replace NA value with the group value
Try ave. It applies a function to groups. Have a look at ?ave for details, e.g.:
df$med_card_new <- ave(df$med_card, df$hhold_no, FUN=function(x)unique(x[!is.na(x)]))
# person_id hhold_no med_card med_card_new
#1 1 1 1 1
#2 2 1 1 1
#3 3 1 NA 1
#4 4 1 NA 1
#5 5 1 NA 1
#6 6 2 0 0
#7 7 2 0 0
#8 8 2 0 0
#9 9 2 0 0
Please note that this will only work if not all values in a household are NA and the should not differ (e.g. person 1 == 1, person 2 == 0).
That is exactly what na.aggregate (link) in the zoo package does:
library(zoo)
transform(df, med_card_new = na.aggregate(med_card, by = hhold_no))
This uses mean; however, you can specify any function you like. For example, if you prefer to return an NA if all items in a group are NA (rather than NaN which is what mean would return if given a zero length vector) then
meanNA <- function(x, ...) if (all(is.na(x))) NA else mean(x, ...)
transform(df, med_card_new = na.aggregate(med_card, by = hhold_no, FUN = meanNA))
data.table solution
library(data.table)
setDT(df)[, med_card2 := unique(med_card[!is.na(med_card)]), by = hhold_no]
# person_id hhold_no med_card med_card2
# 1: 1 1 1 1
# 2: 2 1 1 1
# 3: 3 1 NA 1
# 4: 4 1 NA 1
# 5: 5 1 NA 1
# 6: 6 2 0 0
# 7: 7 2 0 0
# 8: 8 2 0 0
# 9: 9 2 0 0
# 10: 10 3 NA 1
# 11: 11 3 NA 1
# 12: 12 3 NA 1
# 13: 13 3 1 1
# 14: 14 3 1 1
# 15: 15 4 1 1
# 16: 16 4 1 1
# 17: 17 5 1 1
# 18: 18 5 1 1
# 19: 19 5 NA 1
# 20: 20 5 NA 1