覆盖left_join dplyr以更新数据

Martin 发表于 Dev

马丁

我的问题与此类似，但是我在LHS中还有其他应保留的列https://stackoverflow.com/a/35642948/9285732

y是val1的x更新值的的子集。在此，我想覆盖相关值，但保留其余值。x

样本数据：

library(tidyverse)

x <- tibble(name = c("hans", "dieter", "bohlen", "hans", "dieter", "alf"), 
            location = c(1,1,1,2,2,3), 
            val1 = 1:6, val2 = 1:6, val3 = 1:6)
y <- tibble(name = c("hans", "dieter", "hans"), 
            location = c(2,2,1), 
            val1 = 10)

> x
# A tibble: 6 x 5
  name   location  val1  val2  val3
  <chr>     <dbl> <int> <int> <int>
1 hans          1     1     1     1
2 dieter        1     2     2     2
3 bohlen        1     3     3     3
4 hans          2     4     4     4
5 dieter        2     5     5     5
6 alf           3     6     6     6

> y
# A tibble: 3 x 3
  name   location  val1
  <chr>     <dbl> <dbl>
1 hans          2    10
2 dieter        2    10
3 hans          1    10

> # desired output
> out
# A tibble: 6 x 5
  name   location  val1  val2  val3
  <chr>     <dbl> <dbl> <int> <int>
1 hans          1    10     1     1
2 dieter        1     2     2     2
3 bohlen        1     3     3     3
4 hans          2    10     4     4
5 dieter        2    10     5     5
6 alf           3     6     6     6

我编写了一个可以执行我想要的功能的函数，但是它很麻烦。我想知道是否有我不知道的更优雅的方法甚至是dplyr函数。

overwrite_join <- function(x, y, by = NULL){

  bycols     <- which(colnames(x) %in% by) 
  commoncols <- which(colnames(x) %in% colnames(y))
  extracols  <- which(!(colnames(x) %in% colnames(y)))

  x1 <- anti_join(x, y, by = by) %>% 
    bind_rows(y) %>%
    select(commoncols) %>% 
    left_join(x %>% select(bycols, extracols), by = by)

  out <- x %>% select(by) %>% 
    left_join(x1, by = by)

  return(out)
}

overwrite_join(t1, t2, by = c("name", "location"))

它的

您可以按照以下方式进行操作

> x %>%
    left_join(y = y, by = c("name", "location")) %>%
    within(., val1.x <- ifelse(!is.na(val1.y), val1.y, val1.x)) %>%
    select(-val1.y)
# # A tibble: 6 x 5
#   name   location val1.x  val2  val3
#   <chr>     <dbl>  <dbl> <int> <int>
# 1 hans          1     10     1     1
# 2 dieter        1      2     2     2
# 3 bohlen        1      3     3     3
# 4 hans          2     10     4     4
# 5 dieter        2     10     5     5
# 6 alf           3      6     6     6

然后重命名val1.x。

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