1441

给定两个数据框:

df1 = data.frame(CustomerId = c(1:6), Product = c(rep("Toaster", 3), rep("Radio", 3)))
df2 = data.frame(CustomerId = c(2, 4, 6), State = c(rep("Alabama", 2), rep("Ohio", 1)))

df1
#  CustomerId Product
#           1 Toaster
#           2 Toaster
#           3 Toaster
#           4   Radio
#           5   Radio
#           6   Radio

df2
#  CustomerId   State
#           2 Alabama
#           4 Alabama
#           6    Ohio

我该如何做数据库风格,即sql 风格,连接?也就是说,我如何得到:

  • and的内连接: 仅返回左表在右表中具有匹配键的行。df1df2
  • and的外连接:返回两个表中的 所有行,从左侧连接在右侧表中具有匹配键的记录。df1df2
  • 和 返回左表中的所有行以及右表中具有匹配键的任何行的左外连接(或简称左连接)df1df2
  • 右外连接df1并返回右表中的df2
    所有行,以及左表中具有匹配键的任何行。

额外学分:

如何执行 SQL 样式的选择语句?

4

13 回答 13

1554

通过使用merge函数及其可选参数:

内连接: merge(df1, df2)将适用于这些示例,因为 R 会通过公共变量名称自动连接帧,但您很可能希望指定merge(df1, df2, by = "CustomerId")以确保仅匹配所需的字段。如果匹配的变量在不同的数据框中具有不同的名称,您也可以使用by.xandby.y

外联: merge(x = df1, y = df2, by = "CustomerId", all = TRUE)

左外: merge(x = df1, y = df2, by = "CustomerId", all.x = TRUE)

右外: merge(x = df1, y = df2, by = "CustomerId", all.y = TRUE)

交叉连接: merge(x = df1, y = df2, by = NULL)

与内部连接一样,您可能希望将“CustomerId”作为匹配变量显式传递给 R。 我认为最好明确说明要合并的标识符;如果输入 data.frames 意外更改并且以后更易于阅读,则更安全。

您可以通过给出by一个向量来合并多个列,例如by = c("CustomerId", "OrderId").

如果要合并的列名不同,您可以指定,例如,by.x = "CustomerId_in_df1", by.y = "CustomerId_in_df2"CustomerId_in_df1一个数据框中CustomerId_in_df2的列名和第二个数据框中的列名。(如果您需要在多列上合并,这些也可以是向量。)

于 2009-08-19T15:15:41.570 回答
249

我建议查看Gabor Grothendieck 的 sqldf 包,它允许您在 SQL 中表达这些操作。

library(sqldf)

## inner join
df3 <- sqldf("SELECT CustomerId, Product, State 
              FROM df1
              JOIN df2 USING(CustomerID)")

## left join (substitute 'right' for right join)
df4 <- sqldf("SELECT CustomerId, Product, State 
              FROM df1
              LEFT JOIN df2 USING(CustomerID)")

我发现 SQL 语法比它的 R 语法更简单、更自然(但这可能只是反映了我的 RDBMS 偏见)。

有关连接的更多信息,请参阅Gabor 的 sqldf GitHub

于 2009-08-20T17:54:49.390 回答
228

内部连接有data.table方法,它非常节省时间和内存(对于一些较大的 data.frames 是必需的):

library(data.table)

dt1 <- data.table(df1, key = "CustomerId") 
dt2 <- data.table(df2, key = "CustomerId")

joined.dt1.dt.2 <- dt1[dt2]

merge也适用于 data.tables (因为它是通用的并且调用merge.data.table

merge(dt1, dt2)

在 stackoverflow 上记录的 data.table:
如何进行 data.table 合并操作
将外键上的 SQL 连接转换为 R data.table 语法
用于合并较大 data.frames 的有效替代方法 R
如何使用 data.table 进行基本的左外连接在 R?

另一个选项是plyrjoin中的函数

library(plyr)

join(df1, df2,
     type = "inner")

#   CustomerId Product   State
# 1          2 Toaster Alabama
# 2          4   Radio Alabama
# 3          6   Radio    Ohio

选项typeinner, left, right, full.

From ?join:与 不同merge,[ join] 保留 x 的顺序,无论使用何种连接类型。

于 2012-03-11T06:24:15.003 回答
223

您也可以使用 Hadley Wickham 的出色 dplyr包进行连接。

library(dplyr)

#make sure that CustomerId cols are both type numeric
#they ARE not using the provided code in question and dplyr will complain
df1$CustomerId <- as.numeric(df1$CustomerId)
df2$CustomerId <- as.numeric(df2$CustomerId)

变异连接:使用 df2 中的匹配将列添加到 df1

#inner
inner_join(df1, df2)

#left outer
left_join(df1, df2)

#right outer
right_join(df1, df2)

#alternate right outer
left_join(df2, df1)

#full join
full_join(df1, df2)

过滤连接:过滤掉df1中的行,不要修改列

semi_join(df1, df2) #keep only observations in df1 that match in df2.
anti_join(df1, df2) #drops all observations in df1 that match in df2.
于 2014-02-06T21:35:06.437 回答
104

在R Wiki上有一些很好的例子。我会在这里偷几个:

合并方法

由于您的键名相同,因此进行内部连接的简短方法是 merge():

merge(df1, df2)

可以使用“all”关键字创建完整的内部连接(两个表中的所有记录):

merge(df1, df2, all=TRUE)

df1 和 df2 的左外连接:

merge(df1, df2, all.x=TRUE)

df1 和 df2 的右外连接:

merge(df1, df2, all.y=TRUE)

您可以翻转它们,拍打它们并摩擦它们以获得您询问的另外两个外部连接:)

下标法

使用下标方法在左侧与 df1 的左外连接将是:

df1[,"State"]<-df2[df1[ ,"Product"], "State"]

可以通过修改左外连接下标示例来创建外连接的其他组合。(是的,我知道这相当于说“我会把它作为练习留给读者......”)

于 2009-08-19T15:15:10.930 回答
90

2014 年新增:

特别是如果您对一般的数据操作(包括排序、过滤、子集、汇总等)也感兴趣,您绝对应该看看dplyr,它带有各种功能,旨在促进您专门处理数据框的工作和某些其他数据库类型。它甚至提供了相当精细的 SQL 接口,甚至提供了将(大多数)SQL 代码直接转换为 R 的函数。

dplyr 包中与加入相关的四个函数是(引用):

  • inner_join(x, y, by = NULL, copy = FALSE, ...): 返回 x 中在 y 中有匹配值的所有行,以及 x 和 y 中的所有列
  • left_join(x, y, by = NULL, copy = FALSE, ...): 返回 x 的所有行,以及 x 和 y 的所有列
  • semi_join(x, y, by = NULL, copy = FALSE, ...):返回 x 中在 y 中有匹配值的所有行,只保留 x 中的列。
  • anti_join(x, y, by = NULL, copy = FALSE, ...): 从 x 中返回所有在 y 中没有匹配值的行,只保留 x 中的列

这一切都非常详细。

可以通过 选择列select(df,"column")。如果这对您来说还不够 SQL-ish,那么sql()您可以在其中输入 SQL 代码原样的函数,它会执行您指定的操作,就像您一直在用 R 编写的一样(有关更多信息,请参阅到dplyr/databases 小插图)。例如,如果应用正确,sql("SELECT * FROM hflights")将从“hflights”dplyr 表(一个“tbl”)中选择所有列。

于 2014-01-29T17:43:26.480 回答
89

更新用于连接数据集的 data.table 方法。请参阅以下每种连接类型的示例。有两种方法,一种是[.data.table在将第二个 data.table 作为第一个参数传递给子集时,另一种方法是使用merge调度到快速 data.table 方法的函数。

df1 = data.frame(CustomerId = c(1:6), Product = c(rep("Toaster", 3), rep("Radio", 3)))
df2 = data.frame(CustomerId = c(2L, 4L, 7L), State = c(rep("Alabama", 2), rep("Ohio", 1))) # one value changed to show full outer join

library(data.table)

dt1 = as.data.table(df1)
dt2 = as.data.table(df2)
setkey(dt1, CustomerId)
setkey(dt2, CustomerId)
# right outer join keyed data.tables
dt1[dt2]

setkey(dt1, NULL)
setkey(dt2, NULL)
# right outer join unkeyed data.tables - use `on` argument
dt1[dt2, on = "CustomerId"]

# left outer join - swap dt1 with dt2
dt2[dt1, on = "CustomerId"]

# inner join - use `nomatch` argument
dt1[dt2, nomatch=NULL, on = "CustomerId"]

# anti join - use `!` operator
dt1[!dt2, on = "CustomerId"]

# inner join - using merge method
merge(dt1, dt2, by = "CustomerId")

# full outer join
merge(dt1, dt2, by = "CustomerId", all = TRUE)

# see ?merge.data.table arguments for other cases

下面的基准测试基于 R、sqldf、dplyr 和 data.table。
基准测试未键入/未索引的数据集。基准测试是在 50M-1 行数据集上执行的,连接列上有 50M-2 个公共值,因此可以测试每个场景(内、左、右、全),并且执行连接仍然不是一件容易的事。它是一种很好地强调连接算法的连接类型。时间为sqldf:0.4.11, dplyr:0.7.8, data.table:1.12.0.

# inner
Unit: seconds
   expr       min        lq      mean    median        uq       max neval
   base 111.66266 111.66266 111.66266 111.66266 111.66266 111.66266     1
  sqldf 624.88388 624.88388 624.88388 624.88388 624.88388 624.88388     1
  dplyr  51.91233  51.91233  51.91233  51.91233  51.91233  51.91233     1
     DT  10.40552  10.40552  10.40552  10.40552  10.40552  10.40552     1
# left
Unit: seconds
   expr        min         lq       mean     median         uq        max 
   base 142.782030 142.782030 142.782030 142.782030 142.782030 142.782030     
  sqldf 613.917109 613.917109 613.917109 613.917109 613.917109 613.917109     
  dplyr  49.711912  49.711912  49.711912  49.711912  49.711912  49.711912     
     DT   9.674348   9.674348   9.674348   9.674348   9.674348   9.674348       
# right
Unit: seconds
   expr        min         lq       mean     median         uq        max
   base 122.366301 122.366301 122.366301 122.366301 122.366301 122.366301     
  sqldf 611.119157 611.119157 611.119157 611.119157 611.119157 611.119157     
  dplyr  50.384841  50.384841  50.384841  50.384841  50.384841  50.384841     
     DT   9.899145   9.899145   9.899145   9.899145   9.899145   9.899145     
# full
Unit: seconds
  expr       min        lq      mean    median        uq       max neval
  base 141.79464 141.79464 141.79464 141.79464 141.79464 141.79464     1
 dplyr  94.66436  94.66436  94.66436  94.66436  94.66436  94.66436     1
    DT  21.62573  21.62573  21.62573  21.62573  21.62573  21.62573     1

请注意,您可以使用其他类型的连接执行data.table
-连接时更新- 如果您想从另一个表查找值到您的主表
-连接时聚合- 如果您想聚合您正在加入的键,您没有实现所有连接结果
-重叠连接- 如果您想按范围合并
-滚动连接- 如果您希望合并能够通过向前或向后滚动来匹配前/后行的​​值
-非等连接- 如果您的连接条件不相等

重现代码:

library(microbenchmark)
library(sqldf)
library(dplyr)
library(data.table)
sapply(c("sqldf","dplyr","data.table"), packageVersion, simplify=FALSE)

n = 5e7
set.seed(108)
df1 = data.frame(x=sample(n,n-1L), y1=rnorm(n-1L))
df2 = data.frame(x=sample(n,n-1L), y2=rnorm(n-1L))
dt1 = as.data.table(df1)
dt2 = as.data.table(df2)

mb = list()
# inner join
microbenchmark(times = 1L,
               base = merge(df1, df2, by = "x"),
               sqldf = sqldf("SELECT * FROM df1 INNER JOIN df2 ON df1.x = df2.x"),
               dplyr = inner_join(df1, df2, by = "x"),
               DT = dt1[dt2, nomatch=NULL, on = "x"]) -> mb$inner

# left outer join
microbenchmark(times = 1L,
               base = merge(df1, df2, by = "x", all.x = TRUE),
               sqldf = sqldf("SELECT * FROM df1 LEFT OUTER JOIN df2 ON df1.x = df2.x"),
               dplyr = left_join(df1, df2, by = c("x"="x")),
               DT = dt2[dt1, on = "x"]) -> mb$left

# right outer join
microbenchmark(times = 1L,
               base = merge(df1, df2, by = "x", all.y = TRUE),
               sqldf = sqldf("SELECT * FROM df2 LEFT OUTER JOIN df1 ON df2.x = df1.x"),
               dplyr = right_join(df1, df2, by = "x"),
               DT = dt1[dt2, on = "x"]) -> mb$right

# full outer join
microbenchmark(times = 1L,
               base = merge(df1, df2, by = "x", all = TRUE),
               dplyr = full_join(df1, df2, by = "x"),
               DT = merge(dt1, dt2, by = "x", all = TRUE)) -> mb$full

lapply(mb, print) -> nul
于 2015-12-11T09:23:09.520 回答
38

dplyr 自 0.4 起实现了所有这些连接,包括outer_join,但值得注意的是,对于 0.4 之前的前几个版本,它曾经不提供outer_join,因此有很多非常糟糕的 hacky 解决方法用户代码漂浮了很长一段时间。之后(你仍然可以在那个时期的 SO、Kaggle 答案、github 中找到这样的代码。因此这个答案仍然有用。)

加入相关的发布亮点

v0.5 (6/2016)

  • 处理 POSIXct 类型、时区、重复项、不同因素级别。更好的错误和警告。
  • 新的 suffix 参数来控制后缀重复的变量名称接收 (#1296)

v0.4.0 (1/2015)

  • 实现右连接和外连接(#96)
  • 变异连接,将新变量从另一个表中的匹配行添加到另一个表中。过滤连接,它根据一个表中的观察值是否匹配另一个表中的观察值来过滤它们。

v0.3 (10/2014)

  • 现在可以通过每个表中的不同变量 left_join: df1 %>% left_join(df2, c("var1" = "var2"))

v0.2 (5/2014)

  • *_join() 不再对列名重新排序 (#324)

v0.1.3 (4/2014)

根据 hadley 在该问题中的评论的解决方法:

  • right_join (x,y) 在行方面与 left_join(y,x) 相同,只是列的顺序不同。使用 select(new_column_order) 轻松解决
  • outer_join基本上是 union(left_join(x, y), right_join(x, y)) - 即保留两个数据帧中的所有行。
于 2014-04-13T10:39:03.113 回答
31

对于具有基数的左连接0..*:0..1或具有基数的右连接的情况,可以将连接器(表)0..1:0..*中的单边列直接分配到连接器(表)上,从而避免创建一个全新的数据表。这需要将加入者的键列匹配到加入者中,并为分配相应地对加入者的行进行索引+排序。0..10..*

如果键是单个列,那么我们可以使用单个调用来match()进行匹配。这就是我将在此答案中介绍的情况。

这是一个基于 OP 的示例,除了我添加了df2一个 id 为 7 的额外行来测试连接器中不匹配键的情况。这实际上是df1左连接df2

df1 <- data.frame(CustomerId=1:6,Product=c(rep('Toaster',3L),rep('Radio',3L)));
df2 <- data.frame(CustomerId=c(2L,4L,6L,7L),State=c(rep('Alabama',2L),'Ohio','Texas'));
df1[names(df2)[-1L]] <- df2[match(df1[,1L],df2[,1L]),-1L];
df1;
##   CustomerId Product   State
## 1          1 Toaster    <NA>
## 2          2 Toaster Alabama
## 3          3 Toaster    <NA>
## 4          4   Radio Alabama
## 5          5   Radio    <NA>
## 6          6   Radio    Ohio

在上面我硬编码了一个假设,即键列是两个输入表的第一列。我认为,总的来说,这不是一个不合理的假设,因为,如果你有一个带有关键列的 data.frame,如果它没有被设置为 data.frame 的第一列,那就太奇怪了一开始。您可以随时重新排序列以使其如此。这种假设的一个有利结果是键列的名称不必是硬编码的,尽管我认为它只是用另一个假设替换了一个假设。简洁是整数索引的另一个优点,也是速度。在下面的基准测试中,我将更改实现以使用字符串名称索引来匹配竞争实现。

我认为如果您有多个表要与单个大表连接,那么这是一个特别合适的解决方案。为每次合并重复重建整个表是不必要且低效的。

另一方面,如果您出于某种原因需要通过此操作保持被加入者保持不变,则不能使用此解决方案,因为它直接修改了被加入者。尽管在这种情况下,您可以简单地制作副本并在副本上执行就地分配。


作为旁注,我简要研究了多列键的可能匹配解决方案。不幸的是,我找到的唯一匹配解决方案是:

  • 低效的串联。例如match(interaction(df1$a,df1$b),interaction(df2$a,df2$b)),或与paste().
  • 低效的笛卡尔连词,例如outer(df1$a,df2$a,`==`) & outer(df1$b,df2$b,`==`)
  • base Rmerge()和等效的基于包的合并函数,它们总是分配一个新表来返回合并结果,因此不适合基于就地分配的解决方案。

例如,请参阅Matching multiple columns on different data frames and getting other column as result将两列与另外两列匹配,Matching on multiple columns,以及我最初提出就地解决方案的这个问题的欺骗,Combine R 中具有不同行数的两个数据帧


基准测试

我决定做自己的基准测试,看看就地分配方法与这个问题中提供的其他解决方案相比如何。

测试代码:

library(microbenchmark);
library(data.table);
library(sqldf);
library(plyr);
library(dplyr);

solSpecs <- list(
    merge=list(testFuncs=list(
        inner=function(df1,df2,key) merge(df1,df2,key),
        left =function(df1,df2,key) merge(df1,df2,key,all.x=T),
        right=function(df1,df2,key) merge(df1,df2,key,all.y=T),
        full =function(df1,df2,key) merge(df1,df2,key,all=T)
    )),
    data.table.unkeyed=list(argSpec='data.table.unkeyed',testFuncs=list(
        inner=function(dt1,dt2,key) dt1[dt2,on=key,nomatch=0L,allow.cartesian=T],
        left =function(dt1,dt2,key) dt2[dt1,on=key,allow.cartesian=T],
        right=function(dt1,dt2,key) dt1[dt2,on=key,allow.cartesian=T],
        full =function(dt1,dt2,key) merge(dt1,dt2,key,all=T,allow.cartesian=T) ## calls merge.data.table()
    )),
    data.table.keyed=list(argSpec='data.table.keyed',testFuncs=list(
        inner=function(dt1,dt2) dt1[dt2,nomatch=0L,allow.cartesian=T],
        left =function(dt1,dt2) dt2[dt1,allow.cartesian=T],
        right=function(dt1,dt2) dt1[dt2,allow.cartesian=T],
        full =function(dt1,dt2) merge(dt1,dt2,all=T,allow.cartesian=T) ## calls merge.data.table()
    )),
    sqldf.unindexed=list(testFuncs=list( ## note: must pass connection=NULL to avoid running against the live DB connection, which would result in collisions with the residual tables from the last query upload
        inner=function(df1,df2,key) sqldf(paste0('select * from df1 inner join df2 using(',paste(collapse=',',key),')'),connection=NULL),
        left =function(df1,df2,key) sqldf(paste0('select * from df1 left join df2 using(',paste(collapse=',',key),')'),connection=NULL),
        right=function(df1,df2,key) sqldf(paste0('select * from df2 left join df1 using(',paste(collapse=',',key),')'),connection=NULL) ## can't do right join proper, not yet supported; inverted left join is equivalent
        ##full =function(df1,df2,key) sqldf(paste0('select * from df1 full join df2 using(',paste(collapse=',',key),')'),connection=NULL) ## can't do full join proper, not yet supported; possible to hack it with a union of left joins, but too unreasonable to include in testing
    )),
    sqldf.indexed=list(testFuncs=list( ## important: requires an active DB connection with preindexed main.df1 and main.df2 ready to go; arguments are actually ignored
        inner=function(df1,df2,key) sqldf(paste0('select * from main.df1 inner join main.df2 using(',paste(collapse=',',key),')')),
        left =function(df1,df2,key) sqldf(paste0('select * from main.df1 left join main.df2 using(',paste(collapse=',',key),')')),
        right=function(df1,df2,key) sqldf(paste0('select * from main.df2 left join main.df1 using(',paste(collapse=',',key),')')) ## can't do right join proper, not yet supported; inverted left join is equivalent
        ##full =function(df1,df2,key) sqldf(paste0('select * from main.df1 full join main.df2 using(',paste(collapse=',',key),')')) ## can't do full join proper, not yet supported; possible to hack it with a union of left joins, but too unreasonable to include in testing
    )),
    plyr=list(testFuncs=list(
        inner=function(df1,df2,key) join(df1,df2,key,'inner'),
        left =function(df1,df2,key) join(df1,df2,key,'left'),
        right=function(df1,df2,key) join(df1,df2,key,'right'),
        full =function(df1,df2,key) join(df1,df2,key,'full')
    )),
    dplyr=list(testFuncs=list(
        inner=function(df1,df2,key) inner_join(df1,df2,key),
        left =function(df1,df2,key) left_join(df1,df2,key),
        right=function(df1,df2,key) right_join(df1,df2,key),
        full =function(df1,df2,key) full_join(df1,df2,key)
    )),
    in.place=list(testFuncs=list(
        left =function(df1,df2,key) { cns <- setdiff(names(df2),key); df1[cns] <- df2[match(df1[,key],df2[,key]),cns]; df1; },
        right=function(df1,df2,key) { cns <- setdiff(names(df1),key); df2[cns] <- df1[match(df2[,key],df1[,key]),cns]; df2; }
    ))
);

getSolTypes <- function() names(solSpecs);
getJoinTypes <- function() unique(unlist(lapply(solSpecs,function(x) names(x$testFuncs))));
getArgSpec <- function(argSpecs,key=NULL) if (is.null(key)) argSpecs$default else argSpecs[[key]];

initSqldf <- function() {
    sqldf(); ## creates sqlite connection on first run, cleans up and closes existing connection otherwise
    if (exists('sqldfInitFlag',envir=globalenv(),inherits=F) && sqldfInitFlag) { ## false only on first run
        sqldf(); ## creates a new connection
    } else {
        assign('sqldfInitFlag',T,envir=globalenv()); ## set to true for the one and only time
    }; ## end if
    invisible();
}; ## end initSqldf()

setUpBenchmarkCall <- function(argSpecs,joinType,solTypes=getSolTypes(),env=parent.frame()) {
    ## builds and returns a list of expressions suitable for passing to the list argument of microbenchmark(), and assigns variables to resolve symbol references in those expressions
    callExpressions <- list();
    nms <- character();
    for (solType in solTypes) {
        testFunc <- solSpecs[[solType]]$testFuncs[[joinType]];
        if (is.null(testFunc)) next; ## this join type is not defined for this solution type
        testFuncName <- paste0('tf.',solType);
        assign(testFuncName,testFunc,envir=env);
        argSpecKey <- solSpecs[[solType]]$argSpec;
        argSpec <- getArgSpec(argSpecs,argSpecKey);
        argList <- setNames(nm=names(argSpec$args),vector('list',length(argSpec$args)));
        for (i in seq_along(argSpec$args)) {
            argName <- paste0('tfa.',argSpecKey,i);
            assign(argName,argSpec$args[[i]],envir=env);
            argList[[i]] <- if (i%in%argSpec$copySpec) call('copy',as.symbol(argName)) else as.symbol(argName);
        }; ## end for
        callExpressions[[length(callExpressions)+1L]] <- do.call(call,c(list(testFuncName),argList),quote=T);
        nms[length(nms)+1L] <- solType;
    }; ## end for
    names(callExpressions) <- nms;
    callExpressions;
}; ## end setUpBenchmarkCall()

harmonize <- function(res) {
    res <- as.data.frame(res); ## coerce to data.frame
    for (ci in which(sapply(res,is.factor))) res[[ci]] <- as.character(res[[ci]]); ## coerce factor columns to character
    for (ci in which(sapply(res,is.logical))) res[[ci]] <- as.integer(res[[ci]]); ## coerce logical columns to integer (works around sqldf quirk of munging logicals to integers)
    ##for (ci in which(sapply(res,inherits,'POSIXct'))) res[[ci]] <- as.double(res[[ci]]); ## coerce POSIXct columns to double (works around sqldf quirk of losing POSIXct class) ----- POSIXct doesn't work at all in sqldf.indexed
    res <- res[order(names(res))]; ## order columns
    res <- res[do.call(order,res),]; ## order rows
    res;
}; ## end harmonize()

checkIdentical <- function(argSpecs,solTypes=getSolTypes()) {
    for (joinType in getJoinTypes()) {
        callExpressions <- setUpBenchmarkCall(argSpecs,joinType,solTypes);
        if (length(callExpressions)<2L) next;
        ex <- harmonize(eval(callExpressions[[1L]]));
        for (i in seq(2L,len=length(callExpressions)-1L)) {
            y <- harmonize(eval(callExpressions[[i]]));
            if (!isTRUE(all.equal(ex,y,check.attributes=F))) {
                ex <<- ex;
                y <<- y;
                solType <- names(callExpressions)[i];
                stop(paste0('non-identical: ',solType,' ',joinType,'.'));
            }; ## end if
        }; ## end for
    }; ## end for
    invisible();
}; ## end checkIdentical()

testJoinType <- function(argSpecs,joinType,solTypes=getSolTypes(),metric=NULL,times=100L) {
    callExpressions <- setUpBenchmarkCall(argSpecs,joinType,solTypes);
    bm <- microbenchmark(list=callExpressions,times=times);
    if (is.null(metric)) return(bm);
    bm <- summary(bm);
    res <- setNames(nm=names(callExpressions),bm[[metric]]);
    attr(res,'unit') <- attr(bm,'unit');
    res;
}; ## end testJoinType()

testAllJoinTypes <- function(argSpecs,solTypes=getSolTypes(),metric=NULL,times=100L) {
    joinTypes <- getJoinTypes();
    resList <- setNames(nm=joinTypes,lapply(joinTypes,function(joinType) testJoinType(argSpecs,joinType,solTypes,metric,times)));
    if (is.null(metric)) return(resList);
    units <- unname(unlist(lapply(resList,attr,'unit')));
    res <- do.call(data.frame,c(list(join=joinTypes),setNames(nm=solTypes,rep(list(rep(NA_real_,length(joinTypes))),length(solTypes))),list(unit=units,stringsAsFactors=F)));
    for (i in seq_along(resList)) res[i,match(names(resList[[i]]),names(res))] <- resList[[i]];
    res;
}; ## end testAllJoinTypes()

testGrid <- function(makeArgSpecsFunc,sizes,overlaps,solTypes=getSolTypes(),joinTypes=getJoinTypes(),metric='median',times=100L) {

    res <- expand.grid(size=sizes,overlap=overlaps,joinType=joinTypes,stringsAsFactors=F);
    res[solTypes] <- NA_real_;
    res$unit <- NA_character_;
    for (ri in seq_len(nrow(res))) {

        size <- res$size[ri];
        overlap <- res$overlap[ri];
        joinType <- res$joinType[ri];

        argSpecs <- makeArgSpecsFunc(size,overlap);

        checkIdentical(argSpecs,solTypes);

        cur <- testJoinType(argSpecs,joinType,solTypes,metric,times);
        res[ri,match(names(cur),names(res))] <- cur;
        res$unit[ri] <- attr(cur,'unit');

    }; ## end for

    res;

}; ## end testGrid()

这是基于我之前演示的 OP 的示例基准:

## OP's example, supplemented with a non-matching row in df2
argSpecs <- list(
    default=list(copySpec=1:2,args=list(
        df1 <- data.frame(CustomerId=1:6,Product=c(rep('Toaster',3L),rep('Radio',3L))),
        df2 <- data.frame(CustomerId=c(2L,4L,6L,7L),State=c(rep('Alabama',2L),'Ohio','Texas')),
        'CustomerId'
    )),
    data.table.unkeyed=list(copySpec=1:2,args=list(
        as.data.table(df1),
        as.data.table(df2),
        'CustomerId'
    )),
    data.table.keyed=list(copySpec=1:2,args=list(
        setkey(as.data.table(df1),CustomerId),
        setkey(as.data.table(df2),CustomerId)
    ))
);
## prepare sqldf
initSqldf();
sqldf('create index df1_key on df1(CustomerId);'); ## upload and create an sqlite index on df1
sqldf('create index df2_key on df2(CustomerId);'); ## upload and create an sqlite index on df2

checkIdentical(argSpecs);

testAllJoinTypes(argSpecs,metric='median');
##    join    merge data.table.unkeyed data.table.keyed sqldf.unindexed sqldf.indexed      plyr    dplyr in.place         unit
## 1 inner  644.259           861.9345          923.516        9157.752      1580.390  959.2250 270.9190       NA microseconds
## 2  left  713.539           888.0205          910.045        8820.334      1529.714  968.4195 270.9185 224.3045 microseconds
## 3 right 1221.804           909.1900          923.944        8930.668      1533.135 1063.7860 269.8495 218.1035 microseconds
## 4  full 1302.203          3107.5380         3184.729              NA            NA 1593.6475 270.7055       NA microseconds

在这里,我对随机输入数据进行基准测试,在两个输入表之间尝试不同的比例和不同的键重叠模式。此基准仍然限于单列整数键的情况。同样,为了确保就地解决方案适用于同一表的左连接和右连接,所有随机测试数据都使用0..1:0..1基数。这是通过在生成第二个 data.frame 的 key 列时对第一个 data.frame 的 key 列进行采样而不替换来实现的。

makeArgSpecs.singleIntegerKey.optionalOneToOne <- function(size,overlap) {

    com <- as.integer(size*overlap);

    argSpecs <- list(
        default=list(copySpec=1:2,args=list(
            df1 <- data.frame(id=sample(size),y1=rnorm(size),y2=rnorm(size)),
            df2 <- data.frame(id=sample(c(if (com>0L) sample(df1$id,com) else integer(),seq(size+1L,len=size-com))),y3=rnorm(size),y4=rnorm(size)),
            'id'
        )),
        data.table.unkeyed=list(copySpec=1:2,args=list(
            as.data.table(df1),
            as.data.table(df2),
            'id'
        )),
        data.table.keyed=list(copySpec=1:2,args=list(
            setkey(as.data.table(df1),id),
            setkey(as.data.table(df2),id)
        ))
    );
    ## prepare sqldf
    initSqldf();
    sqldf('create index df1_key on df1(id);'); ## upload and create an sqlite index on df1
    sqldf('create index df2_key on df2(id);'); ## upload and create an sqlite index on df2

    argSpecs;

}; ## end makeArgSpecs.singleIntegerKey.optionalOneToOne()

## cross of various input sizes and key overlaps
sizes <- c(1e1L,1e3L,1e6L);
overlaps <- c(0.99,0.5,0.01);
system.time({ res <- testGrid(makeArgSpecs.singleIntegerKey.optionalOneToOne,sizes,overlaps); });
##     user   system  elapsed
## 22024.65 12308.63 34493.19

我编写了一些代码来创建上述结果的对数图。我为每个重叠百分比生成了一个单独的图。它有点混乱,但我喜欢在同一个图中表示所有解决方案类型和连接类型。

我使用样条插值来显示每个解决方案/连接类型组合的平滑曲线,用单独的 pch 符号绘制。连接类型由 pch 符号捕获,使用点表示内部,左右尖括号表示左右,菱形表示完整。解决方案类型由图例中所示的颜色捕获。

plotRes <- function(res,titleFunc,useFloor=F) {
    solTypes <- setdiff(names(res),c('size','overlap','joinType','unit')); ## derive from res
    normMult <- c(microseconds=1e-3,milliseconds=1); ## normalize to milliseconds
    joinTypes <- getJoinTypes();
    cols <- c(merge='purple',data.table.unkeyed='blue',data.table.keyed='#00DDDD',sqldf.unindexed='brown',sqldf.indexed='orange',plyr='red',dplyr='#00BB00',in.place='magenta');
    pchs <- list(inner=20L,left='<',right='>',full=23L);
    cexs <- c(inner=0.7,left=1,right=1,full=0.7);
    NP <- 60L;
    ord <- order(decreasing=T,colMeans(res[res$size==max(res$size),solTypes],na.rm=T));
    ymajors <- data.frame(y=c(1,1e3),label=c('1ms','1s'),stringsAsFactors=F);
    for (overlap in unique(res$overlap)) {
        x1 <- res[res$overlap==overlap,];
        x1[solTypes] <- x1[solTypes]*normMult[x1$unit]; x1$unit <- NULL;
        xlim <- c(1e1,max(x1$size));
        xticks <- 10^seq(log10(xlim[1L]),log10(xlim[2L]));
        ylim <- c(1e-1,10^((if (useFloor) floor else ceiling)(log10(max(x1[solTypes],na.rm=T))))); ## use floor() to zoom in a little more, only sqldf.unindexed will break above, but xpd=NA will keep it visible
        yticks <- 10^seq(log10(ylim[1L]),log10(ylim[2L]));
        yticks.minor <- rep(yticks[-length(yticks)],each=9L)*1:9;
        plot(NA,xlim=xlim,ylim=ylim,xaxs='i',yaxs='i',axes=F,xlab='size (rows)',ylab='time (ms)',log='xy');
        abline(v=xticks,col='lightgrey');
        abline(h=yticks.minor,col='lightgrey',lty=3L);
        abline(h=yticks,col='lightgrey');
        axis(1L,xticks,parse(text=sprintf('10^%d',as.integer(log10(xticks)))));
        axis(2L,yticks,parse(text=sprintf('10^%d',as.integer(log10(yticks)))),las=1L);
        axis(4L,ymajors$y,ymajors$label,las=1L,tick=F,cex.axis=0.7,hadj=0.5);
        for (joinType in rev(joinTypes)) { ## reverse to draw full first, since it's larger and would be more obtrusive if drawn last
            x2 <- x1[x1$joinType==joinType,];
            for (solType in solTypes) {
                if (any(!is.na(x2[[solType]]))) {
                    xy <- spline(x2$size,x2[[solType]],xout=10^(seq(log10(x2$size[1L]),log10(x2$size[nrow(x2)]),len=NP)));
                    points(xy$x,xy$y,pch=pchs[[joinType]],col=cols[solType],cex=cexs[joinType],xpd=NA);
                }; ## end if
            }; ## end for
        }; ## end for
        ## custom legend
        ## due to logarithmic skew, must do all distance calcs in inches, and convert to user coords afterward
        ## the bottom-left corner of the legend will be defined in normalized figure coords, although we can convert to inches immediately
        leg.cex <- 0.7;
        leg.x.in <- grconvertX(0.275,'nfc','in');
        leg.y.in <- grconvertY(0.6,'nfc','in');
        leg.x.user <- grconvertX(leg.x.in,'in');
        leg.y.user <- grconvertY(leg.y.in,'in');
        leg.outpad.w.in <- 0.1;
        leg.outpad.h.in <- 0.1;
        leg.midpad.w.in <- 0.1;
        leg.midpad.h.in <- 0.1;
        leg.sol.w.in <- max(strwidth(solTypes,'in',leg.cex));
        leg.sol.h.in <- max(strheight(solTypes,'in',leg.cex))*1.5; ## multiplication factor for greater line height
        leg.join.w.in <- max(strheight(joinTypes,'in',leg.cex))*1.5; ## ditto
        leg.join.h.in <- max(strwidth(joinTypes,'in',leg.cex));
        leg.main.w.in <- leg.join.w.in*length(joinTypes);
        leg.main.h.in <- leg.sol.h.in*length(solTypes);
        leg.x2.user <- grconvertX(leg.x.in+leg.outpad.w.in*2+leg.main.w.in+leg.midpad.w.in+leg.sol.w.in,'in');
        leg.y2.user <- grconvertY(leg.y.in+leg.outpad.h.in*2+leg.main.h.in+leg.midpad.h.in+leg.join.h.in,'in');
        leg.cols.x.user <- grconvertX(leg.x.in+leg.outpad.w.in+leg.join.w.in*(0.5+seq(0L,length(joinTypes)-1L)),'in');
        leg.lines.y.user <- grconvertY(leg.y.in+leg.outpad.h.in+leg.main.h.in-leg.sol.h.in*(0.5+seq(0L,length(solTypes)-1L)),'in');
        leg.sol.x.user <- grconvertX(leg.x.in+leg.outpad.w.in+leg.main.w.in+leg.midpad.w.in,'in');
        leg.join.y.user <- grconvertY(leg.y.in+leg.outpad.h.in+leg.main.h.in+leg.midpad.h.in,'in');
        rect(leg.x.user,leg.y.user,leg.x2.user,leg.y2.user,col='white');
        text(leg.sol.x.user,leg.lines.y.user,solTypes[ord],cex=leg.cex,pos=4L,offset=0);
        text(leg.cols.x.user,leg.join.y.user,joinTypes,cex=leg.cex,pos=4L,offset=0,srt=90); ## srt rotation applies *after* pos/offset positioning
        for (i in seq_along(joinTypes)) {
            joinType <- joinTypes[i];
            points(rep(leg.cols.x.user[i],length(solTypes)),ifelse(colSums(!is.na(x1[x1$joinType==joinType,solTypes[ord]]))==0L,NA,leg.lines.y.user),pch=pchs[[joinType]],col=cols[solTypes[ord]]);
        }; ## end for
        title(titleFunc(overlap));
        readline(sprintf('overlap %.02f',overlap));
    }; ## end for
}; ## end plotRes()

titleFunc <- function(overlap) sprintf('R merge solutions: single-column integer key, 0..1:0..1 cardinality, %d%% overlap',as.integer(overlap*100));
plotRes(res,titleFunc,T);

R-merge-benchmark-single-column-integer-key-optional-one-to-one-99

R-merge-benchmark-single-column-integer-key-optional-one-to-one-50

R-merge-benchmark-single-column-integer-key-optional-one-to-one-1


这是第二个大型基准测试,它在关键列的数量和类型以及基数方面更加繁重。对于这个基准,我使用三个关键列:一个字符、一个整数和一个逻辑,对基数没有限制(即0..*:0..*)。(一般来说,由于浮点比较的复杂性,不建议使用双精度值或复值定义键列,而且基本上没有人使用原始类型,更不用说键列了,所以我没有将这些类型包含在键中列。另外,为了提供信息,我最初尝试通过包含一个 POSIXct 键列来使用四个键列,但是由于某种原因,POSIXct 类型在解决方案中表现不佳sqldf.indexed,可能是由于浮点比较异常,所以我删除它。)

makeArgSpecs.assortedKey.optionalManyToMany <- function(size,overlap,uniquePct=75) {

    ## number of unique keys in df1
    u1Size <- as.integer(size*uniquePct/100);

    ## (roughly) divide u1Size into bases, so we can use expand.grid() to produce the required number of unique key values with repetitions within individual key columns
    ## use ceiling() to ensure we cover u1Size; will truncate afterward
    u1SizePerKeyColumn <- as.integer(ceiling(u1Size^(1/3)));

    ## generate the unique key values for df1
    keys1 <- expand.grid(stringsAsFactors=F,
        idCharacter=replicate(u1SizePerKeyColumn,paste(collapse='',sample(letters,sample(4:12,1L),T))),
        idInteger=sample(u1SizePerKeyColumn),
        idLogical=sample(c(F,T),u1SizePerKeyColumn,T)
        ##idPOSIXct=as.POSIXct('2016-01-01 00:00:00','UTC')+sample(u1SizePerKeyColumn)
    )[seq_len(u1Size),];

    ## rbind some repetitions of the unique keys; this will prepare one side of the many-to-many relationship
    ## also scramble the order afterward
    keys1 <- rbind(keys1,keys1[sample(nrow(keys1),size-u1Size,T),])[sample(size),];

    ## common and unilateral key counts
    com <- as.integer(size*overlap);
    uni <- size-com;

    ## generate some unilateral keys for df2 by synthesizing outside of the idInteger range of df1
    keys2 <- data.frame(stringsAsFactors=F,
        idCharacter=replicate(uni,paste(collapse='',sample(letters,sample(4:12,1L),T))),
        idInteger=u1SizePerKeyColumn+sample(uni),
        idLogical=sample(c(F,T),uni,T)
        ##idPOSIXct=as.POSIXct('2016-01-01 00:00:00','UTC')+u1SizePerKeyColumn+sample(uni)
    );

    ## rbind random keys from df1; this will complete the many-to-many relationship
    ## also scramble the order afterward
    keys2 <- rbind(keys2,keys1[sample(nrow(keys1),com,T),])[sample(size),];

    ##keyNames <- c('idCharacter','idInteger','idLogical','idPOSIXct');
    keyNames <- c('idCharacter','idInteger','idLogical');
    ## note: was going to use raw and complex type for two of the non-key columns, but data.table doesn't seem to fully support them
    argSpecs <- list(
        default=list(copySpec=1:2,args=list(
            df1 <- cbind(stringsAsFactors=F,keys1,y1=sample(c(F,T),size,T),y2=sample(size),y3=rnorm(size),y4=replicate(size,paste(collapse='',sample(letters,sample(4:12,1L),T)))),
            df2 <- cbind(stringsAsFactors=F,keys2,y5=sample(c(F,T),size,T),y6=sample(size),y7=rnorm(size),y8=replicate(size,paste(collapse='',sample(letters,sample(4:12,1L),T)))),
            keyNames
        )),
        data.table.unkeyed=list(copySpec=1:2,args=list(
            as.data.table(df1),
            as.data.table(df2),
            keyNames
        )),
        data.table.keyed=list(copySpec=1:2,args=list(
            setkeyv(as.data.table(df1),keyNames),
            setkeyv(as.data.table(df2),keyNames)
        ))
    );
    ## prepare sqldf
    initSqldf();
    sqldf(paste0('create index df1_key on df1(',paste(collapse=',',keyNames),');')); ## upload and create an sqlite index on df1
    sqldf(paste0('create index df2_key on df2(',paste(collapse=',',keyNames),');')); ## upload and create an sqlite index on df2

    argSpecs;

}; ## end makeArgSpecs.assortedKey.optionalManyToMany()

sizes <- c(1e1L,1e3L,1e5L); ## 1e5L instead of 1e6L to respect more heavy-duty inputs
overlaps <- c(0.99,0.5,0.01);
solTypes <- setdiff(getSolTypes(),'in.place');
system.time({ res <- testGrid(makeArgSpecs.assortedKey.optionalManyToMany,sizes,overlaps,solTypes); });
##     user   system  elapsed
## 38895.50   784.19 39745.53

使用上面给出的相同绘图代码生成的绘图:

titleFunc <- function(overlap) sprintf('R merge solutions: character/integer/logical key, 0..*:0..* cardinality, %d%% overlap',as.integer(overlap*100));
plotRes(res,titleFunc,F);

R-merge-benchmark-assorted-key-optional-many-to-many-99

R-merge-benchmark-assorted-key-optional-many-to-many-50

R-merge-benchmark-assorted-key-optional-many-to-many-1

于 2016-06-30T18:11:26.883 回答
28

在加入两个数据帧时,每个数据帧有大约 100 万行,一个有 2 列,另一个有大约 20 列,我惊讶地发现merge(..., all.x = TRUE, all.y = TRUE)它的速度更快dplyr::full_join()。这是 dplyr v0.4

合并大约需要 17 秒,full_join 大约需要 65 秒。

不过有些食物,因为我通常默认使用 dplyr 来执行操作任务。

于 2015-02-26T18:11:13.897 回答
12
  1. 使用merge函数我们可以选择左表或右表的变量,就像我们熟悉的 SQL 中的 select 语句一样(EX : Select a.* ...or Select b.* from .....)
  2. 我们必须添加额外的代码,这些代码将从新加入的表中获取子集。

    • SQL:- select a.* from df1 a inner join df2 b on a.CustomerId=b.CustomerId

    • 回复:-merge(df1, df2, by.x = "CustomerId", by.y = "CustomerId")[,names(df1)]

同样的方法

  • SQL:-select b.* from df1 a inner join df2 b on a.CustomerId=b.CustomerId

  • 回复:-merge(df1, df2, by.x = "CustomerId", by.y = "CustomerId")[,names(df2)]

于 2015-08-26T09:57:46.753 回答
11

对于所有列的内部连接,您还可以使用fintersectdata.table -packageintersectdplyr -package作为merge不指定by-columns 的替代方法。这将给出两个数据帧之间相等的行:

merge(df1, df2)
#   V1 V2
# 1  B  2
# 2  C  3

dplyr::intersect(df1, df2)
#   V1 V2
# 1  B  2
# 2  C  3

data.table::fintersect(setDT(df1), setDT(df2))
#    V1 V2
# 1:  B  2
# 2:  C  3

示例数据:

df1 <- data.frame(V1 = LETTERS[1:4], V2 = 1:4)
df2 <- data.frame(V1 = LETTERS[2:3], V2 = 2:3)
于 2017-09-11T11:35:31.030 回答
8

更新加入。另一个重要的 SQL 样式连接是“更新连接”,其中一个表中的列使用另一个表更新(或创建)。

修改 OP 的示例表...

sales = data.frame(
  CustomerId = c(1, 1, 1, 3, 4, 6), 
  Year = 2000:2005,
  Product = c(rep("Toaster", 3), rep("Radio", 3))
)
cust = data.frame(
  CustomerId = c(1, 1, 4, 6), 
  Year = c(2001L, 2002L, 2002L, 2002L),
  State = state.name[1:4]
)

sales
# CustomerId Year Product
#          1 2000 Toaster
#          1 2001 Toaster
#          1 2002 Toaster
#          3 2003   Radio
#          4 2004   Radio
#          6 2005   Radio

cust
# CustomerId Year    State
#          1 2001  Alabama
#          1 2002   Alaska
#          4 2002  Arizona
#          6 2002 Arkansas

假设我们想将客户的状态从添加cust到购买表sales,忽略年份列。使用基础 R,我们可以识别匹配的行,然后将值复制到:

sales$State <- cust$State[ match(sales$CustomerId, cust$CustomerId) ]

# CustomerId Year Product    State
#          1 2000 Toaster  Alabama
#          1 2001 Toaster  Alabama
#          1 2002 Toaster  Alabama
#          3 2003   Radio     <NA>
#          4 2004   Radio  Arizona
#          6 2005   Radio Arkansas

# cleanup for the next example
sales$State <- NULL

从这里可以看出,match从客户表中选择第一个匹配行。


使用多列更新连接。当我们只加入单个列并且对第一个匹配感到满意时,上述方法效果很好。假设我们希望客户表中的测量年份与销售年份相匹配。

正如@bgoldst 的回答所提到的,对于这种情况,matchwithinteraction可能是一种选择。更直接地说,可以使用 data.table:

library(data.table)
setDT(sales); setDT(cust)

sales[, State := cust[sales, on=.(CustomerId, Year), x.State]]

#    CustomerId Year Product   State
# 1:          1 2000 Toaster    <NA>
# 2:          1 2001 Toaster Alabama
# 3:          1 2002 Toaster  Alaska
# 4:          3 2003   Radio    <NA>
# 5:          4 2004   Radio    <NA>
# 6:          6 2005   Radio    <NA>

# cleanup for next example
sales[, State := NULL]

滚动更新加入。或者,我们可能希望采用发现客户的最后状态:

sales[, State := cust[sales, on=.(CustomerId, Year), roll=TRUE, x.State]]

#    CustomerId Year Product    State
# 1:          1 2000 Toaster     <NA>
# 2:          1 2001 Toaster  Alabama
# 3:          1 2002 Toaster   Alaska
# 4:          3 2003   Radio     <NA>
# 5:          4 2004   Radio  Arizona
# 6:          6 2005   Radio Arkansas

上面的三个示例都侧重于创建/添加新列。有关更新/修改现有列的示例,请参阅相关的 R 常见问题解答。

于 2018-09-04T16:30:15.407 回答