在 R 中按组补充缺失的日期（年月）和值

Question

I am trying to fill the missing date and rate. For the missing date, I wish to refill yr_month between 2019 (January) and 2021 (January). For the rate, I wish to refill any missing value as zero.

The close example I find is here - How to add only missing Dates in Dataframe but my challenge is that I have one more column by groups. It means each patient will have 2 years of data.

What's the best way I can do this on an efficient way?

# A tibble: 10 x 3
# Groups:   clinic, yr_month [10]
   clinic   yr_month   rate
   <chr>    <chr>     <dbl>
 1 patient1 2019-01  0.528 
 2 patient1 2019-04  0.528 
 3 patient1 2020-05  0.528 
 4 patient1 2021-01  1.06  
 5 patient2 2019-01  0.0671
 6 patient2 2019-02  0.436 
 7 patient2 2019-03  0.805 
 8 patient2 2019-04  0.671 
 9 patient2 2019-05  0.268 
10 patient2 2019-06  0.101 

dput

structure(list(clinic = c("patient1", "patient1", "patient1", 
"patient1", "patient2", "patient2", "patient2", "patient2", "patient2", 
"patient2"), yr_month = c("2019-01", "2019-04", "2020-05", "2021-01", 
"2019-01", "2019-02", "2019-03", "2019-04", "2019-05", "2019-06"
), rate = c(0.527704485488127, 0.527704485488127, 0.527704485488127, 
1.05540897097625, 0.0671163461861136, 0.436256250209739, 0.805396154233364, 
0.671163461861136, 0.268465384744455, 0.10067451927917)), row.names = c(NA, 
-10L), groups = structure(list(clinic = c("patient1", "patient1", 
"patient1", "patient1", "patient2", "patient2", "patient2", "patient2", 
"patient2", "patient2"), yr_month = c("2019-01", "2019-04", "2020-05", 
"2021-01", "2019-01", "2019-02", "2019-03", "2019-04", "2019-05", 
"2019-06"), .rows = structure(list(1L, 2L, 3L, 4L, 5L, 6L, 7L, 
    8L, 9L, 10L), ptype = integer(0), class = c("vctrs_list_of", 
"vctrs_vctr", "list"))), row.names = c(NA, 10L), class = c("tbl_df", 
"tbl", "data.frame"), .drop = TRUE), class = c("grouped_df", 
"tbl_df", "tbl", "data.frame"))

Expected output:

# Groups:   clinic, yr_month 
   clinic   yr_month   rate
   <chr>    <chr>     <dbl>
 1 patient1 2019-01      0
 1 patient1 2019-02      0 
 1 patient1 2019-03      0
 1 patient1 2019-04  0.528
 1 patient1 2019-05      0
 1 patient1 2019-06      0 
 1 patient1 2019-07      0
 1 patient1 2019-08      0 
 1 patient1 2019-09      0
...
25 patient2 2019-01  0.0671
26 patient2 2019-02  0.436 
27 patient2 2019-03  0.805 
28 patient2 2019-04  0.671 
29 patient2 2019-05  0.268 
30 patient2 2019-06  0.101 
...
48 patient2 2021-01      0  
 

Answer 1

Using tidyverse and zoo you could try the following. You can use group_by to consider each patient within a clinic. For your dates, you can use as.yearmon from zoo to use a year-month format. Then, using complete from tidyr you can fill in the missing rate values with 0 for missing months.

library(tidyverse)
library(zoo)

df %>%
  group_by(clinic) %>%
  mutate(yr_month = as.yearmon(yr_month)) %>%
  arrange(yr_month) %>%
  complete(yr_month = seq(first(yr_month), last(yr_month), by = 1 / 12), fill = list(rate = 0)) 

Answer 2

first we create an empty data set for your given month year combinations:

library(dplyr)
library(tidyr)
library(purrr)
month_df <- tibble(
  month = rep(c(1:12), 3),
  year = c(rep(2019, 12), rep(2020, 12), rep(2021, 12))
  ) %>% 
  mutate(month_ind = ifelse(nchar(month) == 1, paste0("0", month), month),
         yr_month = paste0(year, "-", month_ind)) %>% 
  select(yr_month)

After that, given df is your data:

df %>% ungroup() %>% 
  group_split(clinic) %>% 
  map(., right_join, month_df) %>% 
  map(., fill, clinic) %>% 
  bind_rows() %>% 
  arrange(clinic, yr_month) %>% 
  mutate(rate = ifelse(is.na(rate), 0 , rate))

output is:

# A tibble: 72 x 3
   clinic   yr_month  rate
   <chr>    <chr>    <dbl>
 1 patient1 2019-01  0.528
 2 patient1 2019-02  0    
 3 patient1 2019-03  0    
 4 patient1 2019-04  0.528
 5 patient1 2019-05  0    
 6 patient1 2019-06  0    
 7 patient1 2019-07  0    
 8 patient1 2019-08  0    
 9 patient1 2019-09  0    
10 patient1 2019-10  0    
# ... with 62 more rows