连接2个数据表同时汇总其中一个数据的最快方法
问题
我有2个数据表
运动：
library(data.table)
consEx = data.table(
begin = as.POSIXct(c("2019-04-01 00:00:10"," 2019-04-07 10:00:00","2019-04-10 23:00:00","2019-04-12 20:00:00","2019-04-15 10:00:00",
"2019-04-20 10:00:00","2019-04-22 13:30:00","2019-04-10 15:30:00","2019-04-12 21:30:00","2019-04-15 20:00:00")),
end = as.POSIXct(c("2019-04-01 20:00:00","2019-04-07 15:00:00","2019-04-11 10:00:00", "2019-04-12 23:30:00","2019-04-15 15:00:00",
"2019-04-21 12:00:00","2019-04-22 17:30:00","2019-04-10 20:00:00","2019-04-13 05:00:00", "2019-04-15 12:30:00")),
carId = c(1,1,1,2,2,3,3,4,4,5),
tripId = c(1:10)
)
和警报：
alertsEx = data.table(
timestamp = as.POSIXct(c("2019-04-01 10:00:00","2019-04-01 10:30:00","2019-04-01 15:00:00","2019-04-15 13:00:00","2019-04-22 14:00:00",
"2019-04-22 15:10:00","2019-04-22 15:40:00","2019-04-10 16:00:00","2019-04-10 17:00:00","2019-04-13 04:00:00")),
type = c("T1","T2","T1",'T3',"T1","T1","T3","T2","T2","T1"),
carId = c(1,1,1,2,3,3,3,4,4,4),
additionalInfo1 = rnorm(10,mean=10,sd=4)
)
运动表记录了一个周期begin- end汽车在该周期内运动。警报表显示在警报发生在汽车，包含type，timestamp和carId
我需要加入这2个表，并通过汇总警报数据type。当在同一时期内有多个相同类型的警报时，我需要additionalInfo1用平均值来进行汇总。
当前方法
我目前正在通过遍历consEx并将函数应用于每一行的方式来执行此操作，该函数返回包含所需计算的列表
findAlerts = function(begin,end,carId_2){
saida = alertsEx[timestamp >= begin & timestamp <= end & carId == carId_2,]
totals = nrow(saida)
saida = split(saida,by="type")
resultsList = list(
"totals" = 0,
"t1Count" = 0,
"t1Mean" = 0,
"t2Count" = 0,
"t2Mean" = 0,
"t3Count" = 0,
"t3Mean" = 0)
resultsList[["totals"]] = totals
types = names(saida)
if("T1" %in% types){
resultsList[["t1Count"]] = nrow(saida[["T1"]])
resultsList[["t1Mean"]] = mean(saida[["T1"]]$additionalInfo1)
}
if("T2" %in% types){
resultsList[["t2Count"]] = nrow(saida[["T2"]])
resultsList[["t2Mean"]] = mean(saida[["T2"]]$additionalInfo1)
}
if("T3" %in% types){
resultsList[["t3Count"]] = nrow(saida[["T3"]])
resultsList[["t3Mean"]] = mean(saida[["T3"]]$additionalInfo1)
}
return(resultsList)
}
for(i in 1:nrow(consEx)){
aux = findAlerts(consEx$begin[i],consEx$end[i],consEx$carId[i])
consEx[i,names(aux) := aux]
}
预期产量
给出预期的输出：
begin end carId tripId totals t1Count t1Mean t2Count t2Mean t3Count t3Mean
1: 2019-04-01 00:00:10 2019-04-01 20:00:00 1 1 3 2 10.123463 1 14.479288 0 0.000000
2: 2019-04-07 10:00:00 2019-04-07 15:00:00 1 2 0 0 0.000000 0 0.000000 0 0.000000
3: 2019-04-10 23:00:00 2019-04-11 10:00:00 1 3 0 0 0.000000 0 0.000000 0 0.000000
4: 2019-04-12 20:00:00 2019-04-12 23:30:00 2 4 0 0 0.000000 0 0.000000 0 0.000000
5: 2019-04-15 10:00:00 2019-04-15 15:00:00 2 5 1 0 0.000000 0 0.000000 1 6.598062
6: 2019-04-20 10:00:00 2019-04-21 12:00:00 3 6 0 0 0.000000 0 0.000000 0 0.000000
7: 2019-04-22 13:30:00 2019-04-22 17:30:00 3 7 3 2 7.610410 0 0.000000 1 10.218593
8: 2019-04-10 15:30:00 2019-04-10 20:00:00 4 8 2 0 0.000000 2 9.703278 0 0.000000
9: 2019-04-12 21:30:00 2019-04-13 05:00:00 4 9 1 1 7.095564 0 0.000000 0 0.000000
10: 2019-04-15 20:00:00 2019-04-15 12:30:00 5 10 0 0 0.000000 0 0.000000 0 0.000000
但这对于我的原始数据来说太慢了，原始数据有1300万行consEx和200万警报。还有几种不同的类型，我需要计算两个指标的平均值，最小值和最大值。
有没有更快的方法来编写函数或循环？
另外，用python编写几乎相同的代码是否会带来更好的结果？我正在考虑重写它，但这需要很长时间，而且我不确定它是否可以解决问题
谢谢
PS：我尝试过的另一种方法是遍历警报并为每个警报分配一个tripIdfrom consEx，然后汇总结果表，但这似乎更慢。
该答案提出了两种data.table方法，它们都使用非等值联接将警报匹配到行程并dcast()进行重整，但是在聚合方法上的区别在于consEx。
第一个变量创建一个新的result数据集，constEx保持不变。
第二个变体constEx在适当位置进行修改。这类似于Alexis的解决方案，但更加简洁并使用了新setnafill()功能（开发版本为1.12.3的版本提供data.table）
形式1：创建新的结果数据集
agg <- consEx[alertsEx, on = .(carId, begin <= timestamp, end >= timestamp),
.(tripId, type, additionalInfo1)][
, .(Count = .N, Mean = mean(additionalInfo1)),
by = .(tripId, type)][
, totals := sum(Count), by = tripId][]
result <- dcast(agg[consEx, on = "tripId"], ... ~ type,
value.var = c("Count", "Mean"), fill = 0)[
, c("Count_NA", "Mean_NA") := NULL][
is.na(totals), totals := 0]
setcolorder(result, names(consEx))
result
Run Code Online (Sandbox Code Playgroud)begin end carId tripId totals Count_T1 Count_T2 Count_T3 Mean_T1 Mean_T2 Mean_T3 1: 2019-04-01 00:00:10 2019-04-01 20:00:00 1 1 3 2 1 0 12.654609 12.375862 0.000000 2: 2019-04-07 10:00:00 2019-04-07 15:00:00 1 2 0 0 0 0 0.000000 0.000000 0.000000 3: 2019-04-10 23:00:00 2019-04-11 10:00:00 1 3 0 0 0 0 0.000000 0.000000 0.000000 4: 2019-04-12 20:00:00 2019-04-12 23:30:00 2 4 0 0 0 0 0.000000 0.000000 0.000000 5: 2019-04-15 10:00:00 2019-04-15 15:00:00 2 5 1 0 0 1 0.000000 0.000000 9.316815 6: 2019-04-20 10:00:00 2019-04-21 12:00:00 3 6 0 0 0 0 0.000000 0.000000 0.000000 7: 2019-04-22 13:30:00 2019-04-22 17:30:00 3 7 3 2 0 1 8.586061 0.000000 11.498512 8: 2019-04-10 15:30:00 2019-04-10 20:00:00 4 8 2 0 2 0 0.000000 8.696356 0.000000 9: 2019-04-12 21:30:00 2019-04-13 05:00:00 4 9 1 1 0 0 9.343681 0.000000 0.000000 10: 2019-04-15 20:00:00 2019-04-15 12:30:00 5 10 0 0 0 0 0.000000 0.000000 0.000000
请注意，发布的不同答案之间的方式可能会有所不同additionalInfo1，rnorm(10, mean = 10, sd = 4)而无需set.seed()事先致电即可创建 。调用set.seed()会创建可复制的随机数。
变体1的说明
第一部分通过非等额联接为每个警报找到匹配的行程，并通过tripId和计算总计，type以及每个行程的总数：
agg
Run Code Online (Sandbox Code Playgroud)tripId type Count Mean totals 1: 1 T1 2 12.654609 3 2: 1 T2 1 12.375862 3 3: 5 T3 1 9.316815 1 4: 7 T1 2 8.586061 3 5: 7 T3 1 11.498512 3 6: 8 T2 2 8.696356 2 7: 9 T1 1 9.343681 1
请注意，tripId假定它是中的唯一键consEx。
在下一步中，将聚合右连接到 consEx
agg[consEx, on = "tripId"]
Run Code Online (Sandbox Code Playgroud)tripId type Count Mean totals begin end carId 1: 1 T1 2 12.654609 3 2019-04-01 00:00:10 2019-04-01 20:00:00 1 2: 1 T2 1 12.375862 3 2019-04-01 00:00:10 2019-04-01 20:00:00 1 3: 2 <NA> NA NA NA 2019-04-07 10:00:00 2019-04-07 15:00:00 1 4: 3 <NA> NA NA NA 2019-04-10 23:00:00 2019-04-11 10:00:00 1 5: 4 <NA> NA NA NA 2019-04-12 20:00:00 2019-04-12 23:30:00 2 6: 5 T3 1 9.316815 1 2019-04-15 10:00:00 2019-04-15 15:00:00 2 7: 6 <NA> NA NA NA 2019-04-20 10:00:00 2019-04-21 12:00:00 3 8: 7 T1 2 8.586061 3 2019-04-22 13:30:00 2019-04-22 17:30:00 3 9: 7 T3 1 11.498512 3 2019-04-22 13:30:00 2019-04-22 17:30:00 3 10: 8 T2 2 8.696356 2 2019-04-10 15:30:00 2019-04-10 20:00:00 4 11: 9 T1 1 9.343681 1 2019-04-12 21:30:00 2019-04-13 05:00:00 4 12: 10 <NA> NA NA NA 2019-04-15 20:00:00 2019-04-15 12:30:00 5
使用可以立即将输出从长格式更改为宽格式dcast()。
最后，通过删除多余的列，将其替换NA为0并更改列顺序来清理结果。
形式2：更新到位
agg <- consEx[alertsEx, on = .(carId, begin <= timestamp, end >= timestamp),
.(tripId, type, additionalInfo1)][
, .(Count = .N, Mean = mean(additionalInfo1)),
by = .(tripId, type)][
, totals := sum(Count), by = tripId][]
wide <- dcast(agg, ... ~ type, value.var = c("Count", "Mean"), fill = 0)
consEx[wide, on = "tripId", (names(wide)) := mget(paste0("i.", names(wide)))]
setnafill(consEx, fill = 0) # data.table version 1.12.3+
consEx
Run Code Online (Sandbox Code Playgroud)begin end carId tripId totals Count_T1 Count_T2 Count_T3 Mean_T1 Mean_T2 Mean_T3 1: 2019-04-01 00:00:10 2019-04-01 20:00:00 1 1 3 2 1 0 12.654609 12.375862 0.000000 2: 2019-04-07 10:00:00 2019-04-07 15:00:00 1 2 0 0 0 0 0.000000 0.000000 0.000000 3: 2019-04-10 23:00:00 2019-04-11 10:00:00 1 3 0 0 0 0 0.000000 0.000000 0.000000 4: 2019-04-12 20:00:00 2019-04-12 23:30:00 2 4 0 0 0 0 0.000000 0.000000 0.000000 5: 2019-04-15 10:00:00 2019-04-15 15:00:00 2 5 1 0 0 1 0.000000 0.000000 9.316815 6: 2019-04-20 10:00:00 2019-04-21 12:00:00 3 6 0 0 0 0 0.000000 0.000000 0.000000 7: 2019-04-22 13:30:00 2019-04-22 17:30:00 3 7 3 2 0 1 8.586061 0.000000 11.498512 8: 2019-04-10 15:30:00 2019-04-10 20:00:00 4 8 2 0 2 0 0.000000 8.696356 0.000000 9: 2019-04-12 21:30:00 2019-04-13 05:00:00 4 9 1 1 0 0 9.343681 0.000000 0.000000 10: 2019-04-15 20:00:00 2019-04-15 12:30:00 5 10 0 0 0 0 0.000000 0.000000 0.000000
变型2的说明
第一部分与变体1中的相同。
在下一步中，将聚集体从长格式更改为宽格式。
然后，执行更新联接，其中通过的匹配行在适当位置进行consEx更新。表达方式wide
(names(wide)) := mget(paste0("i.", names(wide)))
是捷径
`:=`(totals = i.totals, Count_T1 = i.Count_T1, ..., Mean_T3 = i.Mean_T3)
其中i.前缀是的列，wide用于避免歧义。
在consEx没有匹配项的行中，附加列包含NA。因此 ，最后setnafill()用于将所有NA值替换为0。
基准测试
在撰写本文时，由以下人员发布了6种不同的解决方案
5种R溶液使用比较press()并mark()从bench包。
样本数据集被参数化以改变行数。正如consEx某些解决方案所修改的那样，每次基准测试运行都从新副本开始。
要使代码与一起使用，需要进行一些小的更改bench。最显着的是，... ~ typein中的快捷方式公式在dcast()内部调用时会导致错误bench::mark()，必须替换为明确命名LHS所有变量的版本。
正如OP指出的那样，他的版本相当慢，第一次基准测试仅使用1000行consEx和50、200和1000行alertsEx
bm <- bench::press(
n_trips = 1E3,
n_alerts = 1E3/c(20, 5, 1),
{
n_cars <- 5
n_periods <- round(n_trips / n_cars)
n_types = 3
types = sprintf("T%02i", 1:n_types)
consEx0 <- data.table(
tripId = 1:n_trips,
carId = rep(1:n_cars, each = n_periods),
begin = seq.POSIXt(as.POSIXct("2000-01-01"), length.out = n_periods, by = "14 days") %>%
rep(n_cars) %>%
`mode<-`("double")
)[, end := begin + 7*24*60*60]
set.seed(1)
idx <- sample(n_trips, n_alerts, TRUE)
alertsEx <- consEx0[
idx,
.(timestamp = begin + runif(n_alerts, 1, 7*24*60*60 - 1),
type = sample(types, n_alerts, TRUE),
carId,
additionalInfo1 = rnorm(n_alerts, mean = 10, sd = 4))]
bench::mark(
fino = {
consEx <- copy(consEx0)
findAlerts = function(begin,end,carId_2){
saida = alertsEx[timestamp >= begin & timestamp <= end & carId == carId_2,]
totals = nrow(saida)
saida = split(saida,by="type")
resultsList = list(
"totals" = 0,
"t1Count" = 0,
"t1Mean" = 0,
"t2Count" = 0,
"t2Mean" = 0,
"t3Count" = 0,
"t3Mean" = 0)
resultsList[["totals"]] = totals
types = names(saida)
if("T1" %in% types){
resultsList[["t1Count"]] = nrow(saida[["T1"]])
resultsList[["t1Mean"]] = mean(saida[["T1"]]$additionalInfo1)
}
if("T2" %in% types){
resultsList[["t2Count"]] = nrow(saida[["T2"]])
resultsList[["t2Mean"]] = mean(saida[["T2"]]$additionalInfo1)
}
if("T3" %in% types){
resultsList[["t3Count"]] = nrow(saida[["T3"]])
resultsList[["t3Mean"]] = mean(saida[["T3"]]$additionalInfo1)
}
return(resultsList)
}
for(i in 1:nrow(consEx)){
aux = findAlerts(consEx$begin[i],consEx$end[i],consEx$carId[i])
consEx[i,names(aux) := aux]
}
consEx[]
},
dave = {
# library(dplyr)
# library(tidyr)
consEx <- copy(consEx0)
#split the consEx down to separate carId
splitcons <- split(consEx, consEx$carId)
alertsEx$tripId <- NA
#loop and only search the cons that match the alerts
alertsEx$tripId <- apply(alertsEx, 1, function(a) {
#retrive the list assicoated to the correct car
cons <- splitcons[[a[["carId"]]]]
alerttime <- as.POSIXct(a[["timestamp"]])
#find the trip which contains the alerttime
tripId <-
which((alerttime >= cons$begin) & (alerttime <= cons$end))
#identify and return the tripId
cons$tripId[tripId]
})
#Referenced:
#/sf/ask/2141446611/
#/sf/ask/2472504821/
alertsEx2 <-
alertsEx %>%
dplyr::group_by(carId, tripId, type) %>%
dplyr::summarize(mean = mean(additionalInfo1), count = dplyr::n())
alerttable <-
tidyr::gather(alertsEx2, variable, val, -(carId:type), na.rm = TRUE) %>%
tidyr::unite(temp, type, variable) %>%
tidyr::spread(temp, val, fill = 0)
consEx %>%
dplyr::left_join(alerttable, by = c("tripId", "carId"))
},
alexis = {
consEx <- copy(consEx0)
joined <- consEx[alertsEx,
.(carId, tripId, type, additionalInfo1),
on = .(carId, begin <= timestamp, end >= timestamp)]
aggregated <- joined[, .(typeCount = .N, typeMean = mean(additionalInfo1)), by = .(carId, tripId, type)]
totals <- aggregated[, .(totals = sum(typeCount)), by = .(carId, tripId)]
long <- dcast(aggregated, carId + tripId ~ type, value.var = c("typeCount", "typeMean"), sep = "", fill = 0)
replaceNA <- function(x) { replace(x, is.na(x), 0) }
consEx[, `:=`(as.character(outer(types, c("Count", "Mean"), paste0)),
lapply(long[consEx,
as.character(outer(types, c("typeCount", "typeMean"),
function(a, b) { paste0(b, a) })),
with = FALSE,
on = .(carId, tripId)],
replaceNA))]
consEx[, totals := lapply(totals[consEx, x.totals, on = .(carId, tripId)], replaceNA)]
setcolorder(consEx, c("carId", "tripId", "begin", "end"))
consEx
},
uwe1 = {
consEx <- copy(consEx0)
agg <- consEx[alertsEx, on = .(carId, begin <= timestamp, end >= timestamp),
.(tripId, type, additionalInfo1)][
, .(Count = .N, Mean = mean(additionalInfo1)),
by = .(tripId, type)][
, totals := sum(Count), by = tripId][]
result <- dcast(agg[consEx, on = "tripId"],
tripId + carId + begin+ end + totals ~ type,
value.var = c("Count", "Mean"), fill = 0)[
, c("Count_NA", "Mean_NA") := NULL][
is.na(totals), totals := 0][]
# setcolorder(result, names(consEx))
result
},
uwe2 = {
consEx <- copy(consEx0)
agg <- consEx[alertsEx, on = .(carId, begin <= timestamp, end >= timestamp),
.(tripId, type, additionalInfo1)][
, .(Count = .N, Mean = mean(additionalInfo1)),
by = .(tripId, type)][
, totals := sum(Count), by = tripId][]
wide <- dcast(agg, tripId + totals ~ type, value.var = c("Count", "Mean"), fill = 0)
consEx[wide, on = "tripId", (names(wide)) := mget(paste0("i.", names(wide)))]
setnafill(consEx, fill = 0) # data.table version 1.12.3+
consEx
},
check = FALSE
)
}
)
library(ggplot2)
autoplot(bm)
请注意对数时标。
该图表证实了OP的假设，即他的方法比其他任何解决方案都慢得多。1000行花了3秒，而OP的生产数据集有1200万行。Dave2e方法的运行时间随着警报数量的增加而增加。
另外，OP的方法对分配的内存要求最高。它最多可分配190 MB，而data.table版本只需要1 MByt。
bm
Run Code Online (Sandbox Code Playgroud)# A tibble: 15 x 15 expression n_trips n_alerts min median `itr/sec` mem_alloc `gc/sec` n_itr n_gc total_time result memory time <bch:expr> <dbl> <dbl> <bch:tm> <bch:tm> <dbl> <bch:byt> <dbl> <int> <dbl> <bch:tm> <list> <list> <lis> 1 fino 1000 50 2.95s 2.95s 0.339 149.25MB 3.39 1 10 2.95s <data~ <Rpro~ <bch~ 2 dave 1000 50 21.52ms 22.61ms 41.6 779.46KB 5.94 21 3 505.01ms <data~ <Rpro~ <bch~ 3 alexis 1000 50 21.98ms 23.09ms 41.2 1007.84KB 3.93 21 2 509.5ms <data~ <Rpro~ <bch~ 4 uwe1 1000 50 14.47ms 15.45ms 62.3 919.12KB 1.95 32 1 513.94ms <data~ <Rpro~ <bch~ 5 uwe2 1000 50 14.14ms 14.73ms 64.2 633.2KB 1.95 33 1 513.91ms <data~ <Rpro~ <bch~ 6 fino 1000 200 3.09s 3.09s 0.323 155.12MB 4.53 1 14 3.09s <data~ <Rpro~ <bch~ 7 dave 1000 200 53.4ms 62.11ms 11.0 1.7MB 5.49 8 4 728.02ms <data~ <Rpro~ <bch~ 8 alexis 1000 200 22.42ms 23.34ms 41.5 1.03MB 3.77 22 2 530.25ms <data~ <Rpro~ <bch~ 9 uwe1 1000 200 14.72ms 15.76ms 62.2 953.07KB 0 32 0 514.46ms <data~ <Rpro~ <bch~ 10 uwe2 1000 200 14.49ms 15.17ms 63.6 695.63KB 1.99 32 1 503.4ms <data~ <Rpro~ <bch~ 11 fino 1000 1000 3.6s 3.6s 0.278 187.32MB 3.61 1 13 3.6s <data~ <Rpro~ <bch~ 12 dave 1000 1000 242.46ms 243.7ms 4.07 6.5MB 6.78 3 5 737.06ms <data~ <Rpro~ <bch~ 13 alexis 1000 1000 24.32ms 25.78ms 37.6 1.21MB 3.95 19 2 505.84ms <data~ <Rpro~ <bch~ 14 uwe1 1000 1000 16.04ms 16.8ms 56.8 1.05MB 1.96 29 1 510.23ms <data~ <Rpro~ <bch~ 15 uwe2 1000 1000 15.69ms 16.41ms 54.8 938.74KB 3.92 28 2 510.63ms <data~ <Rpro~ <bch~
因此，在第二个基准测试中省略了Fino的方法，因为问题大小较大，分别为10k和100k行，consEx而alertsExresp分别为2k和20k行。
由于Dave2e的方法在20k行中的速度比最快的方法慢100倍，因此在第三次运行中被忽略了。这将模拟OP的12M行consEx和2M行的生产数据集，其中alertsEx：
print(bm, n = Inf)
# A tibble: 27 x 15 expression n_trips n_alerts min median `itr/sec` mem_alloc `gc/sec` n_itr n_gc total_time result memory time <bch:expr> <dbl> <dbl> <bch:tm> <bch:tm> <dbl> <bch:byt> <dbl> <int> <dbl> <bch:tm> <list> <list> <lis> 1 alexis 1.00e5 20000 747.87ms 747.87ms 1.34 38.18MB 6.69 1 5 747.87ms <data~ <Rpro~ <bch~ 2 uwe1 1.00e5 20000 124.52ms 138.89ms 7.32 37.88MB 3.66 4 2 546.17ms <data~ <Rpro~ <bch~ 3 uwe2 1.00e5 20000 72.29ms 73.76ms 11.3 16.63MB 1.88 6 1 533.26ms <data~ <Rpro~ <bch~ 4 alexis 1.00e6 20000 7.51s 7.51s 0.133 335.44MB 3.33 1 25 7.51s <data~ <Rpro~ <bch~ 5 uwe1 1.00e6 20000 995.31ms 995.31ms 1.00 346.94MB 2.01 1 2 995.31ms <data~ <Rpro~ <bch~ 6 uwe2 1.00e6 20000 714.19ms 714.19ms 1.40 89.27MB 1.40 1 1 714.19ms <data~ <Rpro~ <bch~ 7 alexis 1.00e7 20000 1.67m 1.67m 0.00999 3.21GB 0.340 1 34 1.67m <data~ <Rpro~ <bch~ 8 uwe1 1.00e7 20000 3.44m 3.44m 0.00485 3.41GB 0.00969 1 2 3.44m <data~ <Rpro~ <bch~ 9 uwe2 1.00e7 20000 42.51s 42.51s 0.0235 810.3MB 0 1 0 42.51s <data~ <Rpro~ <bch~ 10 alexis 1.00e5 200000 15.38s 15.38s 0.0650 73.22MB 0.0650 1 1 15.38s <data~ <Rpro~ <bch~ 11 uwe1 1.00e5 200000 1.34s 1.34s 0.747 63.81MB 0 1 0 1.34s <data~ <Rpro~ <bch~ 12 uwe2 1.00e5 200000 1.47s 1.47s 0.681 58.98MB 0 1 0 1.47s <data~ <Rpro~ <bch~ 13 alexis 1.00e6 200000 2.91m 2.91m 0.00573 375.93MB 0.0115 1 2 2.91m <data~ <Rpro~ <bch~ 14 uwe1 1.00e6 200000 9.72s 9.72s 0.103 371.69MB 0 1 0 9.72s <data~ <Rpro~ <bch~ 15 uwe2 1.00e6 200000 888.67ms 888.67ms 1.13 161.82MB 0 1 0 888.67ms <data~ <Rpro~ <bch~ 16 alexis 1.00e7 200000 6.29m 6.29m 0.00265 3.15GB 0.0928 1 35 6.29m <data~ <Rpro~ <bch~ 17 uw