问题
In a data table I want to perform a forward and backward gap-filling procedure over a period of 3 days in both directions.
# Example data:
library(data.table)
library(zoo)
dt <- data.table(Value = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, 0.1359223, NA, NA, NA, NA, 0.0000000, 0.0000000, 0.0000000, 0.0000000, 0.0000000, NA))
> dt
Value
1: NA
2: NA
3: NA
4: NA
5: NA
6: NA
7: NA
8: NA
9: NA
10: 0.1359223
11: NA
12: NA
13: NA
14: NA
15: 0.0000000
16: 0.0000000
17: 0.0000000
18: 0.0000000
19: 0.0000000
20: NA
Therefore I want to create two new columns, one with forward replacement of the NAs and one with the backward replacement.
# desired output
Value forward backward
1: NA NA NA
2: NA NA NA
3: NA NA NA
4: NA NA NA
5: NA NA NA
6: NA NA NA
7: NA NA 0.1359223
8: NA NA 0.1359223
9: NA NA 0.1359223
10: 0.1359223 0.1359223 0.1359223
11: NA 0.1359223 NA
12: NA 0.1359223 0.0000000
13: NA 0.1359223 0.0000000
14: NA NA 0.0000000
15: 0.0000000 0.0000000 0.0000000
16: 0.0000000 0.0000000 0.0000000
17: 0.0000000 0.0000000 0.0000000
18: 0.0000000 0.0000000 0.0000000
19: 0.0000000 0.0000000 0.0000000
20: NA 0.0000000 NA
The forward replacement works fine with the following code:
dt$forward <- NA
r <- rle(is.na(dt$Value))
dt$forward <- na.locf(dt$Value, fromLast = F, na.rm = F)
is.na(dt$forward) <- sequence(r$lengths) > 3 & rep(r$values, r$lengths)
But I don´t know how to modify that code to do the backward replacement. How can I get around this? Thank you!
回答1:
Hacky, but why not just flip your column?
Code
# Using your result as basis
dt$Value <- rev(dt$Value)
dt$backward <- NA
r <- rle(is.na(dt$Value))
dt$backward <- na.locf(dt$Value, fromLast = F, na.rm = F)
is.na(dt$backward) <- sequence(r$lengths) > 3 & rep(r$values, r$lengths)
dt$Value <- rev(dt$Value)
dt$backward <- rev(dt$backward)
Result
> dt
Value forward backward
1: NA NA NA
2: NA NA NA
3: NA NA NA
4: NA NA NA
5: NA NA NA
6: NA NA NA
7: NA NA 0.1359223
8: NA NA 0.1359223
9: NA NA 0.1359223
10: 0.1359223 0.1359223 0.1359223
11: NA 0.1359223 NA
12: NA 0.1359223 0.0000000
13: NA 0.1359223 0.0000000
14: NA NA 0.0000000
15: 0.0000000 0.0000000 0.0000000
16: 0.0000000 0.0000000 0.0000000
17: 0.0000000 0.0000000 0.0000000
18: 0.0000000 0.0000000 0.0000000
19: 0.0000000 0.0000000 0.0000000
20: NA 0.0000000 NA
来源:https://stackoverflow.com/questions/53231287/backward-replacement-of-nas-in-time-series-only-to-a-limited-number-of-observati