Fill a column's blank spaces contingent on a second column in R -

i'd appreciate one. have similar data below.

df$a df$b 1    . 1    . 1    . 1    6 2    . 2    . 2    7 

what need fill in df$b each value corresponds end of run of values in df$a. example below.

df$a df$b 1    6 1    6 1    6 1    6 2    7 2    7 2    7 

any welcome.

it seems me missing values denoted .. better read dataset na.strings="." missing values na. current dataset, 'b' column character/factor class (depending upon whether used stringsasfactors=false/true (default) in read.table/read.csv.

using data.table, convert data.frame data.table (setdt(df1)), change 'character' class 'numeric' (b:= as.numeric(b)). result in coercing . na (a warning appear). grouped "a", change "b" values last element (b:= b[.n])

library(data.table) setdt(df1)[,b:= as.numeric(b)][,b:=b[.n] , = a] #   b #1: 1 6 #2: 1 6 #3: 1 6 #4: 1 6 #5: 2 7 #6: 2 7 #7: 2 7 

or dplyr

library(dplyr) df1 %>%      group_by(a) %>%      mutate(b= as.numeric(tail(b,1))) 

or using ave base r

df1$b <- with(df1, as.numeric(ave(b, a, fun=function(x) tail(x,1)))) 

data

df1 <- structure(list(a = c(1l, 1l, 1l, 1l, 2l, 2l, 2l), b = c(".",  ".", ".", "6", ".", ".", "7")), .names = c("a", "b"),  class = "data.frame", row.names = c(na, -7l))