错误，使用逗号作为分隔符拆分为新行时。

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英文:

Error while splitting into new row with comma as delimiter

问题

我有以下的数据框：

  1. temp = structure(list(pid = c("s1", "s1", "s1"), LEFT_GENE = c("PTPRO", "EPS8", "DPY19L2,AC084357.2,AC027667.1"), RIGHT_GENE = c("", "FOx,D", "DPY19L2P2,S100A11P1")), row.names = c(1L, 2L, 3L), class = "data.frame")

我想要将以逗号分隔的每个项拆分为新的行，并创建新的组合。例如，最后一行应该创建6个额外的新行。然而，我遇到了一个我不理解的错误。

  1. temp %>%
  2.   separate_rows(LEFT_GENE:RIGHT_GENE, sep=",") %>%
  3.   data.frame(stringsAsFactors = FALSE)

错误消息是：

  1. Error in `fn()`:
  2. ! In row 3, can't recycle input of size 3 to size 2.
  3. Run `rlang::last_error()` to see where the error occurred.

然而，错误似乎来自第3行，因为前两行正常工作。

  1. temp[1:2, ] %>%
  2.   separate_rows(LEFT_GENE:RIGHT_GENE, sep=",") %>%
  3.   data.frame(stringsAsFactors = FALSE)

有人知道问题是什么吗？

英文:

I have the following dataframe

  1. temp = structure(list(pid = c("s1", "s1", "s1"), LEFT_GENE = c("PTPRO", "EPS8", "DPY19L2,AC084357.2,AC027667.1"
  2. ), RIGHT_GENE = c("", "FOx,D", "DPY19L2P2,S100A11P1")), row.names = c(1L, 2L, 3L), class = "data.frame")
  5.   pid                     LEFT_GENE          RIGHT_GENE
  6. 1  s1                         PTPRO                    
  7. 2  s1                          EPS8                 FOx, D
  8. 3  s1 DPY19L2,AC084357.2,AC027667.1 DPY19L2P2,S100A11P1

I want to split each item delimited with a comma into a new row and create new combination.
For example, the last row should create 6 new additional rows. However I'm getting this error I don't understand.

  1. temp %>%
  2.   separate_rows(LEFT_GENE:RIGHT_GENE, sep=",") %>%  
  3.   data.frame ( stringsAsFactors = F)
  5. Error in `fn()`:
  6. ! In row 3, can't recycle input of size 3 to size 2.
  7. Run `rlang::last_error()` to see where the error occurred.

However the error seems to be coming from row 3 since rows 1:2 works fine

  1. > temp[1:2, 
  2. +      ] %>%
  3. +   separate_rows(LEFT_GENE:RIGHT_GENE, sep=",") %>%  
  4. +   data.frame ( stringsAsFactors = F)
  5.   pid LEFT_GENE RIGHT_GENE
  6. 1  s1     PTPRO           
  7. 2  s1      EPS8        FOx
  8. 3  s1      EPS8          D

Does anyone know what the issue is?

答案1

得分: 3

  1. 你只能一次分开一列
  3.      temp %>%
  4.        separate_rows(RIGHT_GENE)%>%
  5.        separate_rows(LEFT_GENE)
  7.     # A tibble: 9 × 3
  8.       pid   LEFT_GENE  RIGHT_GENE 
  9.       <chr> <chr>      <chr>      
  10.     1 s1    PTPRO      ""         
  11.     2 s1    EPS8       "FOx"      
  12.     3 s1    EPS8       "D"        
  13.     4 s1    DPY19L2    "DPY19L2P2"
  14.     5 s1    AC084357.2 "DPY19L2P2"
  15.     6 s1    AC027667.1 "DPY19L2P2"
  16.     7 s1    DPY19L2    "S100A11P1"
  17.     8 s1    AC084357.2 "S100A11P1"
  18.     9 s1    AC027667.1 "S100A11P1"
英文:

You can only separate one column at a time

  1.  temp %>%
  2.    separate_rows(RIGHT_GENE)%>%
  3.    separate_rows(LEFT_GENE)
  5. # A tibble: 9 × 3
  6.   pid   LEFT_GENE  RIGHT_GENE 
  7.   <chr> <chr>      <chr>      
  8. 1 s1    PTPRO      ""         
  9. 2 s1    EPS8       "FOx"      
  10. 3 s1    EPS8       "D"        
  11. 4 s1    DPY19L2    "DPY19L2P2"
  12. 5 s1    AC084357.2 "DPY19L2P2"
  13. 6 s1    AC027667.1 "DPY19L2P2"
  14. 7 s1    DPY19L2    "S100A11P1"
  15. 8 s1    AC084357.2 "S100A11P1"
  16. 9 s1    AC027667.1 "S100A11P1"

答案2

得分: 1

如果我们需要6行，一个选项是

  1. library(dplyr)
  2. library(tidyr)
  3. library(stringr)
  4. library(purrr)
  5. temp %>%
  6.   mutate(across(ends_with("_GENE"), ~ strsplit(.x,  split = ",")), 
  7.   cnt = pmax(lengths(LEFT_GENE), lengths(RIGHT_GENE))) %>%
  8.   mutate(across(ends_with("_GENE"),
  9.     ~ map2(.x, cnt, ~ `length<-`(.x, .y))) %>% 
  10.   select(-cnt) %>% 
  11.   unnest_longer(where(is.list))

-输出

  1. # A tibble: 6 × 3
  2.   pid   LEFT_GENE  RIGHT_GENE
  3.   <chr> <chr>      <chr>     
  4. 1 s1    PTPRO      <NA>      
  5. 2 s1    EPS8       FOx       
  6. 3 s1    <NA>       D         
  7. 4 s1    DPY19L2    DPY19L2P2 
  8. 5 s1    AC084357.2 S100A11P1 
  9. 6 s1    AC027667.1 <NA>

如果NA应该被前一个非NA替代，请在末尾添加fill:

  1. ...
  2. %>% fill(ends_with("_GENE"))
  3. # A tibble: 6 × 3
  4.   pid   LEFT_GENE  RIGHT_GENE
  5.   <chr> <chr>      <chr>     
  6. 1 s1    PTPRO      <NA>      
  7. 2 s1    EPS8       FOx       
  8. 3 s1    EPS8       D         
  9. 4 s1    DPY19L2    DPY19L2P2 
  10. 5 s1    AC084357.2 S100A11P1 
  11. 6 s1    AC027667.1 S100A11P1
英文:

If we need 6 rows, an option is

  1. library(dplyr)
  2. library(tidyr)
  3. library(stringr)
  4. library(purrr)
  5. temp %&gt;% 
  6.   mutate(across(ends_with(&quot;_GENE&quot;), ~ strsplit(.x,  split = &quot;,&quot;)), 
  7.   cnt = pmax(lengths(LEFT_GENE), lengths(RIGHT_GENE))) %&gt;% 
  8.   mutate(across(ends_with(&quot;_GENE&quot;),
  9.     ~ map2(.x, cnt, ~ `length&lt;-`(.x, .y)))) %&gt;%
  10.   select(-cnt) %&gt;%
  11.   unnest_longer(where(is.list))

-output

  1. # A tibble: 6 &#215; 3
  2.   pid   LEFT_GENE  RIGHT_GENE
  3.   &lt;chr&gt; &lt;chr&gt;      &lt;chr&gt;     
  4. 1 s1    PTPRO      &lt;NA&gt;      
  5. 2 s1    EPS8       FOx       
  6. 3 s1    &lt;NA&gt;       D         
  7. 4 s1    DPY19L2    DPY19L2P2 
  8. 5 s1    AC084357.2 S100A11P1 
  9. 6 s1    AC027667.1 &lt;NA&gt;

If the NAs should be replaced by the previous non-NA, add fill at the end

  1. ...
  2. %&gt;% fill(ends_with(&quot;_GENE&quot;))
  3. # A tibble: 6 &#215; 3
  4.   pid   LEFT_GENE  RIGHT_GENE
  5.   &lt;chr&gt; &lt;chr&gt;      &lt;chr&gt;     
  6. 1 s1    PTPRO      &lt;NA&gt;      
  7. 2 s1    EPS8       FOx       
  8. 3 s1    EPS8       D         
  9. 4 s1    DPY19L2    DPY19L2P2 
  10. 5 s1    AC084357.2 S100A11P1 
  11. 6 s1    AC027667.1 S100A11P1

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  • 本文由 发表于 2023年2月14日 02:04:34
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