Question

我有两个具有相同结构的数据帧 - 都有两个ID列和25个字符串数据列。我想加入这两个并在ID匹配时连接数据列中的字符串。所以，例如：

df_1：

id_1    id_2    col_1    col2    ...    col_25
a1      b1      A        A       ...    <NA>
a1      b2      A        <NA>    ...    A
a2      b1      <NA>     <NA>    ...    A

df_2：

id_1    id_2    col_1    col2    ...    col_25
a1      b1      B        <NA>    ...    <NA>
a1      b2      <NA>     B       ...    B
a1      b3      B        <NA>    ...    B

合并后，这应该给出

df_combined：

id_1    id_2    col_1    col2    ...    col_25
a1      b1      A, B     A       ...    <NA>
a1      b2      A        B       ...    A, B
a1      b3      B        <NA>    ...    B
a2      b1      <NA>     <NA>    ...    A

当我尝试使用连接或合并时，它会重复除ID列之外的所有内容（因此我最终得到50个数据列）。我需要使用其他东西吗？

谢谢！

Answer 1

如果您没有任何空字符串，则可以执行此操作：

library(dplyr)    
bind_rows(df_1,df_2) %>%
  group_by(id_1,id_2) %>%
  summarize_all(~ paste(na.omit(.x),collapse=", ")) %>%
  `[<-`(.=="",value=NA)

使用magrittr，您可以避免不那么漂亮'[<-'并将其替换为inset

library(magrittr)
bind_rows(df_1,df_2) %>%
  group_by(id_1,id_2) %>%
  summarize_all(~ paste(na.omit(.x),collapse=", ")) %>%
  inset(.=="",value=NA)

Answer 2

使用melt()和dcast()重塑数据的替代解决方案：

library(data.table)
rbind(setDT(df_1), setDT(df_2))[
  , melt(.SD, measure.var = patterns("col"), na.rm = TRUE)][
    , dcast(.SD, id_1 + id_2 ~ variable, toString, fill = NA)]

   id_1 id_2 col_1 col2 col_25
1:   a1   b1  A, B    A     NA
2:   a1   b2     A    B   A, B
3:   a1   b3     B   NA      B
4:   a2   b1    NA   NA      A

数据

df_1 <- fread(
  "id_1    id_2    col_1    col2    ...    col_25
a1      b1      A        A       ...    <NA>
a1      b2      A        <NA>    ...    A
a2      b1      <NA>     <NA>    ...    A",
  drop = 5L, na.strings = "<NA>"
)

df_2 <- fread(
  "id_1    id_2    col_1    col2    ...    col_25
a1      b1      B        <NA>    ...    <NA>
a1      b2      <NA>     B       ...    B
a1      b3      B        <NA>    ...    B",
  drop = 5L, na.strings = "<NA>"
)

Answer 3

详细说明@ zx8754评论的想法，并使用dplyr包，

library(dplyr)

df1 %>% 
 bind_rows(df2) %>% 
 mutate_at(vars(-contains('id')), funs(replace(., is.na(.), ''))) %>% 
 group_by(id_1, id_2) %>% 
 summarise_all(funs(trimws(paste(., collapse = ' ')))) %>% 
 mutate_all(funs(replace(., . == '', NA)))

给出，

# A tibble: 4 x 5
# Groups:   id_1 [2]
   id_1  id_2 col_1  col2 col_25
  <chr> <chr> <chr> <chr>  <chr>
1    a1    b1   A B     A   <NA>
2    a1    b2     A     B    A B
3    a1    b3     B  <NA>      B
4    a2    b1  <NA>  <NA>      A

注意：

以上脚本假定您的NAs是实际的NA（不是字符）

您的变量为as.character

数据

dput(df1) structure(list(id_1 = c("a1", "a1", "a2"), id_2 = c("b1", "b2", "b1"), col_1 = c("A", "A", NA), col2 = c("A", NA, NA), col_25 = c(NA, "A", "A")), .Names = c("id_1", "id_2", "col_1", "col2", "col_25" ), row.names = c(NA, -3L), class = "data.frame") > dput(df2) structure(list(id_1 = c("a1", "a1", "a1"), id_2 = c("b1", "b2", "b3"), col_1 = c("B", NA, "B"), col2 = c(NA, "B", NA), col_25 = c(NA, "B", "B")), .Names = c("id_1", "id_2", "col_1", "col2", "col_25" ), row.names = c(NA, -3L), class = "data.frame")

通过连接列连接两个数据帧

3 个答案:

数据