如何解析一系列数字？

huangapple
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英文:

How to parse a range of numbers?

问题

Sure, here's the translated code part:

  1. 我有一些数据看起来像这样：
  3. > dput(range)
  4. structure(list(Range = c("0T-1T", "2T-5T")), class = "data.frame", row.names = c(NA, 
  5. -2L))
  7. 基本上，我想知道是否可以解析这些值，以便最终的数据框看起来像这样：
  9. > dput(range2)
  10. structure(list(Range = c("0T", "1T", "2T", "3T", "4T", "5T")), class = "data.frame", row.names = c(NA, 
  11. -6L))
英文:

I have some data that looks like this:

  1. > dput(range)
  2. structure(list(Range = c("0T-1T", "2T-5T")), class = "data.frame", row.names = c(NA, 
  3. -2L))

Basically, I am wondering if it is possible to parse these values out, so that the final dataframe looks like this:

  1. > dput(range2)
  2. structure(list(Range = c("0T", "1T", "2T", "3T", "4T", "5T")), class = "data.frame", row.names = c(NA, 
  3. -6L))

答案1

得分: 1

A tidyverse solution with tidyr::separate_longer_delim() + tidyr::full_seq():

  1. library(tidyverse)
  3. range %>%
  4.   group_by(grp = Range) %>%
  5.   separate_longer_delim(Range, '-') %>%
  6.   reframe(Range = paste0(full_seq(parse_number(Range), 1), 'T')) %>%
  7.   select(-grp)

# A tibble: 6 × 1

Range

1 0T

2 1T

3 2T

4 3T

5 4T

6 5T

  2. <details>
  3. <summary>英文:</summary>
  5. A `tidyverse` solution with `tidyr::separate_longer_delim()` + `tidyr::full_seq()`:
  7. ```r
  8. library(tidyverse)
  10. range %&gt;%
  11.   group_by(grp = Range) %&gt;%
  12.   separate_longer_delim(Range, &#39;-&#39;) %&gt;%
  13.   reframe(Range = paste0(full_seq(parse_number(Range), 1), &#39;T&#39;)) %&gt;%
  14.   select(-grp)
  16. # # A tibble: 6 &#215; 1
  17. #   Range
  18. #   &lt;chr&gt;
  19. # 1 0T
  20. # 2 1T   
  21. # 3 2T
  22. # 4 3T
  23. # 5 4T
  24. # 6 5T

答案2

得分: 0

可能有更简洁的方法，但这是一种方法：

  1. library(dplyr, warn.conflicts = F)
  2. library(stringr)
  3. library(tidyr)
  4. library(purrr)
  5. structure(list(Range = c("0T-1T", "2T-5T")), class = "data.frame", row.names = c(NA, -2L)) %>%
  6.     mutate(
  7.         Range = str_split(Range, '-'),
  8.         Range = map(Range, str_remove, 'T'),
  9.         Range = map(Range, as.integer),
  10.         Range = map(Range, ~ paste0('T', seq(.x[1], .x[2], 1)))
  11.     ) %>%
  12.     unnest(Range)
  13. #> # A tibble: 6 × 1
  14. #>   Range
  15. #>   <chr>
  16. #> 1 T0   
  17. #> 2 T1   
  18. #> 3 T2   
  19. #> 4 T3   
  20. #> 5 T4   
  21. #> 6 T5

创建于2023年06月08日，使用reprex v2.0.2

假设仅在每个"XT-YT"对内展开，而不是整个范围（这肯定更容易）。

英文:

There is probably a cleaner way but here is one way:

  1. library(dplyr, warn.conflicts = F)
  2. library(stringr)
  3. library(tidyr)
  4. library(purrr)
  5. structure(list(Range = c(&quot;0T-1T&quot;, &quot;2T-5T&quot;)), class = &quot;data.frame&quot;, row.names = c(NA, 
  6.                                                                                  -2L)) %&gt;% 
  7.     mutate(
  8.         Range = str_split(Range, &#39;-&#39;),
  9.         Range = map(Range, str_remove, &#39;T&#39;),
  10.         Range = map(Range, as.integer),
  11.         Range = map(Range, ~ paste0(&#39;T&#39;, seq(.x[1], .x[2], 1)))
  12.     ) %&gt;% 
  13.     unnest(Range)
  14. #&gt; # A tibble: 6 &#215; 1
  15. #&gt;   Range
  16. #&gt;   &lt;chr&gt;
  17. #&gt; 1 T0   
  18. #&gt; 2 T1   
  19. #&gt; 3 T2   
  20. #&gt; 4 T3   
  21. #&gt; 5 T4   
  22. #&gt; 6 T5

<sup>Created on 2023-06-08 with reprex v2.0.2</sup>

Assumes to only expand withing each "XT-YT" pair rather than the full range (which would be easier for sure).

huangapple
  • 本文由 发表于 2023年6月9日 05:15:57
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