How to add a group id without looping?
r
Solution
How about a short `c++` written `for` loop using `Rcpp`. This little function takes a `numeric` vector, i.e. your `ordernum` column and a `threshold` argument (the cumulative sum you want to start a new ID from) and returns a vector of IDs of length equal to the input vector. Should run relatively quickly as it's a `for` loop in `c++`. The code snippet below will install `Rcpp` for you if you haven't already got it installed and will compile the function ready for use. Just copy and paste into R...
if( !require(Rcpp) ) install.packages("Rcpp"); require(Rcpp)
Rcpp::cppFunction( ' NumericVector grpid( NumericVector x , int threshold ){
int n = x.size();
NumericVector out(n);
int tot = 0;
int id = 1;
for( int i = 0; i < n; ++i){
tot += x[i];
out[i] = id;
if( tot >= threshold ){
id += 1;
tot = 0;
}
}
return out;
}')
Then to use the function just use it like any other R function, supplying the relevant arguments:
df$groupid <- grpid( df$ordernum , 30 )
# productid ordernum groupid
#1 p1 10 1
#2 p2 20 1
#3 p3 30 2
#4 p4 5 3
#5 p5 20 3
#6 p6 8 3
BENCHMARKING COMPARISON
OP asked me to benchmark the Rcpp loop against a base R for loop. Here is the code and results. About a 400-fold increase in speed on a vector of 100,000 product ids:
set.seed(1)
x <- sample(30,1e5,repl=T)
for.loop <- quote({
tot <- 0
id <- 1
out <- numeric(length(x))
for( i in 1:length(x) ){
tot <- tot + x[i]
out[i] <- id
if( tot >= 30 ){
tot <- 0
id <- id + 1
}
}
})
rcpp.loop <- quote( out <- grpid(x,30))
require( microbenchmark )
print( bm , unit = "relative" , digits = 2 , "median" )
Unit: relative
expr min lq median uq max neval
eval(rcpp.loop) 1 1 1 1 1 50
eval(for.loop) 533 462 442 428 325 50
Problem
I have dataframe such as: ``` productid ordernum p1 10 p2 20 p3 30 p4 5 p5 20 p6 8 ``` I would like to add another column, which called groupid,it groups the products together in sequence and once the sum(ordernum) reach 30 , assign a new group id, e.g. the result should be ``` productid ordernum groupid p1 10 1 p2 20 1 p3 30 2 p4 5 3 p5 20 3 p6 8 3 ``` It is very easy to do by looping, by how can I achieve this without looping?