在data.frame（或data.table）中，我想用最接近的先前非 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yy这样的函数：

> yy
[1] NA NA NA  2  2  2  2  3  3  3  4  4

我需要对许多（总共约 1 Tb）小型data.frames（约 30-50 Mb）重复此操作，其中一行是 NA 是它的所有条目。解决问题的好方法是什么？

我制作的丑陋解决方案使用此功能：

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
}

该函数fill.NAs的使用如下：

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]
}

输出

> y
[1] NA  2  2  2  2  3  3  3  4  4  4

…这似乎工作。但是，伙计，它丑吗！有什么建议么？