本文介绍了用R中的上一行和下一行平均值替换NA的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
如何快速用前一行和下一行的平均值替换 NA?
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
这样 B 的成绩就是 63.
such that B's grade would be 63.
推荐答案
或者您可以尝试 na.approx 包中的 zoo:替换缺失值 (NAs)通过线性插值"
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
如果您有多个连续的NA，这也适用:
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 是基于 base 函数 approx，可以替代使用:
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
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