如何根据r中的另一列填写缺失的一列

Question

I have a subset of data frame as below. I want to fill the NAs in column "age at disease" so that the age of one individual with disease be same as the sibling (identified from familyID) without disease.

structure(list(id = c(1, 2, 3, 4, 5, 6), 
           familyId = c(1, 1, 2, 2, 3, 3), 
           disease = c(1, 0, 0, 1, 1, 0), 
           `age at disease` = c("40","NA", "NA", "43", "52", "NA")), 
      class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, -6L))

which means that the last column "age at disease" should be: c(40,40,43,43,52,52).

Answer 1

You can use the following code:

library(dplyr)
library(tidyr)
df %>%
  na_if("NA") %>%
  group_by(familyId) %>%
  fill(`age at disease`) %>%
  fill(`age at disease`, .direction = "up")

Output:

# A tibble: 6 × 4
# Groups:   familyId [3]
     id familyId disease `age at disease`
  <dbl>    <dbl>   <dbl> <chr>           
1     1        1       1 40              
2     2        1       0 40              
3     3        2       0 43              
4     4        2       1 43              
5     5        3       1 52              
6     6        3       0 52  

Answer 2

If there is only a single non-NA element per group, we may also do

library(dplyr)
df1 %>%
   type.convert(as.is = TRUE) %>%
   group_by(familyId) %>%
   mutate(`age at disease` = `age at disease`[complete.cases(`age at disease`)][1]) %>% 
   ungroup

-output

# A tibble: 6 × 4
     id familyId disease `age at disease`
  <dbl>    <dbl>   <dbl> <chr>           
1     1        1       1 40              
2     2        1       0 40              
3     3        2       0 43              
4     4        2       1 43              
5     5        3       1 52              
6     6        3       0 52       

Answer 3

Here is another dplyr approach:

df %>%
  group_by(familyId) %>% 
  arrange(`age at disease`,.by_group = TRUE) %>% 
  mutate(`age at disease` = first(`age at disease`))

     id familyId disease `age at disease`
  <dbl>    <dbl>   <dbl> <chr>           
1     1        1       1 40              
2     2        1       0 40              
3     4        2       1 43              
4     3        2       0 43              
5     5        3       1 52              
6     6        3       0 52