我有一个data.table,其中包含3个日期变量:year
,start
,end
。
test <- data.table(year=2001:2003,start=c(2003,2002,2000),end=c(2003,2004,2002),x_desired=c(F,T,F))
O想要创建一个新变量x
,对于每行,指示year
是否在start
和end
定义的范围内。正确的期望结果在变量x_desired
中。
我想这可以通过以下方式完成:
test[,x:=(year %in% start:end)]
,但结果显然不正确。我想逐行定义范围,但不知道如何表达。
答案 0 :(得分:6)
另一种方法
#first, create a x-column with all FALSE
DT[, x := FALSE ]
#update the x-column subset where year is between start and end to TRUE
DT[ year %between% list(start,end), x := TRUE]
应该快速运行...基准很快就会出现
n = 1000000
set.seed(123)
dt <- data.table(year =sample( 2001:2003, n, replace = TRUE),
start=sample( c(2003,2002,2000), n, replace = TRUE),
end =sample( c(2003,2004,2002), n, replace = TRUE) )
microbenchmark::microbenchmark(
wimpel = {
DT <- copy(dt)
DT[, x := FALSE ]
DT[ year %between% list(start,end), x := TRUE]
},
akrun_nrow = {
DT <- copy(dt)
DT[, x := between(year, start, end), 1:nrow(DT)]
},
akrun_map = {
DT <- copy(dt)
DT[, x := unlist(do.call(Map, c(f = between, unname(.SD)))), .SDcols = year:end]
},
akrun_pmap = {
DT <- copy(dt)
DT[, x := purrr::pmap_lgl(.SD[, .(x = year, left = start, right = end)], between)]
},
markus = {
DT <- copy(dt)
DT[, col := mapply(between, year, start, end)]
},
times = 3
)
结果
Unit: milliseconds
expr min lq mean median uq max neval
wimpel 29.98388 30.41861 48.98399 30.85333 58.48404 86.11475 3
akrun_nrow 2741.35268 2755.01860 2944.58975 2768.68453 3046.20829 3323.73206 3
akrun_map 3673.21253 3683.22849 3711.51209 3693.24446 3730.66188 3768.07929 3
akrun_pmap 3281.13335 3291.04689 3406.46131 3300.96043 3469.12528 3637.29013 3
markus 3408.07869 3569.33044 3670.68141 3730.58219 3801.98277 3873.38334 3
似乎有明显的赢家..但也许我在这里错过了什么?
答案 1 :(得分:4)
另一种方式
set(DT, NULL, "x", between(DT$year, DT$start, DT$end))
基准
library(data.table)
setDTthreads(40L)
n = 1e9
set.seed(123)
DT = data.table(year =sample( 2001:2003, n, replace = TRUE),
start=sample( c(2003,2002,2000), n, replace = TRUE),
end =sample( c(2003,2004,2002), n, replace = TRUE) )
d = copy(DT)
system.time({DT[, x := FALSE ]; DT[ year %between% list(start,end), x := TRUE]})
system.time(set(d, NULL, "x", between(DT$year, DT$start, DT$end)))
all.equal(d, DT)
时间
1e6
> system.time({DT[, x := FALSE ]; DT[ year %between% list(start,end), x := TRUE]})
user system elapsed
0.433 0.056 0.053
> system.time(set(d, NULL, "x", between(DT$year, DT$start, DT$end)))
user system elapsed
0.152 0.000 0.025
1e8
> system.time({DT[, x := FALSE ]; DT[ year %between% list(start,end), x := TRUE]})
user system elapsed
3.811 1.889 3.061
> system.time(set(d, NULL, "x", between(DT$year, DT$start, DT$end)))
user system elapsed
2.650 1.112 2.132
1e9
> system.time({DT[, x := FALSE ]; DT[ year %between% list(start,end), x := TRUE]})
user system elapsed
32.073 32.600 27.347
> system.time(set(d, NULL, "x", between(DT$year, DT$start, DT$end)))
user system elapsed
21.798 8.517 18.248
答案 2 :(得分:2)
一个选项是between
test[, x := between(year, start, end), 1:nrow(test)]
test
# year start end x_desired x
#1: 2001 2003 2003 FALSE FALSE
#2: 2002 2002 2004 TRUE TRUE
#3: 2003 2000 2002 FALSE FALSE
test[, x := year >= start & year <= end]
或者另一个选择是Map
test[, x := unlist(do.call(Map, c(f = between, unname(.SD)))), .SDcols = year:end]
或者使用pmap
中的purrr
library(purrr)
test[, x := pmap_lgl(.SD[, .(x = year, left = start, right = end)], between)]
在新选项上添加了基准(使用与@Wimpel的大数据相同的数据集)
microbenchmark(
wimpel = {
DT <- copy(dt)
DT[, x := FALSE ]
DT[ year %between% list(start,end), x := TRUE]
},
akrun = {
DT <- copy(dt)
DT[, x := year >= start & year <= end]
}, times = 3)
# Unit: milliseconds
# expr min lq mean median uq max neval
# wimpel 23.25196 40.72112 49.29130 58.19027 62.31098 66.43168 3
# akrun 19.56071 22.04272 22.96553 24.52473 24.66793 24.81114 3