在R中将模拟结果存储为data.table

Question

I've to do plenty of simulation and it takes lot of time. I think processing time can be reduced through data.table . How can I store the results of mdply(data.frame(prob=seq(from = 0.1, to = 0.9, by = 0.1)), rbinom, n = 5, size = 2) into data.table without saving its output first to data.frame .

library(plyr)
df1 <- mdply(data.frame(prob=seq(from = 0.1, to = 0.9, by = 0.1)), rbinom, n = 5, size = 2)
library(data.table)
dt1 <- data.table(df1)

Edited

I know that I can use setDT(df1) to avoid the creating to dt1 . However, the main problem is about mdply which creates a data.frame which consumes a lot of time.

Answer 1

plyr and data.table are for very similar purposes, so you usually don't need to switch back and forth between the two at all. You can do everything with data.table in this case:

dt = data.table(prob = seq(0.1, 0.9, by = 0.1))
dt = dt[, as.list(rbinom(prob, n = 5, size = 2)), by = prob]
dt
   prob V1 V2 V3 V4 V5
1:  0.1  0  0  0  0  0
2:  0.2  0  0  0  0  1
3:  0.3  1  2  1  0  1
4:  0.4  1  1  2  1  0
5:  0.5  2  2  1  1  1
6:  0.6  1  1  0  0  1
7:  0.7  2  1  2  1  0
8:  0.8  2  1  2  0  1
9:  0.9  2  2  2  2  2

I would add that my hunch is that the fastest way to do this would be to make the matrix first and then assign columns.

> mat = mapply(rbinom, prob = dt$prob, n = 5, size = 2)
> cbind(dt, t(mat))
   prob V1 V2 V3 V4 V5
1:  0.1  0  0  0  0  0
2:  0.2  1  0  0  1  1
3:  0.3  1  1  1  0  0
4:  0.4  1  0  2  1  1
5:  0.5  1  1  1  0  2
6:  0.6  2  0  2  1  1
7:  0.7  1  1  1  2  1
8:  0.8  1  2  1  0  2
9:  0.9  1  1  2  1  1

A very quick test on an 8000 row table shows this is faster:

> dt = data.table(prob = (seq(0.1, 0.9, by = 0.00001)))
> system.time(for(i in 1:10) dt[, as.list(rbinom(prob, n = 5, size = 2)), by = prob])
   user  system elapsed 
   6.14    0.00    6.16 
> system.time(for(i in 1:10) {mat = mapply(rbinom, prob = dt$prob, n = 5, size = 2) ; cbind(dt, t(mat))})
   user  system elapsed 
   2.61    0.00    2.62 

And both are a substantial improvement on the original:

> system.time(for(i in 1:10) {df1 = mdply(df, rbinom, n = 5, size = 2) ; dt1 = data.table(df1)})
   user  system elapsed 
 152.23   46.60  200.07