用最新的非NA值替换NA
在data.frame（或data.table）中，我想用最近的非NA值“填充”NA。一个简单的例子，使用向量（而不是a data.frame）如下：
> y <- c(NA, 2, 2, NA, NA, 3, NA, 4, NA, NA)
我想要一个fill.NAs()允许我构造的函数yy：
> yy
[1] NA NA NA 2 2 2 2 3 3 3 4 4
我需要对许多（总计~1 Tb）小尺寸data.frames（~30-50 Mb）重复此操作，其中一行是NA，其所有条目都是。解决问题的好方法是什么？
我做的丑陋的解决方案使用这个功能：
last <- function (x){
x[length(x)]
}
fill.NAs <- function(isNA){
if (isNA[1] == 1) {
isNA[1:max({which(isNA==0)[1]-1},1)] <- 0 # first is NAs
# can't be forward filled
}
isNA.neg <- isNA.pos <- isNA.diff <- diff(isNA)
isNA.pos[isNA.diff < 0] <- 0
isNA.neg[isNA.diff > 0] <- 0
which.isNA.neg <- which(as.logical(isNA.neg))
if (length(which.isNA.neg)==0) return(NULL) # generates warnings later, but works
which.isNA.pos <- which(as.logical(isNA.pos))
which.isNA <- which(as.logical(isNA))
if (length(which.isNA.neg)==length(which.isNA.pos)){
replacement <- rep(which.isNA.pos[2:length(which.isNA.neg)],
which.isNA.neg[2:max(length(which.isNA.neg)-1,2)] -
which.isNA.pos[1:max(length(which.isNA.neg)-1,1)])
replacement <- c(replacement, rep(last(which.isNA.pos), last(which.isNA) - last(which.isNA.pos)))
} else {
replacement <- rep(which.isNA.pos[1:length(which.isNA.neg)], which.isNA.neg - which.isNA.pos[1:length(which.isNA.neg)])
replacement <- c(replacement, rep(last(which.isNA.pos), last(which.isNA) - last(which.isNA.pos)))
}
replacement
}
该功能fill.NAs使用如下：
y <- c(NA, 2, 2, NA, NA, 3, NA, 4, NA, NA)
isNA <- as.numeric(is.na(y))
replacement <- fill.NAs(isNA)
if (length(replacement)){
which.isNA <- which(as.logical(isNA))
to.replace <- which.isNA[which(isNA==0)[1]:length(which.isNA)]
y[to.replace] <- y[replacement]
}
产量
> y
[1] NA 2 2 2 2 3 3 3 4 4 4
......似乎有效。但是，伙计，这太丑了！有什么建议？