as the number of groups increases:
ave(x, x, FUN = \(v) which(y == v[1])[1:length(v)])
[1]  1  5 NA  2  3 NA

Can anyone recommend an efficient way to achieve this in preferably (but not mandatory) base R?
Larger dataset for benchmarking:
set.seed(5)
x <- sample(5e3, 1e5, replace = TRUE)
y <- sample(x, replace = TRUE)



Answer:
A variant in base using split.
split the indices of both vectors by its value. Subset the second list with the names of the first, that both have the same order. Change NULL to NA and bring the lengths of the second list to those from the first. Reorder the indices of the second list by those of the first.
x <- c(3L, 1L, 2L, 3L, 3L, 2L)
y <- c(3L, 3L, 3L, 3L, 1L, 3L)

a <- split(seq_along(x), x)
b <- split(seq_along(y), y)[names(a)]
b[lengths(b)==0] <- NA
b <- unlist(Map(`length<-`, b, lengths(a)), FALSE, FALSE)
`[<-`(b, unlist(a, FALSE, FALSE), b)
#[1]  1  5 NA  2  3 NA

I tried to exchange the part
b <- split(seq_along(y), y)[names(a)]
b[lengths(b)==0] <- NA

with
b <- list2env(split(seq_along(y), y))
b <- mget(names(a), b, ifnotfound = NA)

But it was not faster.
An RCPP version.
Store the indices of the second vector ín a queue for each unique value in an unordered_map. Iterate over all values of the first vector and take the indices from the queue.
Rcpp::sourceCpp(code=r"(
#include <Rcpp.h>
#include <unordered_map>
#include <queue>

using namespace Rcpp;
// [[Rcpp::export]]
IntegerVector pm(const std::vector<int>& a, const std::vector<int>& b) {
  IntegerVector idx(no_init(a.size()));
  std::unordered_map<int, std::queue<int> > lut;
  for(int i = 0; i < b.size(); ++i) lut[b[i]].push(i);
  for(int i = 0; i < idx.size(); ++i) {
    auto search = lut.find(a[i]);
    if(search != lut.end() && search->second.size() > 0) {
      idx[i] = search->second.front() + 1;
      search->second.pop();
    } else {idx[i] = NA_INTEGER;}
  }
  return idx;
}
)")
pm(x, y)
#[1]  1  5 NA  2  3 NA

A for this case specialized RCPP version.
Create a vector of the length of the maximum value of the first vector and count how many times a value is present. Create another queue vector of the same length and sore there the indices of the values of the second vector until it has reached the number of the first. Iterate over all values of the first vector and take the indices from the queue.
Rcpp::sourceCpp(code=r"(
#include <Rcpp.h>
#include <vector>
#include <array>
#include <queue>
#include <algorithm>

using namespace Rcpp;
// [[Rcpp::export]]
IntegerVector pm2(const std::vector<int>& a, const std::vector<int>& b) {
  IntegerVector idx(no_init(a.size()));
  int max = 1 + *std::max_element(a.begin(), a.end());
  std::vector<int> n(max);
  for(int i = 0; i < a.size(); ++i) ++n[a[i]];
  std::vector<std::queue<int> > lut(max);
  for(int i = 0; i < b.size(); ++i) {
    if(b[i] < max && n[b[i]] > 0) {
      --n[b[i]];
      lut[b[i]].push(i);
    }
  }
  for(int i = 0; i < idx.size(); ++i) {
    auto & P = lut[a[i]];
    if(P.size() > 0) {
      idx[i] = P.front() + 1;
      P.pop();
    } else {idx[i] = NA_INTEGER;}
  }
  return idx;
}
)")
pm2(x,y)
#[1]  1  5 NA  2  3 NA


Benchmark
set.seed(5)
x <- sample(5e3, 1e5, replace = TRUE)
y <- sample(x, replace = TRUE)

library(data.table)

matchall <- function(x, y) {
  data.table(y, rowid(y))[
    data.table(x, rowid(x)), on = .(y = x, V2), which = TRUE
  ]
}

rmatch <- function(x, y) {
  xp <- cbind(seq_along(x), x)[order(x),]
  yp <- cbind(seq_along(y), y)[order(y),]
  result <- numeric(length(x))
  
  xi <- yi <- 1
  Nx <- length(x)
  Ny <- length(y)
  while (xi <= Nx) {
    if (yi > Ny) {
      result[xp[xi,1]] <- NA
      xi <- xi + 1
    } else if (xp[xi,2] == yp[yi,2]) {
      result[xp[xi,1]] = yp[yi,1]
      xi <- xi + 1
      yi <- yi + 1
    } else if (xp[xi,2] < yp[yi,2]) {
      result[xp[xi,1]] <- NA
      xi <- xi + 1
    } else if (xp[xi,2] > yp[yi,2]) {
      yi <- yi + 1
    }
  }
  result  
}

bench::mark(
ave = ave(x, x, FUN = \(v) which(y == v[1])[1:length(v)]),
rmatch = rmatch(x, y),
make.name = match(make.names(x, TRUE), make.names(y, TRUE)),
paste = do.call(match, lapply(list(x, y), \(v) paste(v, ave(v, v, FUN = seq_along)))),
make.unique = match(make.unique(as.character(x)), make.unique(as.character(y))),
split = {a <- split(seq_along(x), x)
  b <- split(seq_along(y), y)[names(a)]
  b[lengths(b)==0] <- NA
  b <- unlist(Map(`length<-`, b, lengths(a)), FALSE, FALSE)
  `[<-`(b, unlist(a, FALSE, FALSE), b)},
data.table = matchall(x, y),
RCPP = pm(x, y),
RCPP2 = pm2(x, y)
)

Result
  expression       min   median `itr/sec` mem_alloc `gc/sec` n_itr  n_gc
  <bch:expr>  <bch:tm> <bch:tm>     <dbl> <bch:byt>    <dbl> <int> <dbl>
1 ave            1.66s    1.66s     0.603    3.73GB    68.7      1   114
2 rmatch      258.29ms 259.35ms     3.86     5.34MB    30.8      2    16
3 make.name   155.69ms 156.82ms     6.37    14.06MB     1.59     4     1
4 paste         93.8ms 102.06ms     9.74    18.13MB     7.79     5     4
5 make.unique  81.67ms   92.8ms    10.4      9.49MB     5.22     6     3
6 split        12.66ms  13.16ms    65.8      7.18MB    16.0     33     8
7 data.table    6.22ms   6.89ms   114.       5.13MB    28.0     57    14
8 RCPP          3.06ms    3.2ms   301.     393.16KB     3.98   151     2
9 RCPP2         1.64ms   1.82ms   514.     393.16KB     8.00   257     4

In this case the C++ version is the fastest and allocates the lowest amount of memory. In case using base the splitB variant is the fastest and rmatch allocates the lowest amount of memory.