Reorder a vector by order of another

Another baseR option is match

vec1[match(vec2, sort(vec2))]
# [1] 0 1 6 4 2 5 9 3 7

edit

Including a benchmark with larger sample size

set.seed(42)
n <- 1e6
vec1 <- seq_len(n)
vec2 <- sample(1:1e7, size = n)

benchmarks <- bench::mark(match = vec1[match(vec2, sort(vec2))],
                          rank = vec1[rank(vec2)],
                          frank = vec1[data.table::frank(vec2)],
                          order_order = vec1[order(order(vec2))],
                          rcpp_order_order = foo(vec1, vec2),
                          iterations = 25)
benchmarks[ , 1:3]

Result

# A tibble: 5 x 3
#  expression            min   median
#  <bch:expr>       <bch:tm> <bch:tm>
#1 match             259.8ms    322ms
#2 rank              825.9ms    876ms
#3 frank              88.6ms    134ms
#4 order_order       110.6ms    139ms
#5 rcpp_order_order  793.5ms    893ms

We can adapt the Rcpp version of order() from this answer (to account for the fact that you do not want to check for duplicates and adding a function to order by an order of an ordering) to make the following Rcpp solution:

#include <Rcpp.h>

Rcpp::IntegerVector order(const Rcpp::NumericVector& x) {
    return Rcpp::match(Rcpp::clone(x).sort(), x);
}

Rcpp::IntegerVector order(const Rcpp::IntegerVector& x) {
    return Rcpp::match(Rcpp::clone(x).sort(), x);
}

// [[Rcpp::export]]
Rcpp::NumericVector foo(const Rcpp::NumericVector x,
                        const Rcpp::NumericVector y) {
    return x[order(order(y))-1];
}

Then we get the expected results:

library(Rcpp)
sourceCpp("foo.cpp")

vec1 <- c(0, 1, 2, 3, 4, 5, 6, 7, 9)
vec2 <- c(1, 2, 7, 5, 3, 6, 80, 4, 8)

foo(vec1, vec2)
# [1] 0 1 6 4 2 5 9 3 7

with decent performance (comparisons are to the R solutions presented by other answers):

benchmarks <- bench::mark(match = vec1[match(vec2, sort(vec2))],
                          rank = vec1[rank(vec2)],
                          order_order = vec1[order(order(vec2))],
                          rcpp_order_order = foo(vec1, vec2),
                          iterations = 10000)
benchmarks[ , 1:3]

# # A tibble: 4 x 3
#   expression            min   median
#   <bch:expr>       <bch:tm> <bch:tm>
# 1 match              28.4µs  31.72µs
# 2 rank               7.99µs   9.84µs
# 3 order_order       26.27µs  30.61µs
# 4 rcpp_order_order   2.51µs   3.23µs

Note that this solution only works if there are no duplicates. (If you might run into duplicates, adding a check is demonstrated in the linked-to answer). Also note that these benchmarks were just done on this data; I don't know for sure how they change at scale.

We could use rank

vec1[rank(vec2)]
#[1] 0 1 6 4 2 5 9 3 7

Or with order

vec1[order(order(vec2))]
#[1] 0 1 6 4 2 5 9 3 7

Or as @markus suggested an option with frank from data.table

library(data.table)
vec1[frank(vec2)]

If I understand you correctly, you want vec1 to follow the same order of vec1. That is, is vec2 is increasing, so should the values of vec1; if vec2 is decreasing, so should vec1 and so on.

sort(vec1)[order(vec2)]