993

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

我该怎么做数据库样式，即sql style, joins？也就是说，我该怎么做：

inner join df1 和 df2 ：
仅返回左表在右表中具有匹配键的行 .
outer join df1 和 df2 ：
返回两个表中的所有行，从左侧连接具有右表中匹配键的记录 .
left outer join (or simply left join) of df1 和 df2
返回左表中的所有行，以及右表中具有匹配键的所有行 .
df1 of df1 和 df2
返回右表中的所有行，以及左表中具有匹配键的所有行 .

额外信用：

如何进行SQL样式选择语句？

13 回答

1

通过使用 merge 函数及其可选参数：

Inner join: merge(df1, df2) 将适用于这些示例，因为R会自动通过公共变量名称连接帧，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数据框中具有不同的名称，也可以使用 by.x 和 by.y 参数 .

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

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

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

Cross join: merge(x = df1, y = df2, by = NULL)

就像内连接一样，您可能希望将“CustomerId”显式传递给R作为匹配变量 . 我认为如果输入data.frames意外地改变并且稍后更容易阅读，那么_865099会更安全 .

您可以通过给 by 向量合并多个列，例如 by = c("CustomerId", "OrderId") .

如果要合并的列名不相同，则可以指定例如 by.x = "CustomerId_in_df1", by.y = "CustomerId_in_df2" ，其中 CustomerId_in_df1 是第一个数据框中列的名称， CustomerId_in_df2 是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 . ）

回复于 2024-04-19T17:04:52+08:00

我建议你查看Gabor Grothendieck's sqldf package，它允许你在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，请参见Gabor's sqldf GitHub .

回复于 2024-04-19T17:04:52+08:00

146
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（因为它是通用的并且调用 merge.data.table ）
```
merge(dt1, dt2)
```
stackoverflow上记录的data.table：
How to do a data.table merge operation
Translating SQL joins on foreign keys to R data.table syntax
Efficient alternatives to merge for larger data.frames R
How to do a basic left outer join with data.table in R?

另一个选项是plyr包中的 join 函数
```
library(plyr)

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

#   CustomerId Product   State
# 1          2 Toaster Alabama
# 2          4   Radio Alabama
# 3          6   Radio    Ohio
```
type 的选项： inner ， left ， right ， full .

来自 ?join ：与 merge 不同，无论使用何种连接类型，[ join ]都会保留x的顺序 .
回复于 2024-04-19T17:04:52+08:00

你可以使用Hadley Wickham的令人敬畏的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.

回复于 2024-04-19T17:04:52+08:00

55
在R Wiki有一些很好的例子 . 我会偷一对夫妇：

Merge Method

由于您的密钥命名相同，因此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，拍打它们并揉搓它们以获得你询问的另外两个外部连接:)

Subscript Method

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“我会把它作为读者的练习......”）
回复于 2024-04-19T17:04:52+08:00
63
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，旨在方便你专门处理数据框架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
这都是here非常详细 .

选择列可以通过 select(df,"column") 完成 . 如果's not SQL-ish enough for you, then there'是 sql() 函数，你可以在其中输入SQL代码，它将执行操作你指定就像你一直在写R一样（有关更多信息，请参阅dplyr/databases vignette） . 例如，如果应用正确， sql("SELECT * FROM hflights") 将选择"hflights" dplyr表中的所有列（"tbl"） .
回复于 2024-04-19T17:04:52+08:00

170

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作为子集的第一个参数，另一种方法是使用 merge 函数调度到快速data.table方法 .

Update on 2016-04-01 - and it isn't April Fools joke!
在1.9.7版本的data.table连接现在能够使用现有索引，这极大地减少了连接的时间 . Below code and benchmark does NOT use data.table indices on join . 如果您正在寻找近实时连接，则应使用data.table索引 .

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=0L, on = "CustomerId"]

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

# inner join
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 . 如果在data.tables或使用sqldf的索引上使用键，则可以获得更好的性能 . Base R和dplyr没有索引或键，因此我没有在基准测试中包含该场景 .
基准测试是在5M-1行数据集上执行的，在连接列上有5M-2个常用值，因此可以测试每个场景（左，右，全，内），并且连接仍然不容易执行 .

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

n = 5e6
set.seed(123)
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)

# inner join
microbenchmark(times = 10L,
               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"),
               data.table = dt1[dt2, nomatch = 0L, on = "x"])
#Unit: milliseconds
#       expr        min         lq      mean     median        uq       max neval
#       base 15546.0097 16083.4915 16687.117 16539.0148 17388.290 18513.216    10
#      sqldf 44392.6685 44709.7128 45096.401 45067.7461 45504.376 45563.472    10
#      dplyr  4124.0068  4248.7758  4281.122  4272.3619  4342.829  4411.388    10
# data.table   937.2461   946.0227  1053.411   973.0805  1214.300  1281.958    10

# left outer join
microbenchmark(times = 10L,
               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")),
               data.table = dt2[dt1, on = "x"])
#Unit: milliseconds
#       expr       min         lq       mean     median         uq       max neval
#       base 16140.791 17107.7366 17441.9538 17414.6263 17821.9035 19453.034    10
#      sqldf 43656.633 44141.9186 44777.1872 44498.7191 45288.7406 47108.900    10
#      dplyr  4062.153  4352.8021  4780.3221  4409.1186  4450.9301  8385.050    10
# data.table   823.218   823.5557   901.0383   837.9206   883.3292  1277.239    10

# right outer join
microbenchmark(times = 10L,
               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"),
               data.table = dt1[dt2, on = "x"])
#Unit: milliseconds
#       expr        min         lq       mean     median        uq       max neval
#       base 15821.3351 15954.9927 16347.3093 16044.3500 16621.887 17604.794    10
#      sqldf 43635.5308 43761.3532 43984.3682 43969.0081 44044.461 44499.891    10
#      dplyr  3936.0329  4028.1239  4102.4167  4045.0854  4219.958  4307.350    10
# data.table   820.8535   835.9101   918.5243   887.0207  1005.721  1068.919    10

# full outer join
microbenchmark(times = 10L,
               base = merge(df1, df2, by = "x", all = TRUE),
               #sqldf = sqldf("SELECT * FROM df1 FULL OUTER JOIN df2 ON df1.x = df2.x"), # not supported
               dplyr = full_join(df1, df2, by = "x"),
               data.table = merge(dt1, dt2, by = "x", all = TRUE))
#Unit: seconds
#       expr       min        lq      mean    median        uq       max neval
#       base 16.176423 16.908908 17.485457 17.364857 18.271790 18.626762    10
#      dplyr  7.610498  7.666426  7.745850  7.710638  7.832125  7.951426    10
# data.table  2.052590  2.130317  2.352626  2.208913  2.470721  2.951948    10

回复于 2024-04-19T17:04:52+08:00

6
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 for the first few releases it used not to offer outer_join, and as a result there was a lot of really bad hacky workaround user code floating around for quite a while (you can still find this in SO and Kaggle answers from that period).

加入相关release highlights：

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)
- Implement right join and outer join (#96)
- 变异连接，它将新变量从另一个表中的匹配行添加到一个表中 . 过滤联接，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结果 .
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implement right_join and outer_join
- Hadley mentioning other advantages here
- 目前有一个次要特征合并，dplyr不是the ability to have separate by.x,by.y columns，例如Python大熊猫呢 .
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 .
回复于 2024-04-19T17:04:52+08:00
73

在连接两个数据帧时，每个有大约100万行，一个有2列，另一个有~20，我惊奇地发现 merge(..., all.x = TRUE, all.y = TRUE) 比 dplyr::full_join() 快 . 这是dplyr v0.4

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 .

回复于 2024-04-19T17:04:52+08:00

对于具有 0..*:0..1 基数的左连接或具有 0..1:0..* 基数的右连接的情况，可以将来自连接器（ 0..1 表）的单边列直接分配到被调用者（ 0..* 表）上，从而避免创建一个全新的数据表 . 这需要将来自被调用者的关键列与加入者匹配，并且索引将相应地为加入者的行排序以进行分配 .

如果键是单列，那么我们可以使用match()的单个调用来进行匹配 . 我将在这个答案中介绍这种情况 .

这里's an example based on the OP, except I' ve为 df2 添加了一个额外的行，其id为7，以测试joiner中非匹配键的情况 . 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，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，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，如果你有几个表想要连接对一个大表 . 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 . 虽然在在这种情况下，您只需制作副本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 paste() 相同的想法 .
低效的笛卡尔连词，例如 outer(df1$a,df2$a,==) & outer(df1$b,df2$b,==) .
base 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赋值的解决方案 .

例如，请参阅Matching multiple columns on different data frames and getting other column as result，match two columns with two other columns，Matching on multiple columns，以及我最初提出的就地解决方案Combine two data frames with different number of rows in R的问题 .

基准测试

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比较 .

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 . 这是在生成第二个data.frame的键列时通过采样而不替换第一个data.frame的键列来实现的 .

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，使用单独的pch符号绘制 . 连接类型由pch符号捕获，左侧和右侧使用内部，左右角括号的点，以及完整的菱形 . 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);

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

在关键列的数量和类型以及基数方面，_865204更重要 . 对于此基准测试，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..* ） . （一般情况下，'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在关键列中包含了那些类型 . 另外，由于某些原因，'s sake, I initially tried to use four key columns by including a POSIXct key column, but the POSIXct type didn'与 sqldf.indexed 解决方案配合得很好，可能是由于浮点比较异常，所以我删除了它 . ）

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

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);

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

回复于 2024-04-19T17:04:52+08:00

1081

对于所有列的内部联接，您还可以使用data.table-package中的 fintersect 或dplyr-package中的 intersect 作为 merge 的替代，而不指定 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)

回复于 2024-04-19T17:04:52+08:00

188
- 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：选择一个 . * ...或者选择b . *来自.....）
- 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)]
回复于 2024-04-19T17:04:52+08:00

Update join. 另一个重要的SQL样式连接是“update join”，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从 cust 添加到购买表 sales ，忽略年份列 . 使用基数R，我们可以识别匹配的行，然后复制值：

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customer表中选择第一个匹配的行 .

Update join with multiple columns. 当我们只加入一个列并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计量年份与销售年份相匹配 .

正如@ bgoldst的答案所提到的那样， match 和 interaction 可能是这种情况的选择 . 更直接的是，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]

Rolling update join. 或者，我们可能想要找到客户所在的最后一个州：

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

上面的三个示例都专注于创建/添加新列 . 有关更新/修改现有列的示例，请参见the related R FAQ .

回复于 2024-04-19T17:04:52+08:00

如何加入（合并）数据框（内部，外部，左侧，右侧）？

13 回答

变异连接：使用df2中的匹配将列添加到df1

过滤联接：过滤掉df1中的行，不要修改列

基准测试

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