使用 microbenchmark 进行基准测试
你可以使用 microbenchmark
包进行亚毫秒精确的表达式评估时序。
在这个例子中, 我们比较了基于特定条件更新组中元素的六个等效 data.table
表达式的速度。
进一步来说:
有 3 列的
data.table
:id
,time
和status
。对于每个 id,我想找到具有最大时间的记录 - 如果对于该记录,如果状态为真,我想在时间> 7 时将其设置为 false
library(microbenchmark)
library(data.table)
set.seed(20160723)
dt <- data.table(id = c(rep(seq(1:10000), each = 10)),
time = c(rep(seq(1:10000), 10)),
status = c(sample(c(TRUE, FALSE), 10000*10, replace = TRUE)))
setkey(dt, id, time) ## create copies of the data so the 'updates-by-reference' don't affect other expressions
dt1 <- copy(dt)
dt2 <- copy(dt)
dt3 <- copy(dt)
dt4 <- copy(dt)
dt5 <- copy(dt)
dt6 <- copy(dt)
microbenchmark(
expression_1 = {
dt1[ dt1[order(time), .I[.N], by = id]$V1, status := status * time < 7 ]
},
expression_2 = {
dt2[,status := c(.SD[-.N, status], .SD[.N, status * time > 7]), by = id]
},
expression_3 = {
dt3[dt3[,.N, by = id][,cumsum(N)], status := status * time > 7]
},
expression_4 = {
y <- dt4[,.SD[.N],by=id]
dt4[y, status := status & time > 7]
},
expression_5 = {
y <- dt5[, .SD[.N, .(time, status)], by = id][time > 7 & status]
dt5[y, status := FALSE]
},
expression_6 = {
dt6[ dt6[, .I == .I[which.max(time)], by = id]$V1 & time > 7, status := FALSE]
},
times = 10L ## specify the number of times each expression is evaluated
)
# Unit: milliseconds
# expr min lq mean median uq max neval
# expression_1 11.646149 13.201670 16.808399 15.643384 18.78640 26.321346 10
# expression_2 8051.898126 8777.016935 9238.323459 8979.553856 9281.93377 12610.869058 10
# expression_3 3.208773 3.385841 4.207903 4.089515 4.70146 5.654702 10
# expression_4 15.758441 16.247833 20.677038 19.028982 21.04170 36.373153 10
# expression_5 7552.970295 8051.080753 8702.064620 8861.608629 9308.62842 9722.234921 10
# expression_6 18.403105 18.812785 22.427984 21.966764 24.66930 28.607064 10
输出结果表明,在这次测试中,expression_3
是最快的。