在R中计算数据框中行内的百分比贡献的方法

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英文:

How to Calculate the Percentage Contribution Within Rows in a Data Frame in R

问题

I apologize for the confusion, but it seems that your request is to provide a translation of the content you provided, without addressing any specific questions or issues in the code. Here is the translated content:

我想知道，我们是否可以计算数据帧中行内值的百分比贡献。

我正在使用的数据帧如下所示：

  1. structure(list(`Row Labels` = c("X1", "X2", "X3", "X4"), `2019-01-01` = c(37, 
  2. 36, 45, 53), `2019-02-01` = c(3, 19, 14, 46), `2019-03-01` = c(28, 
  3. 2, 28, 28), `2019-04-01` = c(48, 70, 18, 16), `2019-05-01` = c(83, 
  4. 71, 58, 26), `2019-06-01` = c(85, 28, 83, 46), `2019-07-01` = c(60, 
  5. 20, 12, 77), `2019-08-01` = c(44, 66, 30, 99), `2019-09-01` = c(21, 
  6. 14, 31, 21), `2019-10-01` = c(26, 72, 72, 16), `2019-11-01` = c(15, 
  7. 96, 23, 100), `2019-12-01` = c(65, 0, 98, 66)), row.names = c(NA, 
  8. -4L), class = c("tbl_df", "tbl", "data.frame"))

我编写的代码如下：

  1. Book1 <- read_excel("X:/X/X/X - X/X/Book1.xlsx")
  3. First_Date <- "2019-01-01"
  4. Last_Date <- "2019-12-01"
  6. Book1 <- Book1 %>%
  7.   mutate(Sum = rowSums(pick(any_of(First_Date):any_of(Last_Date)))) %>%
  8.   mutate(across(pick(any_of(First_Date):any_of(Last_Date), ~./rowSums(pick(any_of(First_Date):any_of(Last_Date)))),.names = "{.col}_%"))

运行此代码时，我收到以下错误：

  1. Error in `mutate()`: ℹ In argument: `across(...)`. Caused by error in `pick()`: ! Formula shorthand must be wrapped in `where()`.
  3. # Bad   data %>% select(~./rowSums(pick(any_of(First_Date):any_of(Last_Date))))
  5. # Good   data %>% select(where(~./rowSums(pick(any_of(First_Date):any_of(Last_Date))))))

以上的代码应该获得以下图像的输出：

在R中计算数据框中行内的百分比贡献的方法

但如果输出以这种格式呈现，我不介意：

在R中计算数据框中行内的百分比贡献的方法

有人可以告诉我在计算贡献时出了什么问题吗？这将会很有帮助。是否有更简单的方法来做这个？

英文:

I wanted to know, if we can calculate the percentage contribution of the values within a row in a data frame.

The data frame I am working with is:

  1. structure(list(`Row Labels` = c(&quot;X1&quot;, &quot;X2&quot;, &quot;X3&quot;, &quot;X4&quot;), `2019-01-01` = c(37, 
  2. 36, 45, 53), `2019-02-01` = c(3, 19, 14, 46), `2019-03-01` = c(28, 
  3. 2, 28, 28), `2019-04-01` = c(48, 70, 18, 16), `2019-05-01` = c(83, 
  4. 71, 58, 26), `2019-06-01` = c(85, 28, 83, 46), `2019-07-01` = c(60, 
  5. 20, 12, 77), `2019-08-01` = c(44, 66, 30, 99), `2019-09-01` = c(21, 
  6. 14, 31, 21), `2019-10-01` = c(26, 72, 72, 16), `2019-11-01` = c(15, 
  7. 96, 23, 100), `2019-12-01` = c(65, 0, 98, 66)), row.names = c(NA, 
  8. -4L), class = c(&quot;tbl_df&quot;, &quot;tbl&quot;, &quot;data.frame&quot;))

The code which I wrote for this is given below:

  1. Book1 &lt;- read_excel(&quot;X:/X/X/X - X/X/Book1.xlsx&quot;)
  3. First_Date &lt;- &quot;2019-01-01&quot;
  4. Last_Date &lt;- &quot;2019-12-01&quot;
  6. Book1 &lt;- Book1 %&gt;% 
  7.   mutate(Sum = rowSums(pick(any_of(First_Date):any_of(Last_Date)))) %&gt;% 
  8.   mutate(across(pick(any_of(First_Date):any_of(Last_Date), ~./rowSums(pick(any_of(First_Date):any_of(Last_Date)))),.names = &quot;{.col}_%&quot;))

When I am running this code, the error I get is:

  1. Error in `mutate()`: ℹ In argument: `across(...)`. Caused by error in `pick()`: ! Formula shorthand must be wrapped in `where()`.
  3.   # Bad   data %&gt;% select(~./rowSums(pick(any_of(First_Date):any_of(Last_Date))))
  5.   # Good   data %&gt;% select(where(~./rowSums(pick(any_of(First_Date):any_of(Last_Date)))))

The code above should get the output of the below image

在R中计算数据框中行内的百分比贡献的方法

but I dont mind if the output is in this format as well

在R中计算数据框中行内的百分比贡献的方法

Can someone let me know what is it that i am getting wrong to find the contribution? It would be helpful. Is there a simpler way of doing it?

答案1

得分: 1

对于使用dplyr的第二个输出，您可以：

  1. library(dplyr)
  3. df %>%
  4.   rowwise() %>%
  5.   mutate(across(where(is.numeric), ~ .x / sum(across(where(is.numeric)))))
  7. # A tibble: 4 × 13
  8. # Rowwise: 
  9.   `Row Labels` `2019-01-01` `2019-02-01` `2019-03-01` `2019-04-01` `2019-05-01` `2019-06-01` `2019-07-01` `2019-08-01` `2019-09-01` `2019-10-01` `2019-11-01`
  10.   <chr>               <dbl>        <dbl>        <dbl>        <dbl>        <dbl>        <dbl>        <dbl>        <dbl>        <dbl>        <dbl>        <dbl>
  11. 1 X1                 0.0718      0.00583      0.0544        0.0932       0.161        0.165        0.117        0.0854       0.0408       0.0505       0.0291
  12. 2 X2                 0.0729      0.0385       0.00405       0.142        0.144        0.0567       0.0405       0.134        0.0283       0.146        0.194 
  13. 3 X3                 0.0879      0.0273       0.0547        0.0352       0.113        0.162        0.0234       0.0586       0.0605       0.141        0.0449
  14. 4 X4                 0.0892      0.0774       0.0471        0.0269       0.0438       0.0774       0.130        0.167        0.0354       0.0269       0.168 
  15. # ℹ 1 more variable: `2019-12-01` <dbl>
  17. 乘以100：
  19. df %>%
  20.   rowwise() %>%
  21.   mutate(across(
  22.     where(is.numeric), ~ .x / sum(across(where(is.numeric))) * 100
  23.                 ))
英文:

For the second output with dplyr you can:

  1. library(dplyr)
  3. df %&gt;% 
  4.   rowwise() %&gt;% 
  5.   mutate(across(where(is.numeric), ~ .x / sum(across(where(is.numeric)))))
  7. # A tibble: 4 &#215; 13
  8. # Rowwise: 
  9.   `Row Labels` `2019-01-01` `2019-02-01` `2019-03-01` `2019-04-01` `2019-05-01` `2019-06-01` `2019-07-01` `2019-08-01` `2019-09-01` `2019-10-01` `2019-11-01`
  10.   &lt;chr&gt;               &lt;dbl&gt;        &lt;dbl&gt;        &lt;dbl&gt;        &lt;dbl&gt;        &lt;dbl&gt;        &lt;dbl&gt;        &lt;dbl&gt;        &lt;dbl&gt;        &lt;dbl&gt;        &lt;dbl&gt;        &lt;dbl&gt;
  11. 1 X1                 0.0718      0.00583      0.0544        0.0932       0.161        0.165        0.117        0.0854       0.0408       0.0505       0.0291
  12. 2 X2                 0.0729      0.0385       0.00405       0.142        0.144        0.0567       0.0405       0.134        0.0283       0.146        0.194 
  13. 3 X3                 0.0879      0.0273       0.0547        0.0352       0.113        0.162        0.0234       0.0586       0.0605       0.141        0.0449
  14. 4 X4                 0.0892      0.0774       0.0471        0.0269       0.0438       0.0774       0.130        0.167        0.0354       0.0269       0.168 
  15. # ℹ 1 more variable: `2019-12-01` &lt;dbl&gt;

Multiply with 100:

  1. df %&gt;%
  2.   rowwise() %&gt;%
  3.   mutate(across(
  4.     where(is.numeric), ~ .x / sum(across(where(is.numeric))) * 100
  5.                 ))

huangapple
  • 本文由 发表于 2023年6月8日 20:04:10
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