Xizam - 25 days ago 12
R Question

# Will defining a start position when doing multiple matches on a sorted vector be faster?

I have a vector with 1 million integers, in ascending order and a vector with a subset of 1000 of these integers, also sorted.

What would be faster? Will the second version become faster if samplevec becomes bigger?

``````samplevec=sort(sample(1:10000000, 1000000))
matchvec=sort(sample(samplevec, 10000))

for (i in matchvec) {
index=match(i, samplevec)
print(index)
}
``````

Or

``````samplevec=sort(sample(1:10000000, 1000000))
matchvec=sort(sample(samplevec, 10000))

previous=1
for (i in matchvec) {
index=match(i, samplevec[previous:length(samplevec)])
previous=index
print(index)
}
``````

It's easy to benchmark. Here are just two points in time. Feel free to pimp this and automate to increase the number of points in time.

``````library(microbenchmark)

set.seed(357)

samplevec = sort(sample(1:1000, 1000))
matchvec = sort(sample(samplevec, 1000))

microbenchmark(
version1 = {
previous=1
for (i in matchvec) {
index=match(i, samplevec[previous:length(samplevec)])
previous=index
}},
version2 = {
for (i in matchvec) {
index = match(i, samplevec)
}}
)

Unit: milliseconds
expr       min        lq      mean    median       uq
version1 10.619105 10.711438 12.057713 10.811051 12.71902
version2  2.419441  2.487062  2.853868  2.506603  2.56024
``````

Here is the second point. This one runs a tad longer.

``````set.seed(357)

samplevec = sort(sample(1:100000, 100000))
matchvec = sort(sample(samplevec, 100000))

microbenchmark(
version1 = {
previous=1
for (i in matchvec) {
index=match(i, samplevec[previous:length(samplevec)])
previous=index
}},
version2 = {
for (i in matchvec) {
index=match(i, samplevec)
}}
)

Unit: seconds
expr       min        lq      mean    median        uq
version1 108.96069 109.61137 110.87308 110.70554 111.61337
version2  15.63668  15.71792  16.20434  15.84646  16.07487
``````