下面是一些使用toString的选项，这个函数可以连接一个字符串向量，使用逗号和空格分隔各个部分。如果不需要逗号，可以使用paste()和collapse参数。

• • 数据表**

# alternative using data.table
library(data.table)
as.data.table(data)[, toString(C), by = list(A, B)]

• • 聚合**不使用包：

# alternative using aggregate from the stats package in the core of R
aggregate(C ~., data, toString)

• • 平方英尺f**

下面是使用SQL函数group_concat和sqldf package的替代方法：

library(sqldf)
sqldf("select A, B, group_concat(C) C from data group by A, B", method = "raw")

• • dplyr**一个dplyr替代品：

library(dplyr)
data %>%
  group_by(A, B) %>%
  summarise(test = toString(C)) %>%
  ungroup()

• • 聚合物**

# plyr
library(plyr)
ddply(data, .(A,B), summarize, C = toString(C))

以下是stringr/tidyverse解决方案：

library(tidyverse)
library(stringr)

data <- data.frame(A = c(rep(111, 3), rep(222, 3)), B = rep(1:2, 3), C = c(5:10))

data %>%
 group_by(A, B) %>%
 summarize(text = str_c(C, collapse = ", "))

# A tibble: 4 x 3
# Groups:   A [2]
      A     B text 
  <dbl> <int> <chr>
1   111     1 5, 7 
2   111     2 6    
3   222     1 9    
4   222     2 8, 10

更改as.character的放置位置：

> out <- ddply(data, .(A, B), summarise, test = list(as.character(C)))
> str(out)
'data.frame':   4 obs. of  3 variables:
 $ A   : num  111 111 222 222
 $ B   : int  1 2 1 2
 $ test:List of 4
  ..$ : chr  "5" "7"
  ..$ : chr "6"
  ..$ : chr "9"
  ..$ : chr  "8" "10"
> out
    A B  test
1 111 1  5, 7
2 111 2     6
3 222 1     9
4 222 2 8, 10

注意，在这种情况下，每个项实际上仍然是一个单独的字符，而不是单个字符串。也就是说，这不是一个看起来像“5，7”的实际字符串，而是两个字符，“5”和“7”，R显示它们之间的逗号。
请与以下内容进行比较：

> out2 <- ddply(data, .(A, B), summarise, test = paste(C, collapse = ", "))
> str(out2)
'data.frame':   4 obs. of  3 variables:
 $ A   : num  111 111 222 222
 $ B   : int  1 2 1 2
 $ test: chr  "5, 7" "6" "9" "8, 10"
> out
    A B  test
1 111 1  5, 7
2 111 2     6
3 222 1     9
4 222 2 8, 10

当然，以R为底的可比溶液为aggregate：

> A1 <- aggregate(C ~ A + B, data, function(x) c(as.character(x)))
> str(A1)
'data.frame':   4 obs. of  3 variables:
 $ A: num  111 222 111 222
 $ B: int  1 1 2 2
 $ C:List of 4
  ..$ 0: chr  "5" "7"
  ..$ 1: chr "9"
  ..$ 2: chr "6"
  ..$ 3: chr  "8" "10"
> A2 <- aggregate(C ~ A + B, data, paste, collapse = ", ")
> str(A2)
'data.frame':   4 obs. of  3 variables:
 $ A: num  111 222 111 222
 $ B: int  1 1 2 2
 $ C: chr  "5, 7" "9" "6" "8, 10"

这里有一个小的改进，以避免重复

# 1. Original data set
data <- data.frame(
  A = c(rep(111, 3), rep(222, 3)), 
  B = rep(1:2, 3), 
  C = c(5:10))

# 2. Add duplicate row
data <- rbind(data, data.table(
  A = 111, B = 1, C = 5
))

# 3. Solution with duplicates
data %>%
  group_by(A, B) %>%
  summarise(test = toString(C)) %>%
  ungroup()

#      A     B test   
#   <dbl> <dbl> <chr>  
# 1   111     1 5, 7, 5
# 2   111     2 6      
# 3   222     1 9      
# 4   222     2 8, 10

# 4. Solution without duplicates
data %>%
  select(A, B, C) %>% unique() %>% 
  group_by(A, B) %>%
  summarise(test = toString(C)) %>%
  ungroup()

#    A     B test 
#   <dbl> <dbl> <chr>
# 1   111     1 5, 7 
# 2   111     2 6    
# 3   222     1 9    
# 4   222     2 8, 10

希望能有用。

使用collapse中的collap

library(collapse)
collap(data, ~ A + B, toString)
    A B     C
1 111 1  5, 7
2 111 2     6
3 222 1     9
4 222 2 8, 10

数据

data <- data.frame(A = c(rep(111, 3), rep(222, 3)), B = rep(1:2, 3), C = c(5:10))

更新的dplyr 1.1.0解决方案，具有与.by的内联分组：

data %>% 
  summarise(test = toString(C), .by = c(A, B))

    A B  test
1 111 1  5, 7
2 111 2     6
3 222 2 8, 10
4 222 1     9

基准：

benchmark <-
  bench::mark(
  data.table = as.data.table(data)[, toString(C), by = list(A, B)],
  aggregate = aggregate(C ~., data, toString),
  sqldf = sqldf("select A, B, group_concat(C) C from data group by A, B", method = "raw"),
  dplyr1.0.0 = data %>%
    group_by(A, B) %>%
    summarise(test = toString(C)) %>%
    ungroup(),
  dplyr1.1.0 = summarise(data, test = toString(C), .by = c(A, B)),
  collapse = collap(data, ~ A + B, toString),
  min_iterations = 30,
  check = FALSE
)

plot(benchmark)