在R中有条件地替换匹配值列表的列数值。

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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：

  1. df1 = data.frame(row = seq(1,6),
  2.                    x = c("a","b","c","d","e","f"))
  4. df2 = data.frame(row = c(5,3,1,15,10),
  5.                  x2 = c("g","h","i","j","k"))
  7. df3 = df1 %>% mutate(x = case_when(
  8.   df1$row == df2$row ~ df2$x2,
  9.   .default = df1$x
  10. ))

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：

  1. df3
  2.   row x
  3. 1   1 i
  4. 2   2 b
  5. 3   3 h
  6. 4   4 d
  7. 5   5 g
  8. 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:

  1. df1 = data.frame(row = seq(1,6),
  2.                    x = c("a","b","c","d","e","f"))
  4. df2 = data.frame(row = c(5,3,1,15,10),
  5.                  x2 = c("g","h","i","j","k"))
  7. df3 = df1 %>% mutate(x = case_when(
  8.   df1$row == df2$row ~ df2$x2,
  9.   .default = df1$x
  10. ))

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:

  1. df3
  2.   row x
  3. 1   1 i
  4. 2   2 b
  5. 3   3 h
  6. 4   4 d
  7. 5   5 g
  8. 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：

  1. library(dplyr)
  2. df1 %>%
  3.     left_join(df2, by = 'row') %>%
  4.     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:

  1. library(dplyr)
  2. df1 %>%
  3.     left_join(df2, by = 'row') %>%
  4.     mutate(x = coalesce(x2, x), .keep = 'unused')
  6.   row x
  7. 1   1 i
  8. 2   2 b
  9. 3   3 h
  10. 4   4 d
  11. 5   5 g
  12. 6   6 f
  14. </details>
  18. # 答案2
  19. **得分**: 1
  21. 我们可以使用 {powerjoin}
  23. ``` r
  24. df1 = data.frame(row = seq(1,6),
  25.                  x = c("a","b","c","d","e","f"))
  27. df2 = data.frame(row = c(5,3,1,15,10),
  28.                  x2 = c("g","h","i","j","k"))
  30. library(powerjoin)
  31. power_left_join(df1, df2 |&gt; dplyr::rename(x = x2), by = "row", conflict = coalesce_yx)
  32. #&gt;   row x
  33. #&gt; 1   1 i
  34. #&gt; 2   2 b
  35. #&gt; 3   3 h
  36. #&gt; 4   4 d
  37. #&gt; 5   5 g
  38. #&gt; 6   6 f

创建于2023年03月17日，使用 reprex v2.0.2

英文:

We might use {powerjoin}

  1. df1 = data.frame(row = seq(1,6),
  2.                  x = c(&quot;a&quot;,&quot;b&quot;,&quot;c&quot;,&quot;d&quot;,&quot;e&quot;,&quot;f&quot;))
  4. df2 = data.frame(row = c(5,3,1,15,10),
  5.                  x2 = c(&quot;g&quot;,&quot;h&quot;,&quot;i&quot;,&quot;j&quot;,&quot;k&quot;))
  7. library(powerjoin)
  8. power_left_join(df1, df2 |&gt; dplyr::rename(x = x2), by = &quot;row&quot;, conflict = coalesce_yx)
  9. #&gt;   row x
  10. #&gt; 1   1 i
  11. #&gt; 2   2 b
  12. #&gt; 3   3 h
  13. #&gt; 4   4 d
  14. #&gt; 5   5 g
  15. #&gt; 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：

  1. df1 %>% 
  2.   rows_update(df2 %>% rename(x = x2), unmatched = "ignore")

结果：

  1. 匹配，按 = "row"
  2.   行 x
  3. 1   1 i
  4. 2   2 b
  5. 3   3 h
  6. 4   4 d
  7. 5   5 g
  8. 6   6 f

如果两个表具有相同的行名称，会更简单：

  1. df1 %>% 
  2.   rows_update(df2, unmatched = "ignore")
英文:

With dplyr 1.1.0:

  1. df1 %&gt;%
  2.   rows_update(df2 %&gt;% rename(x = x2), unmatched = &quot;ignore&quot;)

Result

  1. Matching, by = &quot;row&quot;
  2.   row x
  3. 1   1 i
  4. 2   2 b
  5. 3   3 h
  6. 4   4 d
  7. 5   5 g
  8. 6   6 f

If both tables had the same rownames it would be simpler:

  1. df1 %&gt;%
  2.   rows_update(df2, unmatched = &quot;ignore&quot;)

huangapple
  • 本文由 发表于 2023年3月4日 05:57:49
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  • dplyr
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