使用`each`还是`times`来处理R中的向量？

huangapple
go评论96阅读模式
英文:

Using `each` vs `times` for a vector in R?

问题

  1. 在下面的数据集中，
  3.     df_final <- data.frame(month = rep(c(1:12), each =30 * 2),
  4.                                site = rep(c("1","2","3","4","5","6"), times = 60 * 2),
  5.                                quad = rep(c(1:5), times = 72 * 2),
  6.                                seed = rep(c(1:2), times = 360))
  8. .. 每个月（1-12），有6个站点（1-6），有5个象限（1-5） `quad`。因此，一个数据集有12 * 6 * 5 = 360行。在这里，我有两个这样的数据集（360 * 2）。我也可以有1000个这样的数据集。对于每个唯一的数据集，我想分配一个不同的种子。
  11.  [![enter image description here][1]][1]
  13. 输出应该如下所示：
  16.     month site quad seed
  17.        1     1    1     1 
  18.        1     1    2     1
  19.        1     1    3     1
  20.        1     1    4     1
  21.        1     1    5     1
  22.        1     1    1     2 
  23.        1     1    2     2
  24.        1     1    3     2
  25.        1     1    4     2
  26.        1     1    5     2
  28.   我尝试使用 `seed = rep(c(1:2), each = 360)`，但那样并不奏效。我该如何得到这个输出？我希望有一种通用的方法，这样我就可以为1000个种子做同样的事情。
  31.   [1]: https://i.stack.imgur.com/RkeUb.png
英文:

In the following dataset,

  1. df_final &lt;- data.frame(month = rep(c(1:12), each =30 * 2),
  2.                            site = rep(c(&quot;1&quot;,&quot;2&quot;,&quot;3&quot;,&quot;4&quot;,&quot;5&quot;,&quot;6&quot;), times = 60 * 2),
  3.                            quad = rep(c(1:5), times = 72 * 2),
  4.                            seed = rep(c(1:2), times = 360))

.. every month (1-12), has 6 sites (1-6), with 5 quadrants (1-5) quad. So one dataset has 12 * 6 * 5 = 360 rows. Here, I have two such datasets (360 * 2). I could also have 1000 such datasets. For every unqiue dataset, I want to assign a different seed.

使用`each`还是`times`来处理R中的向量？

output should looks like:

  1. month site quad seed
  2.    1     1    1     1 
  3.    1     1    2     1
  4.    1     1    3     1
  5.    1     1    4     1
  6.    1     1    5     1
  7.    1     1    1     2 
  8.    1     1    2     2
  9.    1     1    3     2
  10.    1     1    4     2
  11.    1     1    5     2

I tried using seed = rep(c(1:2), each = 360 but that didn't work as well. How can I get this output? I want a general way to do it, so that I can do the same for 1000 seeds.

答案1

得分: 1

You can use cur_group_id:

  1. library(dplyr)
  2. df_final %>% 
  3.   group_by(month, site) %>% 
  4.   mutate(seed = cur_group_id()) %>% 
  5.   ungroup() %>% arrange(month, site)

或者在列排序后使用 consecutive_id：

  1. library(dplyr) #1.1.0+
  2. df_final %>% 
  3.   arrange(month, site) %>% 
  4.   mutate(seed = consecutive_id(month, site))
英文:

You can use cur_group_id:

  1. library(dplyr)
  2. df_final %&gt;% 
  3.   group_by(month, site) %&gt;% 
  4.   mutate(seed = cur_group_id()) %&gt;% 
  5.   ungroup() %&gt;% arrange(month, site)

Or consecutive_id after arranging the columns:

  1. library(dplyr) #1.1.0+
  2. df_final %&gt;% 
  3.   arrange(month, site) %&gt;% 
  4.   mutate(seed = consecutive_id(month, site))

答案2

得分: 1

使用 data.table

  1. library(data.table)
  2. setDT(df_final)[order(month, site), seed := rleid(month, site)]

或者

  1. setDT(df_final)[, seed := .GRP, .(month, site)]
英文:

Using data.table

  1. library(data.table)
  2. setDT(df_final)[order(month, site), seed := rleid(month, site)]

Or do

  1. setDT(df_final)[, seed := .GRP, .(month, site)]

huangapple
  • 本文由 发表于 2023年3月7日 01:33:47
  • 转载请务必保留本文链接：https://go.coder-hub.com/75654032.html
  • dataframe
  • r
  • vector
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定