| chunk {bit} | R Documentation |
creates a sequence of range indexes using a syntax not completely unlike 'seq'
chunk(...)
## Default S3 method:
chunk(from = NULL, to = NULL, by = NULL, length.out = NULL, along.with = NULL, overlap = 0L, method = c("bbatch", "seq"), ...)
from |
the starting value of the sequence. |
to |
the (maximal) end value of the sequence. |
by |
increment of the sequence |
length.out |
desired length of the sequence. |
along.with |
take the length from the length of this argument. |
overlap |
number of values to overlap (will lower the starting value of the sequence, the first range becomes smaller |
method |
default 'bbatch' will try to balance the chunk size, see bbatch, 'seq' will create chunks like seq |
... |
ignored |
chunk is generic, the default method is described here, other methods that automatically consider RAM needs are provided with package 'ff', see for example chunk.ffdf
chunk.default returns a list of ri objects representing chunks of subscripts
Jens Oehlschlägel
chunk(1, 100, by=30)
chunk(1, 100, by=30, method="seq")
## Not run:
require(foreach)
m <- 10000
k <- 1000
n <- m*k
cat("Four ways to loop from 1 to n. Slowest foreach to fastest chunk is 1700:1 on a dual core notebook with 3GB RAM\n")
z <- 0L; print(k*system.time({it <- icount(m); foreach (i = it) %do% { z <- i; NULL }})); z
z <- 0L; print(system.time({i <- 0L;while (i<n) {i <- i + 1L; z <- i}})); z
z <- 0L; print(system.time(for (i in 1:n) z <- i)); z
z <- 0L; n <- m*k; print(system.time(for (ch in chunk(1, n, by=m)){for (i in ch[1]:ch[2])z <- i})); z
cat("Seven ways to calculate sum(1:n). Slowest foreach to fastest chunk is 61000:1 on a dual core notebook with 3GB RAM\n")
print(k*system.time({it <- icount(m); foreach (i = it, .combine="+") %do% { i }}))
z <- 0; print(k*system.time({it <- icount(m); foreach (i = it) %do% { z <- z + i; NULL }})); z
z <- 0; print(system.time({i <- 0L;while (i<n) {i <- i + 1L; z <- z + i}})); z
z <- 0; print(system.time(for (i in 1:n) z <- z + i)); z
print(system.time(sum(as.double(1:n))))
z <- 0; n <- m*k; print(system.time(for (ch in chunk(1, n, by=m)){for (i in ch[1]:ch[2])z <- z + i})); z
z <- 0; n <- m*k; print(system.time(for (ch in chunk(1, n, by=m)){z <- z + sum(as.double(ch[1]:ch[2]))})); z
## End(Not run)