我想在不带组的data.table中填充一些NA值。请考虑以下data.table提取物，该数据代表时间和距离：
library(data.table)
df <- data.frame(time = seq(7173, 7195, 1), dist = c(31091.33, NA, 31100.00, 31103.27, NA, NA, NA, NA, 31124.98, NA,31132.81, NA, NA, NA, NA, 31154.19, NA, 31161.47, NA, NA, NA, NA, 31182.97))
DT<- data.table(df)
我想要在DT data.table中，使用取决于之前/之后的非NA值的函数来填充NA值。例如，在j中编写一个函数来替换每条指令
DT[2, dist := (31091.33 + (31100-31091.33) / 2)]
然后
DT[5:8, dist := (31103.27 + "something" * (31124.98 - 31103.27) / 5)]
等等...
最佳答案
内联代码说明。例如，您可以使用df[,dist_before := NULL]删除临时列。
library(data.table)
df=data.table(time=seq(7173,7195,1),dist=c(31091.33,NA,31100.00,31103.27,NA,NA,NA,
NA,31124.98,NA,31132.81,NA,NA,NA,NA,31154.19,NA,31161.47,NA,NA,NA,NA,31182.97))
df
#> time dist
#> 1: 7173 31091.33
#> 2: 7174 NA
#> 3: 7175 31100.00
#> 4: 7176 31103.27
#> 5: 7177 NA
#> 6: 7178 NA
#> 7: 7179 NA
#> 8: 7180 NA
#> 9: 7181 31124.98
#> 10: 7182 NA
#> 11: 7183 31132.81
#> 12: 7184 NA
#> 13: 7185 NA
#> 14: 7186 NA
#> 15: 7187 NA
#> 16: 7188 31154.19
#> 17: 7189 NA
#> 18: 7190 31161.47
#> 19: 7191 NA
#> 20: 7192 NA
#> 21: 7193 NA
#> 22: 7194 NA
#> 23: 7195 31182.97
#> time dist
# Carry forward the last non-missing observation
df[,dist_before := nafill(dist, "locf")]
# Bring back the next non-missing dist
df[,dist_after := nafill(dist, "nocb")]
# rleid will create groups based on run-lengths of values within the data.
# This means 4 NA's in a row will be grouped together, for example.
# We then count the missings and add 1, because we want the
# last NA before the next non-missing to be less than the non-missing value.
df[, rle := rleid(dist)][,missings := max(.N + 1 , 2), by = rle][]
#> time dist dist_before dist_after rle missings
#> 1: 7173 31091.33 31091.33 31091.33 1 2
#> 2: 7174 NA 31091.33 31100.00 2 2
#> 3: 7175 31100.00 31100.00 31100.00 3 2
#> 4: 7176 31103.27 31103.27 31103.27 4 2
#> 5: 7177 NA 31103.27 31124.98 5 5
#> 6: 7178 NA 31103.27 31124.98 5 5
#> 7: 7179 NA 31103.27 31124.98 5 5
#> 8: 7180 NA 31103.27 31124.98 5 5
#> 9: 7181 31124.98 31124.98 31124.98 6 2
#> 10: 7182 NA 31124.98 31132.81 7 2
#> 11: 7183 31132.81 31132.81 31132.81 8 2
#> 12: 7184 NA 31132.81 31154.19 9 5
#> 13: 7185 NA 31132.81 31154.19 9 5
#> 14: 7186 NA 31132.81 31154.19 9 5
#> 15: 7187 NA 31132.81 31154.19 9 5
#> 16: 7188 31154.19 31154.19 31154.19 10 2
#> 17: 7189 NA 31154.19 31161.47 11 2
#> 18: 7190 31161.47 31161.47 31161.47 12 2
#> 19: 7191 NA 31161.47 31182.97 13 5
#> 20: 7192 NA 31161.47 31182.97 13 5
#> 21: 7193 NA 31161.47 31182.97 13 5
#> 22: 7194 NA 31161.47 31182.97 13 5
#> 23: 7195 31182.97 31182.97 31182.97 14 2
#> time dist dist_before dist_after rle missings
# .SD[,.I] will get us the row number relative to the group it is in.
# For example, row 5 dist is calculated as
# dist_before + 1 * (dist_after - dist_before)/5
df[is.na(dist), dist := dist_before + .SD[,.I] *
(dist_after - dist_before)/(missings), by = rle]
df[]
#> time dist dist_before dist_after rle missings
#> 1: 7173 31091.33 31091.33 31091.33 1 2
#> 2: 7174 31095.67 31091.33 31100.00 2 2
#> 3: 7175 31100.00 31100.00 31100.00 3 2
#> 4: 7176 31103.27 31103.27 31103.27 4 2
#> 5: 7177 31107.61 31103.27 31124.98 5 5
#> 6: 7178 31111.95 31103.27 31124.98 5 5
#> 7: 7179 31116.30 31103.27 31124.98 5 5
#> 8: 7180 31120.64 31103.27 31124.98 5 5
#> 9: 7181 31124.98 31124.98 31124.98 6 2
#> 10: 7182 31128.90 31124.98 31132.81 7 2
#> 11: 7183 31132.81 31132.81 31132.81 8 2
#> 12: 7184 31137.09 31132.81 31154.19 9 5
#> 13: 7185 31141.36 31132.81 31154.19 9 5
#> 14: 7186 31145.64 31132.81 31154.19 9 5
#> 15: 7187 31149.91 31132.81 31154.19 9 5
#> 16: 7188 31154.19 31154.19 31154.19 10 2
#> 17: 7189 31157.83 31154.19 31161.47 11 2
#> 18: 7190 31161.47 31161.47 31161.47 12 2
#> 19: 7191 31165.77 31161.47 31182.97 13 5
#> 20: 7192 31170.07 31161.47 31182.97 13 5
#> 21: 7193 31174.37 31161.47 31182.97 13 5
#> 22: 7194 31178.67 31161.47 31182.97 13 5
#> 23: 7195 31182.97 31182.97 31182.97 14 2
#> time dist dist_before dist_after rle missings
关于r - 不带组的data.table线性内插NA值，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/58901584/