right_join 并且 mutate 不保留 R 中的索引

Question

我正在将 column_data 映射到 master，如果列值存在于 master 中，则将其保存为 Key
例如：Parent 代表 P 和 Child 代表 C

问题是我正在获取输出，但输出的索引不同

数据

column_data <- c("", "", "", "", "", "", "", "", "", "", "", "", "", "", "", 
"", "P", "C", "C")

master  <- list("Parent" = c("P"),
                "Child" = c("C")
               )

代码

library(dplyr)


df <- data.frame("column" = column_data)
df <-stack(master) %>% 
     type.convert(as.is = TRUE) %>%
     right_join(df, by = c('values' = 'column')) %>%
     mutate(output = coalesce(ind, values))

这应该是输出：

structure(list(values = c("", "", "", "", "", "", "", "", "", 
"", "", "", "", "", "", "", "P", "C", "C"), ind = c(NA, NA, NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, "Parent", 
"Child", "Child"), output = c("", "", "", "", "", "", "", "", 
"", "", "", "", "", "", "", "", "Parent", "Child", "Child")), class = "data.frame", row.names = c(NA, 
-19L))

但是我得到这个作为输出：

structure(list(values = c("P", "C", "C", "", "", "", "", "", 
"", "", "", "", "", "", "", "", "", "", ""), ind = c("Parent", 
"Child", "Child", NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA, NA, NA, NA), output = c("Parent", "Child", "Child", "", 
"", "", "", "", "", "", "", "", "", "", "", "", "", "", "")), row.names = c(NA, 
-19L), class = "data.frame")

Answer 1

使用 dplyr，如果您执行 right_join(x, y)，则结果将包括 x 的匹配行的子集，然后是 y 的不匹配行。

从关于变异连接的 R 文档中，返回的值将是：

An object of the same type as x. The order of the rows and columns of x is preserved as much as possible. The output has the following properties:

For inner_join(), a subset of x rows. For left_join(), all x rows. For right_join(), a subset of x rows, followed by unmatched y rows. For full_join(), all x rows, followed by unmatched y rows.

这就是为什么在结果 data.frame.

的开头有 3 个匹配的行

要获得保留 df 行顺序的所需结果，请按如下方式尝试 left_join：

df2 <- stack(master) %>% 
  type.convert(as.is = TRUE) 

df %>%
  left_join(df2, by = c('column' = 'values')) %>%
  mutate(output = coalesce(ind, column))

输出

   column    ind output
1           <NA>       
2           <NA>       
3           <NA>       
4           <NA>       
5           <NA>       
6           <NA>       
7           <NA>       
8           <NA>       
9           <NA>       
10          <NA>       
11          <NA>       
12          <NA>       
13          <NA>       
14          <NA>       
15          <NA>       
16          <NA>       
17      P Parent Parent
18      C  Child  Child
19      C  Child  Child

right_join 并且 mutate 不保留 R 中的索引

right_join and mutate does not preserve the index in R

r

dataframe

dplyr

tidyverse

tibble