英文:
Replacing column values that conditionally match a list of values in R
问题
我尝试在一个数据框中替换数值，当它与一个远小于其大小的第二个数据框中的标识符匹配时。下面是我尝试的一个示例：
df1 = data.frame(row = seq(1,6),
x = c("a","b","c","d","e","f"))
df2 = data.frame(row = c(5,3,1,15,10),
x2 = c("g","h","i","j","k"))
df3 = df1 %>% mutate(x = case_when(
df1$row == df2$row ~ df2$x2,
.default = df1$x
))
我试图实现这个操作，即当 df1$row 与 df2$row 匹配时，用 df2$x2 中的值替换 df1$x，否则保留 df1$x。预期输出如下：
df3
row x
1 1 i
2 2 b
3 3 h
4 4 d
5 5 g
6 6 f
感谢任何帮助。
英文:
I am trying to replace values in one dataframe when it matches an identifier in a second dataframe of a much smaller size. A toy example of what I've tried:
df1 = data.frame(row = seq(1,6),
x = c("a","b","c","d","e","f"))
df2 = data.frame(row = c(5,3,1,15,10),
x2 = c("g","h","i","j","k"))
df3 = df1 %>% mutate(x = case_when(
df1$row == df2$row ~ df2$x2,
.default = df1$x
))
I am attempting this to read, when df1$row matches df2$row, replace df1$x with the value from df2$x2 and otherwise leave df1$x. The expected output:
df3
row x
1 1 i
2 2 b
3 3 h
4 4 d
5 5 g
6 6 f
Any help appreciated.
答案1
得分: 1
我们可以通过row进行join，然后使用coalesce：
library(dplyr)
df1 %>%
left_join(df2, by = 'row') %>%
mutate(x = coalesce(x2, x), .keep = 'unused')
row x
1 1 i
2 2 b
3 3 h
4 4 d
5 5 g
6 6 f
英文:
We can join by row, then use coalesce:
library(dplyr)
df1 %>%
left_join(df2, by = 'row') %>%
mutate(x = coalesce(x2, x), .keep = 'unused')
row x
1 1 i
2 2 b
3 3 h
4 4 d
5 5 g
6 6 f
</details>
# 答案2
**得分**: 1
我们可以使用 {powerjoin}
``` r
df1 = data.frame(row = seq(1,6),
x = c("a","b","c","d","e","f"))
df2 = data.frame(row = c(5,3,1,15,10),
x2 = c("g","h","i","j","k"))
library(powerjoin)
power_left_join(df1, df2 |> dplyr::rename(x = x2), by = "row", conflict = coalesce_yx)
#> row x
#> 1 1 i
#> 2 2 b
#> 3 3 h
#> 4 4 d
#> 5 5 g
#> 6 6 f
创建于2023年03月17日，使用 reprex v2.0.2
英文:
We might use {powerjoin}
df1 = data.frame(row = seq(1,6),
x = c("a","b","c","d","e","f"))
df2 = data.frame(row = c(5,3,1,15,10),
x2 = c("g","h","i","j","k"))
library(powerjoin)
power_left_join(df1, df2 |> dplyr::rename(x = x2), by = "row", conflict = coalesce_yx)
#> row x
#> 1 1 i
#> 2 2 b
#> 3 3 h
#> 4 4 d
#> 5 5 g
#> 6 6 f
<sup>Created on 2023-03-17 with reprex v2.0.2</sup>
答案3
得分: 0
使用dplyr 1.1.0版本：
df1 %>%
rows_update(df2 %>% rename(x = x2), unmatched = "ignore")
结果：
匹配，按 = "row"
行 x
1 1 i
2 2 b
3 3 h
4 4 d
5 5 g
6 6 f
如果两个表具有相同的行名称，会更简单：
df1 %>%
rows_update(df2, unmatched = "ignore")
英文:
With dplyr 1.1.0:
df1 %>%
rows_update(df2 %>% rename(x = x2), unmatched = "ignore")
Result
Matching, by = "row"
row x
1 1 i
2 2 b
3 3 h
4 4 d
5 5 g
6 6 f
If both tables had the same rownames it would be simpler:
df1 %>%
rows_update(df2, unmatched = "ignore")
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