复制列名并拼接

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英文:

Replicate column names and splice

问题

以下是翻译好的部分：

  1. 我有一个如下的数据框：
  3. df <- data.frame(A = c(2, 0, 1), B = c(0, 3, 2))
  5. #   A B
  6. # 1 2 0
  7. # 2 0 3
  8. # 3 1 2

每个单元格中的数字表示相应列名应重复的次数。重复部分应该由分号(;)拼接成单个字符串。预期的输出如下：

  1. #      A     B
  2. # 1  A;A  <NA>
  3. # 2 <NA> B;B;B
  4. # 3    A   B;B

我正在寻找一种处理更大数据集的高效方法：

  1. set.seed(1234)
  2. df <- as.data.frame(matrix(sample(0:5, 1e4*26, replace = TRUE), 1e4, 26))
  3. names(df) <- LETTERS
  5. #    A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
  6. # 1  3 0 3 5 4 0 3 2 2 1 3 3 4 3 2 4 0 1 4 5 2 5 5 2 0 0
  7. # 2  1 5 1 0 3 3 2 0 1 5 5 2 5 0 2 5 1 1 2 4 5 5 0 5 0 0
  8. # 3  5 5 2 0 1 4 5 4 0 5 5 1 1 1 2 2 4 5 4 5 5 5 0 4 0 0
  9. # ...
  10. # [ reached 'max' / getOption("max.print") -- omitted 9997 rows ]

我更喜欢使用base或tidyverse的解决方案。欢迎使用data.table，尽管我对它不太熟悉。

  2. <details>
  3. <summary>英文:</summary>
  5. I have a dataframe as follows:
  7. ```r
  8. df &lt;- data.frame(A = c(2, 0, 1), B = c(0, 3, 2))
  10. #   A B
  11. # 1 2 0
  12. # 2 0 3
  13. # 3 1 2

The number in each cell indicates the times for which the corresponding column name should repeat. The replicates should be spliced by semicolons(;) to a single string. The expected output turns out to

  1. #      A     B
  2. # 1  A;A  &lt;NA&gt;
  3. # 2 &lt;NA&gt; B;B;B
  4. # 3    A   B;B

I'm searching a efficient way to deal with a much larger dataset:

  1. set.seed(1234)
  2. df &lt;- as.data.frame(matrix(sample(0:5, 1e4*26, replace = TRUE), 1e4, 26))
  3. names(df) &lt;- LETTERS
  5. #    A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
  6. # 1  3 0 3 5 4 0 3 2 2 1 3 3 4 3 2 4 0 1 4 5 2 5 5 2 0 0
  7. # 2  1 5 1 0 3 3 2 0 1 5 5 2 5 0 2 5 1 1 2 4 5 5 0 5 0 0
  8. # 3  5 5 2 0 1 4 5 4 0 5 5 1 1 1 2 2 4 5 4 5 5 5 0 4 0 0
  9. # ...
  10. # [ reached &#39;max&#39; / getOption(&quot;max.print&quot;) -- omitted 9997 rows ]

I prefer base or tidyverse solutions. data.table is welcome but I'm unfamiliar to it though.

答案1

得分: 3

以下是已翻译的代码部分：

在R中的基本选项：

  1. df[] <- mapply(\(x, y) sapply(x, \(z) ifelse(z == 0, NA, paste(rep(y, z), collapse = ";"))),
  2.        df, names(df))
  4. #      A     B
  5. # 1  A;A  <NA>
  6. # 2 <NA> B;B;B
  7. # 3    A   B;B

转换为tidyverse：

  1. library(purrr)
  2. library(dplyr)
  3. df %>%
  4.   imap_dfr(\(x, y) sapply(x, \(z) ifelse(z == 0, NA, paste(rep(y, z), collapse = ";"))))

另一个dplyr选项（但可能较慢）：

  1. df %>%
  2.   rowwise() %>%
  3.   mutate(across(everything(), ~ ifelse(.x == 0, NA, paste(rep(rep(cur_column(), n()), .x), collapse = ";"))))

基准测试结果：

  1. library(microbenchmark)
  2. mb <- microbenchmark(
  3.   darren = {res <- df
  4.   n <- unlist(res)
  5.   res[res > 0] <- sapply(
  6.     split(rep.int(names(df)[col(df)], n), rep.int(seq_along(n), n)),
  7.     paste, collapse = ';'
  8.   )
  9.   res[res == 0] <- NA},
  10.   mael = mapply(\(x, y) sapply(x, \(z) ifelse(z == 0, NA, paste(rep(y, z), collapse = ";"))),
  11.                 df, names(df)),
  12.   setup = {
  13.     set.seed(1234)
  14.     df <- as.data.frame(matrix(sample(0:5, 1e4*26, replace = TRUE), 1e4, 26))
  15.     names(df) <- LETTERS
  16.   },
  17.   times = 10
  18. )
  20. #Unit: seconds
  21. #   expr      min       lq     mean   median       uq      max neval
  22. # darren 1.017067 1.042890 1.116436 1.105676 1.203968 1.227238    10
  23. #   mael 1.426412 1.518794 1.576665 1.581085 1.628206 1.713553    10
英文:

Here is a base R option:

  1. df[] &lt;- mapply(\(x, y) sapply(x, \(z) ifelse(z == 0, NA, paste(rep(y, z), collapse = &quot;;&quot;))),
  2.        df, names(df))
  4. #      A     B
  5. # 1  A;A  &lt;NA&gt;
  6. # 2 &lt;NA&gt; B;B;B
  7. # 3    A   B;B

And converted to tidyverse:

  1. library(purrr)
  2. library(dplyr)
  3. df %&gt;% 
  4.  imap_dfr(\(x, y) sapply(x, \(z) ifelse(z == 0, NA, paste(rep(y, z), collapse = &quot;;&quot;))))

Another dplyr option (but probably slow):

  1. df %&gt;% 
  2.   rowwise() %&gt;% 
  3.   mutate(across(everything(), ~ ifelse(.x == 0, NA, paste(rep(rep(cur_column(), n()), .x), collapse = &quot;;&quot;))))

Benchmark:

  1. library(microbenchmark)
  2. mb &lt;- microbenchmark(
  3.   darren = {res &lt;- df
  4.   n &lt;- unlist(res)
  5.   res[res &gt; 0] &lt;- sapply(
  6.     split(rep.int(names(df)[col(df)], n), rep.int(seq_along(n), n)),
  7.     paste, collapse = &#39;;&#39;
  8.   )
  9.   res[res == 0] &lt;- NA},
  10.   mael = mapply(\(x, y) sapply(x, \(z) ifelse(z == 0, NA, paste(rep(y, z), collapse = &quot;;&quot;))),
  11.                 df, names(df)),
  12.   setup = {
  13.     set.seed(1234)
  14.     df &lt;- as.data.frame(matrix(sample(0:5, 1e4*26, replace = TRUE), 1e4, 26))
  15.     names(df) &lt;- LETTERS
  16.   },
  17.   times = 10
  18. )
  20. #Unit: seconds
  21. #   expr      min       lq     mean   median       uq      max neval
  22. # darren 1.017067 1.042890 1.116436 1.105676 1.203968 1.227238    10
  23. #   mael 1.426412 1.518794 1.576665 1.581085 1.628206 1.713553    10
  25. </details>
  29. # 答案2
  30. **得分**: 2
  32. 一种 `base` 选项：
  34. ```r
  35. res <- df
  36. n <- unlist(res)
  37. res[res > 0] <- sapply(
  38.   split(rep.int(names(df)[col(df)], n), rep.int(seq_along(n), n)),
  39.   paste, collapse = ';'
  40. )
  41. res[res == 0] <- NA
  43. res[1:5, 1:5]
  44. #           A         B       C         D       E
  45. # 1     A;A;A      <NA>   C;C;C D;D;D;D;D E;E;E;E
  46. # 2         A B;B;B;B;B       C      <NA>   E;E;E
  47. # 3 A;A;A;A;A B;B;B;B;B     C;C      <NA>       E
  48. # 4   A;A;A;A   B;B;B;B    <NA>         D E;E;E;E
  49. # 5     A;A;A         B C;C;C;C       D;D   E;E;E
英文:

One base option:

  1. res &lt;- df
  2. n &lt;- unlist(res)
  3. res[res &gt; 0] &lt;- sapply(
  4.   split(rep.int(names(df)[col(df)], n), rep.int(seq_along(n), n)),
  5.   paste, collapse = &#39;;&#39;
  6. )
  7. res[res == 0] &lt;- NA
  9. res[1:5, 1:5]
  10. #           A         B       C         D       E
  11. # 1     A;A;A      &lt;NA&gt;   C;C;C D;D;D;D;D E;E;E;E
  12. # 2         A B;B;B;B;B       C      &lt;NA&gt;   E;E;E
  13. # 3 A;A;A;A;A B;B;B;B;B     C;C      &lt;NA&gt;       E
  14. # 4   A;A;A;A   B;B;B;B    &lt;NA&gt;         D E;E;E;E
  15. # 5     A;A;A         B C;C;C;C       D;D   E;E;E

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  • 本文由 发表于 2023年7月20日 20:37:25
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