我有两个数据帧：
id dates
MUM-1 2015-07-10
MUM-1 2015-07-11
MUM-1 2015-07-12
MUM-2 2014-01-14
MUM-2 2014-01-15
MUM-2 2014-01-16
MUM-2 2014-01-17
和
id dates field1 field2
MUM-1 2015-07-10 1 0
MUM-1 2015-07-12 2 1
MUM-2 2014-01-14 4 3
MUM-2 2014-01-17 0 1
合并数据：
id dates field1 field2
MUM-1 2015-07-10 1 0
MUM-1 2015-07-11 na na
MUM-1 2015-07-12 2 1
MUM-2 2014-01-14 4 3
MUM-2 2014-01-15 na na
MUM-2 2014-01-16 na na
MUM-2 2014-01-17 0 1
代码：merge(x= df1, y= df2, by= 'id', all.x= T)
我正在使用合并，但由于两个数据帧的大小太大，处理时间太长。合并功能有替代方法吗？也许在dplyr？因此它在比较中处理速度很快。两个数据帧都有超过900K行。
答案 0 :(得分：11)
您可以简单地按以下方式加入，而不是将merge与data.table一起使用：
setDT(df1)
setDT(df2)
df2[df1, on = c('id','dates')]
这给出了：
> df2[df1]
id dates field1 field2
1: MUM-1 2015-07-10 1 0
2: MUM-1 2015-07-11 NA NA
3: MUM-1 2015-07-12 2 1
4: MUM-2 2014-01-14 4 3
5: MUM-2 2014-01-15 NA NA
6: MUM-2 2014-01-16 NA NA
7: MUM-2 2014-01-17 0 1
使用dplyr：
library(dplyr)
dplr <- left_join(df1, df2, by=c("id","dates"))
如@Arun在评论中所提到的，对于具有七行的小型数据集，基准测试不是很有意义。因此，我们创建一些更大的数据集：
dt1 <- data.table(id=gl(2, 730, labels = c("MUM-1", "MUM-2")),
dates=c(seq(as.Date("2010-01-01"), as.Date("2011-12-31"), by="days"),
seq(as.Date("2013-01-01"), as.Date("2014-12-31"), by="days")))
dt2 <- data.table(id=gl(2, 730, labels = c("MUM-1", "MUM-2")),
dates=c(seq(as.Date("2010-01-01"), as.Date("2011-12-31"), by="days"),
seq(as.Date("2013-01-01"), as.Date("2014-12-31"), by="days")),
field1=sample(c(0,1,2,3,4), size=730, replace = TRUE),
field2=sample(c(0,1,2,3,4), size=730, replace = TRUE))
dt2 <- dt2[sample(nrow(dt2), 800)]
可以看出，@ Arun的方法稍快一些：
library(rbenchmark)
benchmark(replications = 10, order = "elapsed", columns = c("test", "elapsed", "relative"),
jaap = dt2[dt1, on = c('id','dates')],
pavo = merge(dt1,dt2,by="id",allow.cartesian=T),
dplr = left_join(dt1, dt2, by=c("id","dates")),
arun = dt1[dt2, c("fiedl1", "field2") := .(field1, field2), on=c("id", "dates")])
test elapsed relative
4 arun 0.015 1.000
1 jaap 0.016 1.067
3 dplr 0.037 2.467
2 pavo 1.033 68.867
有关大型数据集的比较，请参阅the answer of @Arun。
答案 1 :(得分：8)
我直接通过引用更新df1，如下所示：
require(data.table) # v1.9.5+
setDT(df1)[df2, c("fiedl1", "field2") :=
.(field1, field2), on=c("id", "dates")]
> df1
# id dates fiedl1 field2
# 1: MUM-1 2015-07-10 1 0
# 2: MUM-1 2015-07-11 NA NA
# 3: MUM-1 2015-07-12 2 1
# 4: MUM-2 2014-01-14 4 3
# 5: MUM-2 2014-01-15 NA NA
# 6: MUM-2 2014-01-16 NA NA
# 7: MUM-2 2014-01-17 0 1
这非常节省内存（并且更快），因为它不会复制整个对象只是添加两列，而是更新到位。
使用比@ Jaap的更新基准更大的数据集更新：
set.seed(1L)
dt1 = CJ(id1 = paste("MUM", 1:1e4, sep = "-"), id2 = sample(1e3L))
dt2 = dt1[sample(nrow(dt1), 1e5L)][, c("field1", "field2") := lapply(c(1e3L, 1e4L), sample, 1e5L, TRUE)][]
# @Jaap's answers
system.time(ans1 <- setDT(dt2)[dt1, on = c('id1','id2')])
# user system elapsed
# 0.209 0.067 0.277
system.time(ans2 <- left_join(setDF(dt1), setDF(dt2), by = c("id1", "id2")))
# user system elapsed
# 119.911 0.530 120.749
# this answer
system.time(ans3 <- setDT(dt1)[dt2, c("field1", "field2") := list(field1, field2), on = c("id1", "id2")])
# user system elapsed
# 0.087 0.013 0.100
sessionInfo()
# R version 3.2.1 (2015-06-18)
# Platform: x86_64-apple-darwin13.4.0 (64-bit)
# Running under: OS X 10.10.4 (Yosemite)
# locale:
# [1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8
# attached base packages:
# [1] stats graphics grDevices utils datasets methods base
# other attached packages:
# [1] data.table_1.9.5 dplyr_0.4.2
# loaded via a namespace (and not attached):
# [1] magrittr_1.5 R6_2.1.0 assertthat_0.1 parallel_3.2.1 DBI_0.3.1
# [6] tools_3.2.1 Rcpp_0.12.0 chron_2.3-45
尽管如此，Dint期望dplyr的速度要慢大约1200倍。
答案 2 :(得分：2)
您可以将两个数据帧都转换为数据表，然后执行合并：
library(data.table)
setDT(df1); setDT(df2)
merge(df1,df2,by="id",allow.cartesian=T)
allow.cartesian部分允许在任何合并元素的键中存在重复值时进行合并（允许新表长度大于orignal元素的最大值，请参阅?data.table。
答案 3 :(得分：1)
我认为目前针对这些案例（大数据集）的最快解决方案是首先设置密钥后的 data.table 合并。
如上所述，您还可以将dplyr＆＃39; left_join与 data.frames 一起使用，但在转换后比较相同的命令会更好 data.frames 到 data.tables 。换句话说，在后台使用带有 data.table 结构的dplyr。
作为一个例子，我将创建两个数据集，然后将它们保存为data.frame，data.table和key，data.table没有密钥。 然后我将执行各种合并并计算时间：
library(data.table)
library(dplyr)
# create and save this dataset as a data.frame and as a data.table
list = seq(1,500000)
random_number = rnorm(500000,10,5)
dataT11 = data.table(list, random_number, key="list") # data.table with a key
dataT12 = data.table(list, random_number) # data.table without key
dataF1 = data.frame(list, random_number)
# create and save this dataset as a data.frame and as a data.table
list = seq(1,500000)
random_number = rnorm(500000,10,5)
dataT21 = data.table(list, random_number, key="list")
dataT22 = data.table(list, random_number)
dataF2 = data.frame(list, random_number)
# check your current data tables (note some have keys)
tables()
# merge the datasets as data.frames and count time
ptm <- proc.time()
dataF3 = merge(dataF1, dataF2, all.x=T)
proc.time() - ptm
# merge the datasets as data.tables by setting the key now and count time
ptm <- proc.time()
dataT3 = merge(dataT12, dataT22, all.x=T, by = "list")
proc.time() - ptm
# merge the datasets as data.tables on the key they have already and count time
ptm <- proc.time()
dataT3 = merge(dataT11, dataT21, all.x=T)
proc.time() - ptm
# merge the datasets as data.tables on the key they have already and count time (alternative)
ptm <- proc.time()
dataT3 = dataT11[dataT21]
proc.time() - ptm
# merge the datasets as data.frames using dplyr and count time
ptm <- proc.time()
dataT3 = dataF1 %>% left_join(dataF2, by="list")
proc.time() - ptm
# merge the datasets as data.tables using dplyr and count time
ptm <- proc.time()
dataT3 = dataT11 %>% left_join(dataT21, by="list")
proc.time() - ptm