Subtracting after and before values of each sample for each numeric column in R

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英文:

Subtracting after and before values of each sample for each numeric column in R

问题

以下是翻译好的部分：

  1. 我有一个类似下面示例结构的数据框
  3. ```R
  4. df <- data.frame(rbind(c("Sample1_x2", 10, 23, 6, 5, "Sample1", "after"),
  5.             c("Sample2_x2", 8, 53, 22, 52, "Sample2", "after"),
  6.             c("Sample1_x1", 12, 2, 44, 15, "Sample1", "before"),
  7.             c("Sample3_x1", 27, 46, 16, 65, "Sample3", "before"),
  8.             c("Sample2_x1", 41, 44, 27, 25, "Sample2", "before"),
  9.             c("Sample3_x2", 5, 38, 9, 29, "Sample3", "after")))
  10. colnames(df) <- c("name", "alpha", "beta", "gamma", "rho", "id", "group")
  11. df <- tibble::column_to_rownames(df, var = "name")
  12. df
  13.            var1 var2 var3 var4      id  group
  14. Sample1_x2   10   23    6    5 Sample1  after
  15. Sample2_x2    8   53   22   52 Sample2  after
  16. Sample1_x1   12    2   44   15 Sample1 before
  17. Sample3_x1   27   46   16   65 Sample3 before
  18. Sample2_x1   41   44   27   25 Sample2 before
  19. Sample3_x2    5   38    9   29 Sample3  after

我想通过计算每个样本的'id'，每个变量列的'after - before'来获得变化的数据框。所需输出如下：

  1.      id   alpha  beta  gamma    rho       
  2. Sample1      -2    21    -38    -10
  3. Sample2     -33     9     -5     27    
  4. Sample3     -22    -8     -7    -36

我尝试使用dplyr::group_by(id, group)，但在mutate()部分无法成功计算每个样本的差异。谢谢您提前的帮助。

  2. <details>
  3. <summary>英文:</summary>
  5. I have a dataframe in a similar structure of the example below

df <- data.frame(rbind(c("Sample1_x2", 10, 23, 6, 5, "Sample1", "after"),
c("Sample2_x2", 8, 53, 22, 52, "Sample2", "after"),
c("Sample1_x1", 12, 2, 44, 15, "Sample1", "before"),
c("Sample3_x1", 27, 46, 16, 65, "Sample3", "before"),
c("Sample2_x1", 41, 44, 27, 25, "Sample2", "before"),
c("Sample3_x2", 5, 38, 9, 29, "Sample3", "after")))
colnames(df) <- c("name", "alpha", "beta", "gamma", "rho", "id", "group")
df <- tibble::column_to_rownames(df, var = "name")
df
var1 var2 var3 var4 id group
Sample1_x2 10 23 6 5 Sample1 after
Sample2_x2 8 53 22 52 Sample2 after
Sample1_x1 12 2 44 15 Sample1 before
Sample3_x1 27 46 16 65 Sample3 before
Sample2_x1 41 44 27 25 Sample2 before
Sample3_x2 5 38 9 29 Sample3 after

  2. I want to get the change dataframe by calculating &#39;after - before&#39; for each sample by `id`, for each variable column (they are not numeric, and each numeric column has different name). Desired output is:
  1.  id   alpha  beta  gamma    rho

Sample1 -2 21 -38 -10
Sample2 -33 9 -5 27
Sample3 -22 -8 -7 -36

  2. I was trying to use `dplyr::group_by(id, group)` but could not succeed in `mutate()` part to calculate the difference for each sample. Thank you in advance.
  4. </details>
  7. # 答案1
  8. **得分**: 3
  10. 你可以在 `data.table` 中使用以下方法：
  12. ```R
  13. library(data.table)
  15. setDT(df)[order(-group), lapply(.SD, \(s) diff(as.numeric(s))), id, .SDcols=1:4]

输出：

  1.         id alpha beta gamma rho
  2. 1: Sample1    -2   21   -38 -10
  3. 2: Sample3   -22   -8    -7 -36
  4. 3: Sample2   -33    9    -5  27

在 dplyr 中，相应的方法如下：

  1. library(dplyr)
  3. df %>%
  4.   arrange(desc(group)) %>%
  5.   reframe(across(-group, ~diff(as.numeric(.x))), .by=id)
英文:

You could use this approach in data.table

  1. library(data.table)
  3. setDT(df)[order(-group), lapply(.SD, \(s) diff(as.numeric(s))), id, .SDcols=1:4]

Output:

  1.         id alpha beta gamma rho
  2. 1: Sample1    -2   21   -38 -10
  3. 2: Sample3   -22   -8    -7 -36
  4. 3: Sample2   -33    9    -5  27

An equivalent approach in dplyr is as follows:

  1. library(dplyr)
  3. df %&gt;% 
  4.   arrange(desc(group)) %&gt;% 
  5.   reframe(across(-group, ~diff(as.numeric(.x))), .by=id)

答案2

得分: 2

这是一个 dplyr 的选项：

  1. library(dplyr)
  3. df %>%
  4.   readr::type_convert() %>%
  5.   summarise(across(where(is.numeric), ~ .x[group == "after"] - .x[group == "before"]),
  6.             .by = id)
  8. #        id alpha beta gamma rho
  9. # 1 Sample1    -2   21   -38 -10
  10. # 2 Sample2   -33    9    -5  27
  11. # 3 Sample3   -22   -8    -7 -36
英文:

This is one dplyr option:

  1. library(dplyr)
  3. df %&gt;%
  4.   readr::type_convert() %&gt;%
  5.   summarise(across(where(is.numeric), ~ .x[group == &quot;after&quot;] - .x[group == &quot;before&quot;]),
  6.             .by = id)
  8. #        id alpha beta gamma rho
  9. # 1 Sample1    -2   21   -38 -10
  10. # 2 Sample2   -33    9    -5  27
  11. # 3 Sample3   -22   -8    -7 -36

答案3

得分: 1

以下是您要翻译的内容：

  1. library(dplyr)
  3. df %>%
  4.   group_by(id) %>%
  5.   arrange(group, .by_group = TRUE) %>%
  6.   type.convert(as.is = TRUE) %>%
  7.   summarise(across(-c(group, name), ~first(.)-last(.))) %>%
  8.   ungroup()
  10. # dplyr >= 1.1.0
  12. df %>%
  13.   arrange(group) %>%
  14.   type.convert(as.is = TRUE) %>%
  15.   summarise(across(-c(group, name), ~first(.)-last(.)), .by=id)
  17. # See @langtang's solution who uses reframe very elegantly instead of summarise!
  19.  id      alpha  beta gamma   rho
  20.   <chr>   <int> <int> <int> <int>
  21. 1 Sample1    -2    21   -38   -10
  22. 2 Sample2   -33     9    -5    27
  23. 3 Sample3   -22    -8    -7   -36

希望这有所帮助！

英文:

Update after clarification (removed first answer):

  1. library(dplyr)
  3. df %&gt;%  
  4.   group_by(id) %&gt;% 
  5.   arrange(group, .by_group = TRUE) %&gt;%
  6.   type.convert(as.is = TRUE) %&gt;% 
  7.   summarise(across(-c(group, name), ~first(.)-last(.))) %&gt;% 
  8.   ungroup()
  11. # dplyr &gt;= 1.1.0
  13. df %&gt;%  
  14.   arrange(group) %&gt;%
  15.   type.convert(as.is = TRUE) %&gt;% 
  16.   summarise(across(-c(group, name), ~first(.)-last(.)), .by=id)
  18. # See @langtang&#39;s solution who uses reframe very elegantly instead of summarise!
  20.  id      alpha  beta gamma   rho
  21.   &lt;chr&gt;   &lt;int&gt; &lt;int&gt; &lt;int&gt; &lt;int&gt;
  22. 1 Sample1    -2    21   -38   -10
  23. 2 Sample2   -33     9    -5    27
  24. 3 Sample3   -22    -8    -7   -36

huangapple
  • 本文由 发表于 2023年5月17日 07:06:57
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  • dataframe
  • dplyr
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