我有纵向数据结构(在 R 中),因为:
dat1 <- data.frame(county = c("a","a","a", "b","b","b", "c","c","c"), year = c(2001, 2002, 2003, 2001, 2002, 2003, 2001, 2002, 2003), count = c(2, NA, 6, 4, NA, 8, 10, NA, 14))
我有一个错过的一年,我想插入 2002 。我做了一个功能:
dat1[dat1$year == 2002,] <- apply(X = data.frame("2001" = dat1[dat1$year == 2001, "count"], "2003" = dat1[dat1$year == 2003, "count"]), MARGIN = 1, FUN = function(x){z <- data.frame(approx(x = x, method = "linear", n = length(2001:2003))$y); return(z[-c(1, nrow(z)),])})
dat1 # the desired result
但只有在我用于插值的两年内没有任何NA
时,它才有效:
dat2 <- dat1
dat2[9,3] <- NA
apply(X = data.frame("2001" = dat2[dat2$year == 2001, "count"], "2003" = dat2[dat2$year == 2003, "count"]), MARGIN = 1, FUN = function(x){z <- data.frame(approx(x = x, method = "linear", n = length(2001:2003))$y); return(z[-c(1, nrow(z)),])})
我知道我可以使用complete.cases()
使其运行但我无法将其插回到原始数据框中,因为行长度会有所不同。
在上一年或下一年的问题中,是否有人能够解决我的NA
问题或更好地解决我的kludge-tastic尝试?
答案 0 :(得分:2)
如果只有NA,则返回NA,如果只有一个非NA,则使用na.aggregate
将单个非NA替换为每个NA,如果一个NA中有多个NA,则使用na.approx
县。
library(zoo)
na_approx <- function(x) {
if (all(is.na(x))) NA
else if (sum(!is.na(x)) == 1) na.aggregate(x)
else na.approx(x, na.rm = FALSE)
}
transform(dat2, count = ave(count, county, FUN = na_approx))
,并提供:
county year count
1 a 2001 2
2 a 2002 4
3 a 2003 6
4 b 2001 4
5 b 2002 6
6 b 2003 8
7 c 2001 10
8 c 2002 10
9 c 2003 10
这是一种可以略微不同的行为,尽管在这个例子中它给出了相同的结果。
na_approx2 <- function(x) {
if (sum(!is.na(x)) > 1) na.approx(x, na.rm = FALSE)
else na.locf(x, na.rm = FALSE)
}
transform(dat2, count = ave(count, county, FUN = na_approx2))