如何在R中从两列创建一个数据框。

huangapple
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英文:

how to create a dataframe from two columns in r

问题

我有以下数据集：

  1. > data.frame(trait = c("Cholesterol", "Cholesterol", "ApoB", "ApoA",
  2. > "TRI", "TRI"), ID = c(1,2,2,1,1,2))
  4.     trait           ID
  5.     Cholesterol     1
  6.     Cholesterol     2
  7.     ApoB            2
  8.     ApoA            1
  9.     TRI             1
  10.     TRI             2

我想创建一个进一步的数据框，如下所示：

  1.     trait        1     2
  2.     Cholesterol  TRUE  TRUE
  3.     ApoB         FALSE TRUE
  4.     ApoA         TRUE  FALSE
  5.     TRI          TRUE  TRUE

你知道如何获得这个第二个数据框吗？

英文:

I have the following dataset:

> data.frame(trait = c("Cholesterol", "Cholesterol", "ApoB", "ApoA",
> "TRI", "TRI"), ID = c(1,2,2,1,1,2))

  1. trait           ID
  2. Cholesterol     1
  3. Cholesterol     2
  4. ApoB            2
  5. ApoA            1
  6. TRI             1
  7. TRI             2

I wanna to create a further dataframe as follow:

  1. trait        1     2
  2. Cholesterol  TRUE  TRUE
  3. ApoB         FALSE TRUE
  4. ApoA         TRUE  FALSE
  5. TRI          TRUE  TRUE

Do you know how to get this second dataframe?

答案1

得分: 3

这是tidyr::pivot_wider()的典型用例。

as.logical会将1或2都强制转换为TRUE，因为values_fill = FALSE，其余的NA值被转换为FALSE。

  1. library(tidyr)
  3. df |>
  4.     pivot_wider(names_from = ID,
  5.                 values_from = ID,
  6.                 values_fn = as.logical,
  7.                 values_fill = FALSE)
  9. # A tibble: 4 × 3
  10.   trait       `1`   `2`  
  11.   <chr>       <lgl> <lgl>
  12. 1 Cholesterol TRUE  TRUE 
  13. 2 ApoB        FALSE TRUE 
  14. 3 ApoA        TRUE  FALSE
  15. 4 TRI         TRUE  TRUE

数据

  1. df <- data.frame(trait = c("Cholesterol", "Cholesterol", "ApoB", "ApoA",
  2.                            "TRI", "TRI"),  ID = c(1,2,2,1,1,2))
英文:

This is the typical case for tidyr::pivot_wider().

as.logical will coerce either 1 or 2 to TRUE, and the remaining NAs are converted to FALSE because of values_fill = FALSE

  1. library(tidyr)
  3. df |&gt;
  4.     pivot_wider(names_from = ID,
  5.                 values_from = ID,
  6.                 values_fn = as.logical,
  7.                 values_fill = FALSE)
  9. # A tibble: 4 &#215; 3
  10.   trait       `1`   `2`  
  11.   &lt;chr&gt;       &lt;lgl&gt; &lt;lgl&gt;
  12. 1 Cholesterol TRUE  TRUE 
  13. 2 ApoB        FALSE TRUE 
  14. 3 ApoA        TRUE  FALSE
  15. 4 TRI         TRUE  TRUE

data

  1. df &lt;- data.frame(trait = c(&quot;Cholesterol&quot;, &quot;Cholesterol&quot;, &quot;ApoB&quot;, &quot;ApoA&quot;,
  2.                            &quot;TRI&quot;, &quot;TRI&quot;),  ID = c(1,2,2,1,1,2))
  4. </details>
  8. # 答案2
  9. **得分**: 1
  11. 你可以尝试 `table`
  12. ```R
  13. (table(df) > 0) %>%
  14.     as.data.frame() %>%
  15.     rownames_to_column(var = "trait")

这将得到

  1.         trait     1     2
  2. 1        ApoA  TRUE FALSE
  3. 2        ApoB FALSE  TRUE
  4. 3 Cholesterol  TRUE  TRUE
  5. 4         TRI  TRUE  TRUE
英文:

You can try table

  1. (table(df) &gt; 0) %&gt;%
  2.     as.data.frame() %&gt;%
  3.     rownames_to_column(var = &quot;trait&quot;)

which gives

  1.         trait     1     2
  2. 1        ApoA  TRUE FALSE
  3. 2        ApoB FALSE  TRUE
  4. 3 Cholesterol  TRUE  TRUE
  5. 4         TRI  TRUE  TRUE

答案3

得分: 0

这是基础R中的一行代码：

  1. cbind(df[1], "1" = df[[2]] == 1, "2" = df[[2]] == 2)
  2. #>         trait     1     2
  3. #> 1 Cholesterol  TRUE FALSE
  4. #> 2 Cholesterol FALSE  TRUE
  5. #> 3        ApoB FALSE  TRUE
  6. #> 4        ApoA  TRUE FALSE
  7. #> 5         TRI  TRUE FALSE
  8. #> 6         TRI FALSE  TRUE

创建于2023年05月28日，使用 reprex v2.0.2

英文:

This is a one-liner in base R:

  1. cbind(df[1], &quot;1&quot; = df[[2]] == 1, &quot;2&quot; = df[[2]] == 2)
  2. #&gt;         trait     1     2
  3. #&gt; 1 Cholesterol  TRUE FALSE
  4. #&gt; 2 Cholesterol FALSE  TRUE
  5. #&gt; 3        ApoB FALSE  TRUE
  6. #&gt; 4        ApoA  TRUE FALSE
  7. #&gt; 5         TRI  TRUE FALSE
  8. #&gt; 6         TRI FALSE  TRUE

<sup>Created on 2023-05-28 with reprex v2.0.2</sup>

答案4

得分: 0

请尝试以下代码：

  1. df <- data.frame(trait = c("胆固醇", "胆固醇", "ApoB", "ApoA", "TRI", "TRI"),
  2.            ID = c(1,2,2,1,1,2)) %>%
  3.   pivot_wider(names_from = ID , values_from = ID, values_fill = FALSE) %>%
  4.   mutate(across(c(`1`,`2`), ~ifelse(.x==0, FALSE, TRUE)))
  7. # 一个 tibble: 4 × 3
  8.   trait       `1`   `2`  
  9.   <chr>       <lgl> <lgl>
  10. 1 胆固醇  TRUE  TRUE 
  11. 2 ApoB        FALSE TRUE 
  12. 3 ApoA        TRUE  FALSE
  13. 4 TRI         TRUE  TRUE

注意：只有代码部分被翻译，其他内容保持不变。

英文:

Please try,

  1. df &lt;- data.frame(trait = c(&quot;Cholesterol&quot;, &quot;Cholesterol&quot;, &quot;ApoB&quot;, &quot;ApoA&quot;, &quot;TRI&quot;, &quot;TRI&quot;), 
  2.            ID = c(1,2,2,1,1,2)) %&gt;% 
  3.   pivot_wider(names_from = ID , values_from = ID, values_fill = FALSE) %&gt;% 
  4.   mutate(across(c(`1`,`2`), ~ifelse(.x==0, FALSE, TRUE)))
  7. # A tibble: 4 &#215; 3
  8.   trait       `1`   `2`  
  9.   &lt;chr&gt;       &lt;lgl&gt; &lt;lgl&gt;
  10. 1 Cholesterol TRUE  TRUE 
  11. 2 ApoB        FALSE TRUE 
  12. 3 ApoA        TRUE  FALSE
  13. 4 TRI         TRUE  TRUE

huangapple
  • 本文由 发表于 2023年5月29日 00:01:11
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  • dplyr
  • r
  • tidyverse
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