如何加入(合并)数据框(内部,外部,左侧,右侧)？

Question

给出两个数据框:

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

我怎样才能做数据库风格,即sql风格,加入？也就是说,我该怎么做:

• 一个内连接的df1和df2:
  只返回行中左表在右表匹配的密钥.
• 一个外连接的df1和df2:
  返回两个表中的所有行,从有右表中的匹配键左连接记录.
• 甲左外连接(或简称为左加入)的df1和df2
  左表中返回所有行,并与匹配的右表键任何行.
• 一个右外连接的df1,并df2
  返回右表中的所有行,任何行与左表中匹配的密钥.

额外信用:

如何进行SQL样式选择语句？

Answer 1

通过使用该merge函数及其可选参数:

内部 merge(df1, df2)联接:将适用于这些示例,因为R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.如果匹配变量在不同的数据框中具有不同的名称,也可以使用by.x和by.y参数.

外部联接: 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)

交叉加入: merge(x = df1, y = df2, by = NULL)

与内连接一样,您可能希望将"CustomerId"显式传递给R作为匹配变量. 我认为最好明确说明要合并的标识符; 如果输入data.frames意外更改并且稍后更容易阅读,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").

如果要合并的列名称不相同,则可以指定,例如,第一个数据框中列的名称by.x = "CustomerId_in_df1", by.y = "CustomerId_in_df2"在哪里,以及第二个数据框中列的名称.(如果需要在多个列上合并,这些也可以是向量.)CustomerId_in_df1CustomerId_in_df2

Answer 2

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等价物更简单,更自然(但这可能只反映了我的RDBMS偏见).

有关连接的更多信息,请参阅Gabor的sqldf GitHub.

Answer 3

内连接有data.table方法,这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(因为它是通用的和调用merge.data.table)

merge(dt1, dt2)

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
为更大的data.frames合并的有效替代方案R
如何使用data.table进行基本的左外连接在R？

另一种选择是join在plyr包中找到的功能

library(plyr)

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

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

为选项type:inner,left,right,full.

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

Answer 4

你也可以使用Hadley Wickham令人敬畏的dplyr软件包进行连接.

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)

过滤联接: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.

Answer 5

在R Wiki上有一些很好的例子.我会偷一对夫妇:

合并方法

由于您的密钥命名相同,因此进行内部联接的简短方法是merge():

merge(df1,df2)

可以使用"all"关键字创建完整的内部联接(来自两个表的所有记录):

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)

你可以翻转它们,拍打它们并按下它们以获得你询问的另外两个外部连接:)

下标方法

使用下标方法在左侧使用df1的左外连接将是:

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

可以通过对左外连接下标示例进行mungling来创建外连接的其他组合.(是的,我知道这相当于说"我会留下它作为读者的练习......")

Answer 6

2014年新增内容:

特别是如果你对一般的数据操作感兴趣(包括排序,过滤,子集,总结等),你一定要看看dplyr,它带有各种功能,所有这些功能都是为了方便你的数据框架工作而设计的.和某些其他数据库类型.它甚至提供了相当精细的SQL接口,甚至还有一个将(大多数)SQL代码直接转换为R的函数.

dplyr包中的四个与连接相关的函数是(引用):

• inner_join(x, y, by = NULL, copy = FALSE, ...):返回x中匹配值的所有行,以及x和y中的所有列
• left_join(x, y, by = NULL, copy = FALSE, ...):返回x中的所有行,以及x和y中的所有列
• semi_join(x, y, by = NULL, copy = FALSE, ...):返回x中所有行,其中y中存在匹配值,仅保留x中的列.
• anti_join(x, y, by = NULL, copy = FALSE, ...):返回x中所有行,其中y中没有匹配的值,只保留x中的列

这一切都在这里非常详细.

选择列可以通过select(df,"column").如果这对你来说不够SQL,那么就有了这个sql()函数,你可以按原样输入SQL代码,它将完成你指定的操作,就像你一直在写R一样(有关更多信息,请参阅到dplyr/databases晕影).例如,如果应用正确,sql("SELECT * FROM hflights")将从"hflights"dplyr表中选择所有列("tbl").

Answer 7

更新data.table方法以加入数据集.请参阅以下每种联接类型的示例.有两种方法,一种是从[.data.table传递第二个data.table作为第一个参数到子集时,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

基准测试基础R,sqldf,dplyr和data.table.
基准测试未加密/未加索引的数据集.基准测试是在50M-1行数据集上执行的,在连接列上有50M-2个常用值,因此可以测试每个场景(内部,左侧,右侧,完整),并且连接仍然不容易执行.它是很好地强调连接算法的连接类型.时序为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,您可以使用以下其他类型的连接data.table:
- 在连接时更新 - 如果要从另一个表中查找值到主表
- 在连接时聚合 - 如果要在加入的密钥上进行聚合,则不需要实现所有连接结果
- 重叠连接 - 如果要按范围合并
- 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一行/后一行中的值
- 非等连接 - 如果您的连接条件不相等

代码重现:

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

Answer 8

dplyr自0.4以来实现了所有这些连接,包括outer_join,但值得注意的是,对于前几个版本,它不使用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(你仍然可以在那个时期的SO和Kaggle答案中找到这个.

与加入相关的发布亮点:

v0.5(6/2016)

• 处理POSIXct类型,时区,重复项,不同因子级别.更好的错误和警告.
• 用于控制后缀重复变量名称接收的新后缀参数(#1296)

v0.4.0(1/2015)

• 实现右连接和外连接(#96)
• 变换连接,它将新变量从另一个表中的匹配行添加到一个表中.过滤连接,根据是否与另一个表中的观察匹配来过滤来自一个表的观察.

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()不再重新排序列名(#324)

v0.1.3(4/2014)

• 有inner_join,left_join,semi_join,anti_join
• outer_join尚未实现,回退是使用base :: merge()(或plyr :: join())
• 还没有实现right_join和outer_join
• 哈德利在这里提到其他优势
• 目前有一个小功能合并,dplyr不具备单独的by.x,by.y列,例如Python pandas.

每个hadley在该问题上的评论的解决方法:

• right_join(x,y)就行而言与left_join(y,x)相同,只是列将是不同的顺序.使用select(new_column_order)轻松解决问题
• 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.

Answer 9

在连接两个数据帧时,每个行数约为1百万行,一行有2列,另一行有〜20行,我惊奇地发现它merge(..., all.x = TRUE, all.y = TRUE)比这更快dplyr::full_join().这与dplyr v0.4有关

合并需要大约17秒,full_join大约需要65秒.

虽然有些食物,因为我通常默认使用dplyr进行操作任务.

Answer 10

For the case of a left join with a 0..*:0..1 cardinality or a right join with a 0..1:0..* cardinality it is possible to assign in-place the unilateral columns from the joiner (the 0..1 table) directly onto the joinee (the 0..* table), and thereby avoid the creation of an entirely new table of data. This requires matching the key columns from the joinee into the joiner and indexing+ordering the joiner's rows accordingly for the assignment.

If the key is a single column, then we can use a single call to match() to do the matching. This is the case I'll cover in this answer.

Here's an example based on the OP, except I've added an extra row to df2 with an id of 7 to test the case of a non-matching key in the joiner. This is effectively df1 left join 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,一般来说,这不是一个不合理的假设,因为,如果你有一个带有键列的data.frame,如果它还没有被设置为data.frame的第一列,那就太奇怪了.一开始.并且您可以随时重新排序列以实现它.这种假设的一个有利结果是,关键列的名称不必是硬编码的,尽管我认为它只是将一个假设替换为另一个假设.Concision是整数索引以及速度的另一个优点.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.

我认为这是一个特别合适的解决方案,如果你有几个表要保持连接对一个大表.为每次合并重复重建整个表将是不必要且低效的.

另一方面,如果您因为任何原因需要通过此操作保持不变,则不能使用此解决方案,因为它直接修改了joinee.虽然在这种情况下您可以简单地复制并在副本上执行就地分配.

作为旁注,我简要介绍了多列密钥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:

• 低效的连接.例如match(interaction(df1$a,df1$b),interaction(df2$a,df2$b)),或与之相同的想法paste().
• 低效的笛卡尔连词,例如outer(df1$a,df2$a,`==`) & outer(df1$b,df2$b,`==`).
• 基本R merge()和等效的基于包的合并函数,它总是分配一个新表来返回合并的结果,因此不适合基于内部分配的解决方案.

例如,看到在不同的数据帧匹配的多个列,并且让其他列结果,与其他两列赛两列,匹配上多列,而这个问题,我本来想出了就地解决方案的欺骗,合并R中具有不同行数的两个数据帧.

标杆

我决定进行自己的基准测试,以了解就地分配方法与此问题中提供的其他解决方案的对比情况.

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()

这是我之前演示的基于OP的示例的基准:

## 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,尝试不同的比例和两个输入表之间的键重叠的不同模式.此基准仍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的键列时不替换第一个data.frame的键列而进行采样来实现的.

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

我写了一些代码来创建上述结果的日志 - 日志图.我为每个重叠百分比生成了一个单独的图.它有点杂乱,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,使用单独的pch符号绘制.连接类型由pch符号捕获,左侧和右侧的内侧,左侧和右侧尖括号使用点,使用完整的菱形.解决方案类型由颜色捕获,如图例中所示.

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);

Here's a second large-scale benchmark that's more heavy-duty, with respect to the number and types of key columns, as well as cardinality. For this benchmark I use three key columns: one character, one integer, and one logical, with no restrictions on cardinality (that is, 0..*:0..*). (In general it's not advisable to define key columns with double or complex values due to floating-point comparison complications, and basically no one ever uses the raw type, much less for key columns, so I haven't included those types in the key columns. Also, for information's sake, I initially tried to use four key columns by including a POSIXct key column, but the POSIXct type didn't play well with the sqldf.indexed solution for some reason, possibly due to floating-point comparison anomalies, so I removed it.)

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

生成的图,使用上面给出的相同绘图代码:

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

Answer 11

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SQL中的select语句一样(EX:选择a.*...或从b中选择b.*)

2. 我们必须添加额外的代码,这些代码将从新连接的表中进行子集化.

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

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

同样的方式

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

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

Answer 12

有关所有列的内部连接,你也可以使用fintersect从data.table -package或intersect从dplyr -package作为替代merge不指定by-columns.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)

Answer 13

更新加入。另一种重要的 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

假设我们要将客户的状态添加cust到购买表中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的行。

使用多列更新连接。当我们只加入一个列并且对第一场比赛感到满意时，上述方法很有效。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，matchwithinteraction可能是这种情况下的一个选择。更直接地，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

以上三个示例都专注于创建/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。