How can I extract a string from between last dash and second to last dash out of a column that contains lists of strings?

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英文:

How can I extract a string from between last dash and second to last dash out of a column that contains lists of strings?

问题

  1. # 以下是翻译好的部分：
  3. 我有一些数据，想要创建一个新列，其中包含在倒数第二个破折号和倒数第一个破折号之间的字符串。但有一个小技巧！我的一些观察结果是"列出的"，我也想从列表项中获取每个目标字符串。
  5. 示例数据如下：
  7. data <- data.frame(
  8.   a = c("1500925OR3-29139-315012", 
  9.         "1500925OR3-2-2913A-315012", 
  10.         "c(\"1500925OR3-200B-315012\", \"1500925OR3-4-2919999-315012\")")
  11. )
  13. 看起来像这样：
  15.                                                            a
  16. 1                                    1500925OR3-29139-315012
  17. 2                                  1500925OR3-2-2913A-315012
  18. 3 c("1500925OR3-200B-315012", "1500925OR3-4-2919999-315012")
  20. 我想要的数据看起来像这样
  22.        a_clean
  23. 1         29139
  24. 2         2913A
  25. 3 200B, 2919999
  27. 我一直在尝试使用正则表达式，但我无法弄清如何获取最后一个破折号之前的字符串。这会捕获最后一个破折号后面的内容...`-[^-]*$`，但显然那不对。
英文:

I have some data and I want to make a new column with the string that is between the last dash and the second to last dash. But there is a twist! Some of my observations are "listed", and I want to get each target string out of the list items as well.

Example data here:

  1. data &lt;- data.frame(
  2.   a = c(&quot;1500925OR3-29139-315012&quot;, 
  3.         &quot;1500925OR3-2-2913A-315012&quot;, 
  4.         &quot;c(\&quot;1500925OR3-200B-315012\&quot;, \&quot;1500925OR3-4-2919999-315012\&quot;)&quot;)
  5. )

looks like:

  1.                                                            a
  2. 1                                    1500925OR3-29139-315012
  3. 2                                  1500925OR3-2-2913A-315012
  4. 3 c(&quot;1500925OR3-200B-315012&quot;, &quot;1500925OR3-4-2919999-315012&quot;)

I want data that looks like this

  1.         a_clean
  2. 1         29139
  3. 2         2913A
  4. 3 200B, 2919999

I've been working on using regex, but I can't figure out how to get the string before the last dash. This grabs the stuff after the last dash...-[^-]*$ but obviously thats not right.

答案1

得分: 3

尝试在sub中使用这个正则表达式，并使用lapply。

  1. dat$b <- lapply(dat$a, \(x) sub('-?.*-(.*)-.*', '\', x, perl=TRUE))
  2. dat
  3. #                                                     a             b
  4. # 1                             1500925OR3-29139-315012         29139
  5. # 2                           1500925OR3-2-2913A-315012         2913A
  6. # 3 1500925OR3-200B-315012, 1500925OR3-4-2919999-315012 200B, 2919999

你提到了一个"list"列，所以我假设你的真实数据看起来是这样的。

数据：

  1. dat <- structure(list(a = list("1500925OR3-29139-315012", "1500925OR3-2-2913A-315012", 
  2.     c("1500925OR3-200B-315012", "1500925OR3-4-2919999-315012"))), row.names = c(NA, -3L), class = "data.frame")
英文:

Try this regex in sub and use lapply.

  1. dat$b &lt;- lapply(dat$a, \(x) sub(&#39;-?.*-(.*)-.*&#39;, &#39;\&#39;, x, perl=TRUE))
  2. dat
  3. #                                                     a             b
  4. # 1                             1500925OR3-29139-315012         29139
  5. # 2                           1500925OR3-2-2913A-315012         2913A
  6. # 3 1500925OR3-200B-315012, 1500925OR3-4-2919999-315012 200B, 2919999

You're talking about a "list" column, so I created one assuming that's what your real data looks like.

Data:

  1. dat &lt;- structure(list(a = list(&quot;1500925OR3-29139-315012&quot;, &quot;1500925OR3-2-2913A-315012&quot;, 
  2.     c(&quot;1500925OR3-200B-315012&quot;, &quot;1500925OR3-4-2919999-315012&quot;
  3.     ))), row.names = c(NA, -3L), class = &quot;data.frame&quot;)

答案2

得分: 2

A tidyverse approach:

  1. library(dplyr)
  2. library(tidyr)
  4. data %>%
  5.   mutate(id = row_number()) %>%
  6.   separate_rows(a, sep = "\\s") %>%
  7.   mutate(b = str_extract(a, "(?<=-)[^-]*(?=-[^-]*$)")) %>%
  8.   summarise(a_clean = toString(b), .by=id) %>%
  9.   select(-id)
  1.  a_clean      
  2.   <chr>        
  3. 1 29139        
  4. 2 2913A        
  5. 3 200B, 2919999
英文:

A tidyverse approach:

  1. library(dplyr)
  2. library(tidyr)
  4. data %&gt;%
  5.   mutate(id = row_number()) %&gt;% 
  6.   separate_rows(a, sep = &quot;\\s&quot;) %&gt;% 
  7.   mutate(b = str_extract(a, &quot;(?&lt;=-)[^-]*(?=-[^-]*$)&quot;)) %&gt;% 
  8.   summarise(a_clean = toString(b), .by=id) %&gt;% 
  9.   select(-id)
  1.  a_clean      
  2.   &lt;chr&gt;        
  3. 1 29139        
  4. 2 2913A        
  5. 3 200B, 2919999

答案3

得分: 2

  1. data.frame(
  2.   a = c(
  3.     "1500925OR3-29139-315012",
  4.     "1500925OR3-2-2913A-315012",
  5.     c("1500925OR3-200B-315012", "1500925OR3-4-2919999-315012")
  6.   ),
  7.   b = c(1:3)
  8. ) %>% separate_rows(a, sep = ',') %>% separate(a,
  9.                                                  c('col1', 'col2', 'col3', 'col4'),
  10.                                                  sep = '-',
  11.                                                  fill = 'left') %>% group_by(b) %>%
  12.   summarise(col3 = str_c(col3, collapse = ","))
  1. # A tibble: 3 x 2
  2.       b col3        
  3.   <int> <chr>       
  4. 1     1 29139       
  5. 2     2 2913A       
  6. 3     3 200B,2919999
英文:

Alternatively,

  1. data.frame(
  2.   a = c(
  3.     &quot;1500925OR3-29139-315012&quot;,
  4.     &quot;1500925OR3-2-2913A-315012&quot;,
  5.     &quot;c(\&quot;1500925OR3-200B-315012\&quot;, \&quot;1500925OR3-4-2919999-315012\&quot;)&quot;
  6.   ),
  7.   b = c(1:3)
  8. ) %&gt;% separate_rows(a, sep = &#39;\\,&#39;) %&gt;% separate(a,
  9.                                                  c(&#39;col1&#39;, &#39;col2&#39;, &#39;col3&#39;, &#39;col4&#39;),
  10.                                                  sep = &#39;\\-&#39;,
  11.                                                  fill = &#39;left&#39;) %&gt;% group_by(b) %&gt;%
  12.   summarise(col3 = str_c(col3, collapse = &quot;,&quot;))
  1. # A tibble: 3 &#215; 2
  2.       b col3        
  3.   &lt;int&gt; &lt;chr&gt;       
  4. 1     1 29139       
  5. 2     2 2913A       
  6. 3     3 200B,2919999

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  • 本文由 发表于 2023年6月26日 00:56:23
  • 转载请务必保留本文链接：https://go.coder-hub.com/76551519.html
  • gsub
  • r
  • regex
  • string
  • stringr
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