用LOCF进行条件插补 [英] Conditional imputation with LOCF

问题描述

我有这个纵向数据的例子.我需要根据之前发生的情况估算0、999或-1值.

I've this example of longitudinal data. I need to impute 0, 999 or -1 values according to what occurs before.

ID = c(1,1,1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,5,6,6,6,6,6,6,6,6) 
Oxy = c(0, 999, 1, 999, 999, 0, 0, 999, 999, 0, 0, -1, 0, 999, 1, 1, -1, 1, 999, -1, 0, -1, 1,0, 999, 0) 
Y = c(2010,2011,2012,2013,2014,2011,2012,2013,2010,2011,2012,2010,2011,
      2012,2010,2011,2012,2013,2014,2015,2016,2017, 2018,2019,2020, 2021) 
Oxy2 = c(0, 999, 1, 1, 1, 0, 0, 999, 999, 0, 0, -1, 0, 999, 1, 1, 1, 1, 999, -1, 0, -1, 1, 1,1,1) 
df = data.frame(ID, Oxy, Y, Oxy2)

基本上，我想从Oxy获取Oxy2.当Oxy的先前值为0或-1时，我需要保留999，并考虑到组ID随着时间的推移，替换所有其他内容，出现在第一个1之后.

Basically, I'd like to get Oxy2 from Oxy. I need to keep 999 when previous values of Oxy are 0 or -1, and replace everything else comes after the first 1 appears, considering the group ID over time.

ID Oxy    Y Oxy2
  1   0 2010    0
  1 999 2011  999
  1   1 2012    1
  1 999 2013    1
  1 999 2014    1
  2   0 2011    0
  2   0 2012    0
  2 999 2013  999
  3 999 2010  999
  3   0 2011    0
  3   0 2012    0
  4  -1 2010   -1
  4   0 2011    0
  4 999 2012  999
  5   1 2010    1
  5   1 2011    1
  5  -1 2012    1
  5   1 2013    1
  6 999 2014  999
  6  -1 2015   -1
  6   0 2016    0
  6  -1 2017   -1
  6   1 2018    1
  6   0 2019    1
  6 999 2020    1
  6   0 2021    1

感谢您的建议.

推荐答案

您可以使用cumsum(Oxy == 1) >= 1标识第一个1之后的行:

You can use cumsum(Oxy == 1) >= 1 to identify which rows come after the first 1:

df %>%
  group_by(ID) %>%
  mutate(OxyFilled = ifelse(cumsum(Oxy == 1) >= 1, 1, Oxy))

输出:

# A tibble: 25 x 5
# Groups:   ID [6]
      ID   Oxy     Y  Oxy2 OxyFilled
   <dbl> <dbl> <dbl> <dbl>     <dbl>
 1     1     0  2010     0         0
 2     1   999  2011   999       999
 3     1     1  2012     1         1
 4     1   999  2013     1         1
 5     1   999  2014     1         1
 6     2     0  2011     0         0
 7     2     0  2012     0         0
 8     2   999  2013   999       999
 9     3   999  2010   999       999
10     3     0  2011     0         0
# … with 15 more rows

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