如何检测具有至少一个错位数字的个体

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英文:

How to detect individuals who have at least one misordered number

问题

这是我的数据集表示：

  1. ID<-c(1,1,1,1,
  2.       2,2,2,
  3.       3,3,
  4.       4,4,4,4)
  5.     
  6. order<-c(44,44,45,41,
  7.          13,21,2013,
  8.          21,10,
  9.          45,65,74,877)
  10.     
  11. mydata<-data.frame(
  12.   ID=ID, order=order
  13. )

变量order应该随着时间增加。通过程序，我应该检测所有在随访时变量order减少的个体（在本例中为ID=1和ID=3）。

英文:

Here is a representation of my dataset:

  1. ID&lt;-c(1,1,1,1,
  2.       2,2,2,
  3.       3,3,
  4.       4,4,4,4)
  6. order&lt;-c(44,44,45,41,
  7.         13,21,2013,
  8.         21,10,
  9.         45,65,74,877)
  11. mydata&lt;-data.frame(
  12.   ID=ID,order=order
  13. )
  16.   ID Time order
  17. 1   1    1    44
  18. 2   1    2    44
  19. 3   1    3    45
  20. 4   1    4    41
  21. 5   2    1    13
  22. 6   2    2    21
  23. 7   2    3  2013
  24. 8   3    1    21
  25. 9   3    2    10
  26. 10  4    1    45
  27. 11  4    2    65
  28. 12  4    3    74
  29. 13  4    4   877

The variable order should increase with time.
With a program, I should detect all the individuals (ID=1 and ID=3 in this case) that have variable order that decreases at a time of follow-up.

答案1

得分: 2

  1. unique(subset(mydata, ave(order, ID, FUN=function(x) c(0, diff(x))) < 0)$ID)
  2. [1] 1 3
  4. or if you want the rows back:
  6. subset(mydata, ave(order, ID, FUN=function(x) c(0, diff(x))) < 0)
  7.   ID order
  8. 4  1    41
  9. 9  3    10
  11. library(dplyr) # version >= 1.1.0
  13. mydata %>%
  14.    filter(order < lag(order), .by=ID)
  16.   ID order
  17. 1  1    41
  18. 2  3    10
英文: 
  1.  unique(subset(mydata, ave(order, ID, FUN=\(x)c(0, diff(x)))&lt;0)$ID)
  2. [1] 1 3

or if you want the rows back:

  1. subset(mydata, ave(order, ID, FUN=\(x)c(0, diff(x)))&lt;0)
  2.   ID order
  3. 4  1    41
  4. 9  3    10
  6. library(dplyr) # version &gt;= 1.1.0
  8. mydata %&gt;% 
  9.    filter(order&lt;lag(order), .by=ID)
  11.   ID order
  12. 1  1    41
  13. 2  3    10

答案2

得分: 2

  1. 根据Roland的评论，请检查`is.unsorted`函数：
  2. ```r
  3. unique(mydata$ID[with(mydata, ave(order, ID, FUN = is.unsorted)) == 1])
  4. #[1] 1 3

或者

  1. subset(mydata, ave(order, ID, FUN = is.unsorted) == 1)
  2. #  ID order
  3. #1  1    44
  4. #2  1    44
  5. #3  1    45
  6. #4  1    41
  7. #8  3    21
  8. #9  3    10

在dplyr中：

  1. library(dplyr) #1.1.0
  2. filter(mydata, is.unsorted(order), .by = ID)

在data.table中：

  1. setDT(mydata)[, .SD[is.unsorted(order)], by = ID]
英文:

As per Roland's comment, check is.unsorted:

  1. unique(mydata$ID[with(mydata, ave(order, ID, FUN = is.unsorted)) == 1])
  2. #[1] 1 3

or

  1. subset(mydata, ave(order, ID, FUN = is.unsorted) == 1)
  2. #  ID order
  3. #1  1    44
  4. #2  1    44
  5. #3  1    45
  6. #4  1    41
  7. #8  3    21
  8. #9  3    10

In dplyr:

  1. library(dplyr) #1.1.0
  2. filter(mydata, is.unsorted(order), .by = ID)

In data.table:

  1. setDT(mydata)[, .SD[is.unsorted(order)], by = ID]

答案3

得分: 1

我们可以检查rank(order)是否对应于row_number()：

  1. library(dplyr)
  3. mydata %>% 
  4.   filter(row_number() != rank(order, ties.method = "first"), .by = ID)
  6. #> # A tibble: 6 × 2
  7. #> # Groups:   ID [2]
  8. #>      ID order
  9. #>   <dbl> <dbl>
  10. #> 1     1    44
  11. #> 2     1    44
  12. #> 3     1    45
  13. #> 4     1    41
  14. #> 5     3    21
  15. #> 6     3    10

要仅获取个体的ID，我们可以在管道中添加 pull(ID) %>% unique()：

  1. mydata %>% 
  2.   filter(row_number() != rank(order, ties.method = "first"), .by = ID) %>% 
  3.   pull(ID) %>% 
  4.   unique()
  5. #> 1 3

数据来自 OP

  1. ID<-c(1,1,1,1,
  2.       2,2,2,
  3.       3,3,
  4.       4,4,4,4)
  6. order<-c(44,44,45,41,
  7.          13,21,2013,
  8.          21,10,
  9.          45,65,74,877)
  11. mydata<-data.frame(
  12.   ID=ID,order=order
  13. )

创建于 2023-03-09，使用 reprex v2.0.2

英文:

We could check if the rank(order) corresponds to the row_number():

  1. library(dplyr)
  3. mydata %&gt;% 
  4.   filter(row_number() != rank(order, ties.method = &quot;first&quot;), .by = ID)
  6. #&gt; # A tibble: 6 &#215; 2
  7. #&gt; # Groups:   ID [2]
  8. #&gt;      ID order
  9. #&gt;   &lt;dbl&gt; &lt;dbl&gt;
  10. #&gt; 1     1    44
  11. #&gt; 2     1    44
  12. #&gt; 3     1    45
  13. #&gt; 4     1    41
  14. #&gt; 5     3    21
  15. #&gt; 6     3    10

To just get the ID of the individuals we can add pull(ID) %&gt;% unique() to the pipe:

  1. mydata %&gt;% 
  2.   filter(row_number() != rank(order, ties.method = &quot;first&quot;), .by = ID) %&gt;% 
  3.   pull(ID) %&gt;% 
  4.   unique()
  5. #&gt; 1 3

Data from OP

  1. ID&lt;-c(1,1,1,1,
  2.       2,2,2,
  3.       3,3,
  4.       4,4,4,4)
  6. order&lt;-c(44,44,45,41,
  7.          13,21,2013,
  8.          21,10,
  9.          45,65,74,877)
  11. mydata&lt;-data.frame(
  12.   ID=ID,order=order
  13. )

<sup>Created on 2023-03-09 with reprex v2.0.2</sup>

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  • 本文由 发表于 2023年3月9日 19:42:20
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