Question

我尝试做类似于what's answered here的事情，这让我获得80％的回报。我有一个带有一个ID列和多个信息列的数据框。我想汇总其他列的所有，以便每个ID只有一行，并且多个条目由例如分号分隔。这是我拥有和想要的一个例子。

HAVE：

     ID  info1          info2
1 id101    one          first
2 id102   twoA second alias A
3 id102   twoB second alias B
4 id103 threeA  third alias A
5 id103 threeB  third alias B
6 id104   four         fourth
7 id105   five          fifth

WANT：

     ID          info1                          info2
1 id101            one                          first
2 id102     twoA; twoB second alias A; second alias B
3 id103 threeA; threeB   third alias A; third alias B
4 id104           four                         fourth
5 id105           five                          fifth

以下是用于生成这些代码的代码：

have <- data.frame(ID=paste0("id", c(101, 102, 102, 103, 103, 104, 105)),
                   info1=c("one", "twoA", "twoB", "threeA", "threeB", "four", "five"), 
                   info2=c("first", "second alias A", "second alias B", "third alias A", "third alias B", "fourth", "fifth"),
                   stringsAsFactors=FALSE)
want <- data_frame(ID=paste0("id", c(101:105)),
                   info1=c("one", "twoA; twoB", "threeA; threeB", "four", "five"), 
                   info2=c("first", "second alias A; second alias B", "third alias A; third alias B", "fourth", "fifth"),
                   stringsAsFactors=FALSE)

This question基本上问了同样的问题，但只有一个＆＃34; info＆＃34;柱。我有多个其他列，并希望为所有这些列执行此操作。

使用dplyr执行此操作的加分点。

Answer 1

以下是使用summarise_each的选项（可以轻松将更改应用于除分组变量之外的所有列）和toString：

require(dplyr)

have %>%
  group_by(ID) %>%
  summarise_each(funs(toString))

#Source: local data frame [5 x 3]
#
#     ID          info1                          info2
#1 id101            one                          first
#2 id102     twoA, twoB second alias A, second alias B
#3 id103 threeA, threeB   third alias A, third alias B
#4 id104           four                         fourth
#5 id105           five                          fifth

或者，如果您希望它以分号分隔，则可以使用：

have %>%
  group_by(ID) %>%
  summarise_each(funs(paste(., collapse = "; ")))

Answer 2

好老aggregate做得很好

aggregate(have[,2:3], by=list(have$ID), paste, collapse=";")

问题是：它是否可以扩展？

Answer 3

这是一个data.table解决方案。

library(data.table)
setDT(have)[, lapply(.SD, paste, collapse = "; "), by = ID]
#       ID          info1                          info2
# 1: id101            one                          first
# 2: id102     twoA; twoB second alias A; second alias B
# 3: id103 threeA; threeB   third alias A; third alias B
# 4: id104           four                         fourth
# 5: id105           five                          fifth

Answer 4

这是 SQL 解决方案^ 1：

library(sqldf)
#Static solution
sqldf("
SELECT ID,
       GROUP_CONCAT(info1,';') as info1,
       GROUP_CONCAT(info2,';') as info2
FROM have
GROUP BY ID")

#Dynamic solution
concat_cols <- colnames(have)[2:ncol(have)]
group_concat <-
  paste(paste0("GROUP_CONCAT(",concat_cols,",';') as ", concat_cols),
        collapse = ",")
sqldf(
  paste("
      SELECT ID,",
      group_concat,"
      FROM have
      GROUP BY ID"))

# Same output for both static and dynamic solutions
#      ID         info1                         info2
# 1 id101           one                         first
# 2 id102     twoA;twoB second alias A;second alias B
# 3 id103 threeA;threeB   third alias A;third alias B
# 4 id104          four                        fourth
# 5 id105          five                         fifth

^ 1 - 可能data.table解决方案在数百万行中表现更好，幸运的是我们还没有那么多基因：）

Answer 5

library(stringr)
library(dplyr)
have %>% tbl_df %>% group_by(ID) %>% summarise_each(funs(str_c(., collapse="; ")))

修改1：因此可能不需要tbl_df，而不是stringr包的str_c，您可以使用paste（在base中）。以上操作是按ID列进行分组，然后将str_c（或paste）函数应用于每个组的每个剩余列。

编辑2：使用data.table包的另一种解决方案：

library(data.table)
dtbl <- as.data.table(have)
dtbl[,lapply(.SD, function(x) paste(x,collapse=";")), by=ID]

以上情况可能会更快，尤其是在设置密钥时：

setkey(dtbl, ID)

“混合”解决方案：您可以对data.tables使用dplyr语法！例如：

dtbl %>% tbl_dt %>%
     group_by(ID) %>% 
     summarise_each(funs(paste(., collapse="; ")))

按ID列折叠所有列

5 个答案: