Data: assume that I have a list of matrixs called S, which could be generated by:
S<-list(c(1:25),c(1:25),c(1:25),c(1:25))
Here is a feasible way which I want to optimize:
for (i in 1:length(S))
{
dim(S[[i]])<-c(5,5)
}
After searching on the net, I tried to use lapply to apply a function over a list, and this is the code I have tried:
mat<-lapply(S, function(x) dim(x)<-c(5,5))
which only returns:
> mat
[[1]]
[1] 5 5
[[2]]
[1] 5 5
[[3]]
[1] 5 5
[[4]]
[1] 5 5
Question: I'm wondering if there is an inbuilt function which could apply a function over a list which doesn't require return, or there are some errors in my code?
Thanks in advance.
Extending your code attempt, you need to include an explicit or implicit return statement:
lapply(S, function(x) { dim(x) <- c(5, 5); return(x) })
lapply(S, function(x) { dim(x) <- c(5, 5); x; })
or faster by recasting every list entry as a matrix:
lapply(S, function(x) matrix(x, 5, 5))
or using purrr::map:
map(S, ~ matrix(., 5, 5))
[Edited by @HunterJiang]
library(microbenchmark)
library(purrr)
library(ggplot2)
N<-30
M<-30
S<-list(c(1:(N*M)),c(1:(N*M)),c(1:(N*M)),c(1:(N*M)))
mb <- microbenchmark(
for_loop = { for (i in 1:length(S)) dim(S[[i]])<-c(N,M) },
dim_plus_return = { S1<-lapply(S, function(x) { dim(x) <- c(N,M); return(x) }) },
cast_matrix = { S1<-lapply(S, function(x) matrix(x, N,M)) },
purrr_map = { S1<-map(S, ~ matrix(.,N,M)) },
set_dim_directly = { S1<-lapply(S, `dim<-`, c(N,M)) }
)
mb
ggplot(mb, aes(expr, log10(time))) +
geom_boxplot() +
labs(y = "Time in log10 nanosec", x = "Method")
When N and M are small, says N=M=30, the speed of methods are:
Unit: microseconds
expr min lq mean median uq max neval
for_loop 2111.950 2236.298 2537.42270 2328.4735 2484.2055 4581.549 100
dim_plus_return 10.264 12.633 32.91945 16.1855 19.3440 1641.794 100
cast_matrix 11.054 13.423 27.40873 16.3830 18.9490 1068.213 100
purrr_map 70.662 77.768 99.41636 93.1640 112.9015 199.748 100
set_dim_directly 5.527 6.909 8.47230 7.8960 9.6720 22.502 100
But when N and M become larger, says N=M=3k, lapply gets slower than it used to be and for loop might be a proper way to do this.
Unit: milliseconds
expr min lq mean median uq max neval
for_loop 2.224456 20.83191 52.76189 41.72521 69.91993 180.9775 100
dim_plus_return 35.930768 37.57671 68.63905 39.31620 74.14185 193.8300 100
cast_matrix 48.220338 51.16917 79.73308 52.37871 87.31804 199.2859 100
purrr_map 49.534089 51.21635 89.11881 61.12987 101.98780 195.1374 100
set_dim_directly 35.151124 37.71112 67.72032 39.91919 74.97617 184.4943 100
Conclusion: S1<-lapply(S, `dim<-`, c(N,M)) fits the small dataset, and for loop could be faster when the dimension of the dataset is very large.