在列表中查找向量元素并在数据框中收集信息。

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英文:

Find elements of a vector in a list and collect information in data frame

问题

从具有基因ID的向量中，我想要确定哪个子列表包含一个或多个ID以及相关的p值。结果应该是一个数据框列表，每个基因ID都有一列，第二列指示ID是否出现为布尔值，第三列包含该基因ID的p值。由于Paul的问题，我意识到一个基因ID可能在多个子列表中，p值将表示其特异性。因此，预期的结果应该是每个子列表的数据框列表，以及指示基因ID是否存在的布尔向量。对编辑表示抱歉。

  1. # 示例数据
  3. the_list <- list(
  4.   a = data.frame(ids = c("ABB", 'SDG', 'SHD', 'DUR'), pval = c(0.01, 0.03, 0.05, 0.05)),
  5.   b = data.frame(ids =c('DYR' ,'LRH' ,'FPR', 'FUR', 'DCTWE', 'IRN', 'DRB'), pval = c(0.01, 0.03, 0.05, 0.05, 4, 5, 6)),
  6.   c = data.frame(ids =c('SYR' ,'SDT', 'DFN' ,'FRQ' ,'DFRR', 'SDR'), pval = c(0.01, 0.03, 0.05, 0.05, 5, 2))
  7. )
  9. the_vector <- c("ABB", 'FUR', 'DFN')
  11. expected_result <- list(
  12.   'a' = data.frame(
  13.     gene_ids = c("ABB", 'FUR', 'DFN'),
  14.     in_list = c(TRUE, FALSE, FALSE),
  15.     pval = c(0.01, NA, NA)
  16.   ),
  17.   'b' = data.frame(
  18.     gene_ids = c("ABB", 'FUR', 'DFN'),
  19.     in_list = c(FALSE, TRUE, FALSE),
  20.     pval = c(NA, 0.05, NA)
  21.   ),
  22.   'c' = data.frame(
  23.     gene_ids = c("ABB", 'FUR', 'DFN'),
  24.     in_list = c(FALSE, FALSE, TRUE),
  25.     pval = c(NA, NA, 0.05)
  26.   )
  27. )
英文:

From a vector with gene ids, I want to identify which sublist contains one or more of the ids and the associated pvalue.
The result should be a list of data frames with a col for each gene ID, a second col indicating if the ID appears as boolean and a third col with the pval of that gene ID)

Thanks to Pauls question i realized that a gene ID might be in several sublists and the pvalue would indicate its specificity. Accordingly, the expected result should be a list for a df for each sublist and a boolean vector indicating if the geneID is present. Sorry for the edit.

example data

  1.    the_list &lt;- list(
  2.   a = data.frame(ids = c(&quot;ABB&quot;, &#39;SDG&#39;, &#39;SHD&#39;, &#39;DUR&#39;), pval = c(0.01, 0.03, 0.05, 0.05)),
  3.   b = data.frame(ids =c(&#39;DYR&#39; ,&#39;LRH&#39; ,&#39;FPR&#39;, &#39;FUR&#39;, &#39;DCTWE&#39;, &#39;IRN&#39;, &#39;DRB&#39;), pval = c(0.01, 0.03, 0.05, 0.05, 4, 5, 6)),
  4.   c = data.frame(ids =c(&#39;SYR&#39; ,&#39;SDT&#39;, &#39;DFN&#39; ,&#39;FRQ&#39; ,&#39;DFRR&#39;, &#39;SDR&#39;), pval = c(0.01, 0.03, 0.05, 0.05, 5, 2))
  5. )
  7. the_vector &lt;- c(&quot;ABB&quot;, &#39;FUR&#39;, &#39;DFN&#39;)
  12.  expected_result &lt;- list(
  13.   &#39;a&#39; = data.frame(
  14.     gene_ids = c(&quot;ABB&quot;, &#39;FUR&#39;, &#39;DFN&#39;),
  15.     in_list = c(T, F, F),
  16.     pval = c(0.01, NA, NA)),
  17.   &#39;b&#39; = data.frame(
  18.     gene_ids = c(&quot;ABB&quot;, &#39;FUR&#39;, &#39;DFN&#39;),
  19.     in_list = c(F, T, F),
  20.     pval = c(NA, 0.05, NA)),
  21.   &#39;c&#39; = data.frame(
  22.     gene_ids = c(&quot;ABB&quot;, &#39;FUR&#39;, &#39;DFN&#39;),
  23.     in_list = c(F, F, T),
  24.     pval = c(NA,  NA, 0.05))
  25.   )

答案1

得分: 4

  1. Base R解决方案：
  3. # 对于每个列表元素：
  4. # res_list => 数据框的列表
  5. res_list <- lapply(
  6.   the_list,
  7.   function(x){
  8.     # 解析匹配元素的索引：
  9.     # idx => 整数向量
  10.     idx <- match(the_vector, x$ids)
  11.     # 解析所需的数据框：
  12.     # 数据框 => 环境
  13.     data.frame(
  14.       gene_ids = the_vector, 
  15.       in_list = !(is.na(idx)),
  16.       p_val = x$pval[idx],
  17.       stringsAsFactors = FALSE,
  18.       row.names = NULL
  19.     )
  20.   }
  21. )
英文:

Base R solution:

  1. # For each list element: 
  2. # res_list =&gt; list of data.frames
  3. res_list &lt;- lapply(
  4.   the_list,
  5.   function(x){
  6.     # Resolve the index of the matched elements: 
  7.     # idx =&gt; integer vector
  8.     idx &lt;- match(the_vector, x$ids)
  9.     # Resolve the required data.frame: 
  10.     # data.frame =&gt; env
  11.     data.frame(
  12.       gene_ids = the_vector, 
  13.       in_list = !(is.na(idx)),
  14.       p_val = x$pval[idx],
  15.       stringsAsFactors = FALSE,
  16.       row.names = NULL
  17.     )
  18.   }
  19. )

答案2

得分: 2

循环遍历数据框列表并合并：

  1. lapply(the_list, function(i){
  2.   res <- merge(data.frame(ids = the_vector), i, all.x = TRUE)
  3.   res$in_list <- !is.na(res$pval)
  4.   res
  5. })
  6. # $a
  7. #   ids pval in_list
  8. # 1 ABB 0.01    TRUE
  9. # 2 DFN   NA   FALSE
  10. # 3 FUR   NA   FALSE
  11. # 
  12. # $b
  13. #   ids pval in_list
  14. # 1 ABB   NA   FALSE
  15. # 2 DFN   NA   FALSE
  16. # 3 FUR 0.05    TRUE
  17. # 
  18. # $c
  19. #   ids pval in_list
  20. # 1 ABB   NA   FALSE
  21. # 2 DFN 0.05    TRUE
  22. # 3 FUR   NA   FALSE
英文:

Loop through the list of data.frames and merge:

  1. lapply(the_list, function(i){
  2.   res &lt;- merge(data.frame(ids = the_vector), i, all.x = TRUE)
  3.   res$in_list &lt;- !is.na(res$pval)
  4.   res
  5.   })
  6. # $a
  7. #   ids pval in_list
  8. # 1 ABB 0.01    TRUE
  9. # 2 DFN   NA   FALSE
  10. # 3 FUR   NA   FALSE
  11. # 
  12. # $b
  13. #   ids pval in_list
  14. # 1 ABB   NA   FALSE
  15. # 2 DFN   NA   FALSE
  16. # 3 FUR 0.05    TRUE
  17. # 
  18. # $c
  19. #   ids pval in_list
  20. # 1 ABB   NA   FALSE
  21. # 2 DFN 0.05    TRUE
  22. # 3 FUR   NA   FALSE

答案3

得分: 1

  1. dplyr::bind_rows(the_list, .id = "in_list") %>%
  2.   filter(ids %in% the_vector)

结果：

  1.   in_list ids pval
  2. 1       a ABB 0.01
  3. 2       b FUR 0.05
  4. 3       c DFN 0.05

编辑 - 也适用于多次匹配 - 即使相同的id以前已经匹配过，也会给出每次匹配的行 - 您可以根据需要按in_list或ids排序。

由于问题已被编辑，这不再符合所需的输出，但我会保留它，因为它简单且包含相同的信息。

英文: 
  1. dplyr::bind_rows(the_list, .id = &quot;in_list&quot;) %&gt;%
  2.   filter(ids %in% the_vector)

gives

  1.   in_list ids pval
  2. 1       a ABB 0.01
  3. 2       b FUR 0.05
  4. 3       c DFN 0.05

Edit - works with multi matches too - will give a row per match, even if the same id has previously matched - you can then order by in_list or ids as appropriate.

This no longer conforms to the desired output since the question was edited, but I will leave it here as it's simple and contains the same information.

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  • 本文由 发表于 2023年6月13日 16:56:56
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