在R中拆分数据框的列。

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英文:

split column in a data frame R

问题

我有一个类似这样的数据框：

物种 时间 位置
Barbar,Barbar 9:30 1
Barbar 10:37 4
Barbar,Pippip 12:03 2
Barbar,Pippip,Hypsav 09:52 5
Pippip,Barbar 07:45 5
Barbar,Pippip 00:00 3

基本上，我应该创建新的行来分割物种列，当在同一情况下有两个标签时。

例如：如果我有这一行：

物种 时间 位置
Barbar,Pippip,Hypsav 09:52 5

我将获得这些行：

物种 时间 位置
Barbar 09:52 5
Pippip 09:52 5
Hypsav 09:52 5

因此，使用第一个数据框，我将获得这种结果：

物种 时间 位置
Barbar 9:30 1
Barbar 9:30 1
Barbar 10:37 4
Barbar 12:03 2
Pippip 12:03 2
Barbar 09:52 5
Pippip 09:52 5
Hypsav 09:52 5
Pippip 07:45 5
Barbar 07:45 5
Barbar 00:00 3
Pippip 00:00 3
英文:

I have a data frame which looks like that :

species time loc
Barbar,Barbar 9:30 1
Barbar 10:37 4
Barbar,Pippip 12:03 2
Barbar,Pippip,Hypsav 09:52 5
Pippip,Barbar 07:45 5
Barbar,Pippip 00:00 3

Basically I whould create new rows to split the species colums when there two tags in rthe same case.

For example : if I had this row :

species time loc
Barbar,Pippip,Hypsav 09:52 5

I whould obtain these rows :

species time loc
Barbar 09:52 5
Pippip 09:52 5
Hypsav 09:52 5

So with the first data frame I would obtain this kind of result :

species time loc
Barbar 9:30 1
Barbar 9:30 1
Barbar 10:37 4
Barbar 12:03 2
Pippip 12:03 2
Barbar 09:52 5
Pippip 09:52 5
Hypsav 09:52 5
Pippip 07:45 5
Barbar 07:45 5
Barbar 00:00 3
Pippip 00:00 3

What can I do to get the result ?

答案1

得分: 2

使用unnest函数

  1. library(dplyr)
  2. library(tidyr)
  4. df %>%
  5.   mutate(species = strsplit(species, ",")) %>%
  6.   unnest(species)

数据

  1. df <- structure(list(species = c("Barbar,Barbar", "Barbar", "Barbar,Pippip", 
  2. "Barbar,Pippip,Hypsav", "Pippip,Barbar", "Barbar,Pippip"), time = c("9:30", 
  3. "10:37", "12:03", "09:52", "07:45", "00:00"), loc = c(1L, 4L, 
  4. 2L, 5L, 5L, 3L)), class = "data.frame", row.names = c(NA, -6L))
英文:

With unnest

  1. library(dplyr)
  2. library(tidyr)
  4. df %&gt;% 
  5.   mutate(species = strsplit(species, &quot;,&quot;)) %&gt;% 
  6.   unnest(species)
  7. # A tibble: 12 &#215; 3
  8.    species time    loc
  9.    &lt;chr&gt;   &lt;chr&gt; &lt;int&gt;
  10.  1 Barbar  9:30      1
  11.  2 Barbar  9:30      1
  12.  3 Barbar  10:37     4
  13.  4 Barbar  12:03     2
  14.  5 Pippip  12:03     2
  15.  6 Barbar  09:52     5
  16.  7 Pippip  09:52     5
  17.  8 Hypsav  09:52     5
  18.  9 Pippip  07:45     5
  19. 10 Barbar  07:45     5
  20. 11 Barbar  00:00     3
  21. 12 Pippip  00:00     3

Data

  1. df &lt;- structure(list(species = c(&quot;Barbar,Barbar&quot;, &quot;Barbar&quot;, &quot;Barbar,Pippip&quot;, 
  2. &quot;Barbar,Pippip,Hypsav&quot;, &quot;Pippip,Barbar&quot;, &quot;Barbar,Pippip&quot;), time = c(&quot;9:30&quot;, 
  3. &quot;10:37&quot;, &quot;12:03&quot;, &quot;09:52&quot;, &quot;07:45&quot;, &quot;00:00&quot;), loc = c(1L, 4L, 
  4. 2L, 5L, 5L, 3L)), class = &quot;data.frame&quot;, row.names = c(NA, -6L))

答案2

得分: 1

以下是代码部分的翻译：

尝试使用逗号作为当前分隔符的以下内容：

  1. library(tidyverse)
  3. data_split <- data %>%
  4.   separate(species, into = c("species1", "species2", "species3"), sep = ",") %>%
  5.   pivot_longer(cols = starts_with("species"), values_to = "species") %>%
  6.   filter(!is.na(species)) %>%
  7.   select(-name)
  8.   print(data_split)

请注意，以上是您提供的代码的翻译部分。

英文:

Try use the following using a , as the present separator

  1. library(tidyverse)
  3. data_split &lt;- data %&gt;%
  4.   separate(species, into = c(&quot;species1&quot;, &quot;species2&quot;, &quot;species3&quot;), sep = &quot;,&quot;) %&gt;%
  5.   pivot_longer(cols = starts_with(&quot;species&quot;), values_to = &quot;species&quot;) %&gt;%
  6.   filter(!is.na(species)) %&gt;%
  7.   select(-name)
  8.   print(data_split)

答案3

得分: 1

或者，您可以使用data.table的方法：

  1. library(data.table)
  3. # 将df转换为data.table
  4. setDT(df)
  6. # 首先拆分物种并重新分配到同一列
  7. # 然后使用“loc”和“time”对物种进行分发
  8. df[, species := strsplit(x = species, split = ","), ][
  9.   , .(species = unlist(species)), by = .(loc, time)]
  11. #    loc  time species
  12. # 1:   1  9:30  Barbar
  13. # 2:   1  9:30  Barbar
  14. # 3:   4 10:37  Barbar
  15. # 4:   2 12:03  Barbar
  16. # 5:   2 12:03  Pippip
  17. # 6:   5 09:52  Barbar
  18. # 7:   5 09:52  Pippip
  19. # 8:   5 09:52  Hypsav
  20. # 9:   5 07:45  Pippip
  21. #10:   5 07:45  Barbar
  22. #11:   3 00:00  Barbar
  23. #12:   3 00:00  Pippip

根据您的一般工作流程或数据大小，您可以评估哪种方法对您最有效。

英文:

Alternatively, you can use the data.table approach:

  1. library(data.table)
  3. # convert df to a data.table
  4. setDT(df)
  6. # at first split the species and reassign it to the same column
  7. # then unlist to distribute the species for every &quot;loc&quot; and &quot;time&quot;
  8. df[,species:=strsplit(x = species, split = &quot;,&quot;),][
  9.   ,.(species = unlist(species)), by=.(loc,time)]
  11. #    loc  time species
  12. # 1:   1  9:30  Barbar
  13. # 2:   1  9:30  Barbar
  14. # 3:   4 10:37  Barbar
  15. # 4:   2 12:03  Barbar
  16. # 5:   2 12:03  Pippip
  17. # 6:   5 09:52  Barbar
  18. # 7:   5 09:52  Pippip
  19. # 8:   5 09:52  Hypsav
  20. # 9:   5 07:45  Pippip
  21. #10:   5 07:45  Barbar
  22. #11:   3 00:00  Barbar
  23. #12:   3 00:00  Pippip

Depending on your general workflow or data size you can evaluate what works best for you.

huangapple
  • 本文由 发表于 2023年4月17日 20:35:33
  • 转载请务必保留本文链接：https://go.coder-hub.com/76035216.html
  • r
  • split
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