使用R中的二项分布来估算缺失值

Question

I have a column with some missing values (q1 = 9) , I would like to impute it based on q1=1(=yes) and q1 =2(=no) binomial distribution like the SPSS script below. I couldn't find the R equivalent function

The SPSS code :

SPSS version :IF  q_1 = 9 x=RV.BINOM(1,0.976) .
if q_1 = 9 & x=1 q_1 = 1.
if q_d1 = 9 & x=0 q_1 = 2.

The column summary is like this

 q_1    n    percent
    1 5868   97.56%
    2  142   2.36%
    9    5   0.08%

Answer 1

You can generate the imputed values with sample .

Missing = which(q1 == 9)
q1[Missing] = sample(2, length(Missing), prob=c(0.976, 0.024))

Answer 2

What about this:

library(tidyverse)

vect1 <- runif(10000, 0, 1)
vect1a <- case_when(
  vect1 < 0.9756 ~ 1,
  vect1 < 0.9756 + 0.0236 ~ 2,
  TRUE ~ 9)
df1 <- tibble(q1 = vect1a)

pct1 <- 0.9756 / (1 - 0.008)
df1a <-  df1 %>% 
  mutate(rand_id = runif(nrow(.), 0, 1),
         q1a = case_when(q1 < 9 ~ q1,
                         rand_id < pct1 ~ 1,
                         TRUE ~ 2))

Mice package can handle impute missing values as well, but not sure it uses binomial distribution.