How could I Replace a NA with mean of its previous and next rows in a fast manner?
name grade
1 A 56
2 B NA
3 C 70
4 D 96
>
Or you may try na.approx from package zoo: "Missing values (NAs) are replaced by linear interpolation"
library(zoo)
x <- c(56, NA, 70, 96)
na.approx(x)
# [1] 56 63 70 96
This also works if you have more than one consecutive NA:
vals <- c(1, NA, NA, 7, NA, 10)
na.approx(vals)
# [1] 1.0 3.0 5.0 7.0 8.5 10.0
na.approx is based on the base function approx, which may be used instead:
vals <- c(1, NA, NA, 7, NA, 10)
xout <- seq_along(vals)
x <- xout[!is.na(vals)]
y <- vals[!is.na(vals)]
approx(x = x, y = y, xout = xout)$y
# [1] 1.0 3.0 5.0 7.0 8.5 10.0