如何用 R 中另一個 tibble 的估算列替換 tibble 中的 NA 列

Question

我想使用df2中的估算值將 de 列替換為df中的NA以獲得df3 。 我可以用left_join和coalesce做到這一點，但我認為這種方法不能很好地概括。 有沒有更好的辦法？

library(tidyverse)

df <- tibble(c = c("a", "a", "a", "b", "b", "b"),
             d = c(1, 2, 3, 1, 2, 3),
             x = c(1, NA, 3, 4, 5,6),
             y = c(1, 2, NA, 4, 5, 6),
             z = c(1, 2, 7, 4, 5, 6))

# I want to replace NA in df by df2

df2 <- tibble(c = c("a", "a", "a"),
             d = c(1, 2, 3),
             x = c(1, 2, 3),
             y = c(1, 2, 2))

# to get

df3 <- tibble(c = c("a", "a", "a", "b", "b", "b"),
             d = c(1, 2, 3, 1, 2, 3),
             x = c(1, 2, 3, 4, 5, 6),
             y = c(1, 2, 2, 4, 5, 6),
             z = c(1, 2, 7, 4, 5, 6))

# is there a better solution than coalesce?

df3 <- df %>% left_join(df2, by = c("c", "d")) %>%
  mutate(x = coalesce(x.x, x.y),
         y = coalesce(y.x, y.y)) %>%
  select(-x.x, -x.y, -y.x, -y.y)
Created on 2021-06-17 by the reprex package (v2.0.0)

Answer 1

我想使用df2的估算值在df中用NA替換 de 列以獲得df3 。 我可以用left_join和coalesce來做到這left_join ，但我認為這種方法不能很好地概括。 有沒有更好的辦法？

library(tidyverse)

df <- tibble(c = c("a", "a", "a", "b", "b", "b"),
             d = c(1, 2, 3, 1, 2, 3),
             x = c(1, NA, 3, 4, 5,6),
             y = c(1, 2, NA, 4, 5, 6),
             z = c(1, 2, 7, 4, 5, 6))

# I want to replace NA in df by df2

df2 <- tibble(c = c("a", "a", "a"),
             d = c(1, 2, 3),
             x = c(1, 2, 3),
             y = c(1, 2, 2))

# to get

df3 <- tibble(c = c("a", "a", "a", "b", "b", "b"),
             d = c(1, 2, 3, 1, 2, 3),
             x = c(1, 2, 3, 4, 5, 6),
             y = c(1, 2, 2, 4, 5, 6),
             z = c(1, 2, 7, 4, 5, 6))

# is there a better solution than coalesce?

df3 <- df %>% left_join(df2, by = c("c", "d")) %>%
  mutate(x = coalesce(x.x, x.y),
         y = coalesce(y.x, y.y)) %>%
  select(-x.x, -x.y, -y.x, -y.y)
Created on 2021-06-17 by the reprex package (v2.0.0)

Answer 2

我想使用df2的估算值在df中用NA替換 de 列以獲得df3 。 我可以用left_join和coalesce來做到這left_join ，但我認為這種方法不能很好地概括。 有沒有更好的辦法？

library(tidyverse)

df <- tibble(c = c("a", "a", "a", "b", "b", "b"),
             d = c(1, 2, 3, 1, 2, 3),
             x = c(1, NA, 3, 4, 5,6),
             y = c(1, 2, NA, 4, 5, 6),
             z = c(1, 2, 7, 4, 5, 6))

# I want to replace NA in df by df2

df2 <- tibble(c = c("a", "a", "a"),
             d = c(1, 2, 3),
             x = c(1, 2, 3),
             y = c(1, 2, 2))

# to get

df3 <- tibble(c = c("a", "a", "a", "b", "b", "b"),
             d = c(1, 2, 3, 1, 2, 3),
             x = c(1, 2, 3, 4, 5, 6),
             y = c(1, 2, 2, 4, 5, 6),
             z = c(1, 2, 7, 4, 5, 6))

# is there a better solution than coalesce?

df3 <- df %>% left_join(df2, by = c("c", "d")) %>%
  mutate(x = coalesce(x.x, x.y),
         y = coalesce(y.x, y.y)) %>%
  select(-x.x, -x.y, -y.x, -y.y)
Created on 2021-06-17 by the reprex package (v2.0.0)

Answer 3

我想使用df2的估算值在df中用NA替換 de 列以獲得df3 。 我可以用left_join和coalesce來做到這left_join ，但我認為這種方法不能很好地概括。 有沒有更好的辦法？

library(tidyverse)

df <- tibble(c = c("a", "a", "a", "b", "b", "b"),
             d = c(1, 2, 3, 1, 2, 3),
             x = c(1, NA, 3, 4, 5,6),
             y = c(1, 2, NA, 4, 5, 6),
             z = c(1, 2, 7, 4, 5, 6))

# I want to replace NA in df by df2

df2 <- tibble(c = c("a", "a", "a"),
             d = c(1, 2, 3),
             x = c(1, 2, 3),
             y = c(1, 2, 2))

# to get

df3 <- tibble(c = c("a", "a", "a", "b", "b", "b"),
             d = c(1, 2, 3, 1, 2, 3),
             x = c(1, 2, 3, 4, 5, 6),
             y = c(1, 2, 2, 4, 5, 6),
             z = c(1, 2, 7, 4, 5, 6))

# is there a better solution than coalesce?

df3 <- df %>% left_join(df2, by = c("c", "d")) %>%
  mutate(x = coalesce(x.x, x.y),
         y = coalesce(y.x, y.y)) %>%
  select(-x.x, -x.y, -y.x, -y.y)
Created on 2021-06-17 by the reprex package (v2.0.0)

Answer 4

我們可以使用 {powerjoin}

library(powerjoin)
power_left_join(df, df2, by = c("c", "d"), conflict = coalesce_xy)
#> # A tibble: 6 × 5
#>   c         d     z     x     y
#>   <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 a         1     1     1     1
#> 2 a         2     2     2     2
#> 3 a         3     7     3     2
#> 4 b         1     4     4     4
#> 5 b         2     5     5     5
#> 6 b         3     6     6     6