在列末尾使用na.locf函数的NA

Question

我使用zoo包和na.locf将列中最后出现的值替换为NA。我的问题是，如果最后一行没有数字，函数就会失败。请看我的例子。

> tail(Dax1Dax2Dax3)
          Time    Res1    Accum1    Res2    Accum2    Res3    Accum3
2   2017-10-18 -109.41 171103.61    <NA>      <NA>    <NA>      <NA>
1   2017-10-20 1174.60 172278.21    <NA>      <NA>    <NA>      <NA>
406 2017-10-23    <NA>      <NA> 1473.25 185076.53    <NA>      <NA>
664 2017-10-25    <NA>      <NA>    <NA>      <NA>  -80.19 188974.10
663 2017-10-26    <NA>      <NA>    <NA>      <NA>  398.16 189372.27
662 2017-11-07    <NA>      <NA>    <NA>      <NA> -348.37 189023.90

require(zoo)

Dax1Dax2Dax3$Accum1 <- na.locf(Dax1Dax2Dax3$Accum1, fromLast = TRUE)
Error in `$<-.data.frame`(`*tmp*`, Accum1, value = c(253L, 253L, 253L,  : 
  replacement has 913 rows, data has 917

> dim(Dax1Dax2Dax3)
[1] 917   7

Answer 1

# 1
Dax1Dax2Dax3$Accum1 <- na.locf(Dax1Dax2Dax3$Accum1, fromLast = TRUE, na.rm = FALSE)

# 2
Dax1Dax2Dax3$Accum1 <- na.locf0(Dax1Dax2Dax3$Accum1, fromLast = TRUE)

# 3
Dax1Dax2Dax3$Accum1 <- na.locf(Dax1Dax2Dax3$Accum1, fromLast = TRUE, rule = 1)

有关详细信息，请参阅?na.locf。

注意：可重现形式的输入为：

Lines <- "
          Time    Res1    Accum1    Res2    Accum2    Res3    Accum3
2   2017-10-18 -109.41 171103.61    <NA>      <NA>    <NA>      <NA>
1   2017-10-20 1174.60 172278.21    <NA>      <NA>    <NA>      <NA>
406 2017-10-23    <NA>      <NA> 1473.25 185076.53    <NA>      <NA>
664 2017-10-25    <NA>      <NA>    <NA>      <NA>  -80.19 188974.10
663 2017-10-26    <NA>      <NA>    <NA>      <NA>  398.16 189372.27
662 2017-11-07    <NA>      <NA>    <NA>      <NA> -348.37 189023.90"

Dax1Dax2Dax3 <- read.table(text = Lines, na.strings = "<NA>")

Answer 2

我们可以使用tidyr包中的fill函数将最后一个值归结为NA。

library(tidyr)
Dax1Dax2Dax3 %>% fill(Res1, Accum1, Res2, Accum2)
#           Time    Res1   Accum1    Res2   Accum2    Res3   Accum3
# 2   2017-10-18 -109.41 171103.6      NA       NA      NA       NA
# 1   2017-10-20 1174.60 172278.2      NA       NA      NA       NA
# 406 2017-10-23 1174.60 172278.2 1473.25 185076.5      NA       NA
# 664 2017-10-25 1174.60 172278.2 1473.25 185076.5  -80.19 188974.1
# 663 2017-10-26 1174.60 172278.2 1473.25 185076.5  398.16 189372.3
# 662 2017-11-07 1174.60 172278.2 1473.25 185076.5 -348.37 189023.9

Answer 3

此外，使用imputeTS包的na.locf函数也是一种选择。它有一个方便的na.remaining参数，可以让您选择如何处理剩余的NAs。这对像这样的情况很有帮助。

library("imputeTS")
Dax1Dax2Dax3$Accum1 <- na.locf(Dax1Dax2Dax3$Accum1, na.remaining = "rev")

以下是na.remaining参数的选项：

• “保留”-使用NAs
• "rm“返回系列-删除剩余的NAs
• ”

“-用总体平均值

• ”rev“替换剩余的NAs-从相反方向执行nocb / locf