I'm trying to fill in NA values with numbers that show exponential growth. Below is a data sample of what I'm trying to do.
library(tidyverse)
expand.grid(X2009H1N1 = "0-17 years",
type = "Cases",
month = seq(as.Date("2009-04-12") , to = as.Date("2010-03-12"), by = "month")) %>%
bind_cols( data.frame(
MidLevelRange = c(0,NA,NA,NA,NA,NA,8000000,16000000,18000000,19000000,19000000,19000000),
lowEst = c(0,NA,NA,NA,NA,NA,5000000,12000000,12000000,13000000,14000000,14000000)
))
I have used %>% arrange(month, X2009H1N1) %>%
group_by(X2009H1N1, type ) %>%
mutate(aprox_MidLevelRange = zoo::na.approx(MidLevelRange, na.rm = FALSE)) but the result does not look exponential to me. Thanks
Have a look at the imputeTS package. It offers plenty of imputation functions for time series. Take a look at this paper to get a good overview of all offered options
In your case using Stineman interpolation ( imputeTS::na_interpolation(x, option ="stine") could maybe be a suitable option.
Here for the example you provided:
x <- expand.grid(
X2009H1N1 = "0-17 years",
type = "Cases",
month = seq(as.Date("2009-04-12"),
to = as.Date("2010-03-12"),
by = "month"
)
) %>%
bind_cols(data.frame(
MidLevelRange = c(0, NA, NA, NA, NA, NA, 8000000, 16000000, 18000000, 19000000, 19000000, 19000000),
lowEst = c(0, NA, NA, NA, NA, NA, 5000000, 12000000, 12000000, 13000000, 14000000, 14000000)
))
x %>%
arrange(month, X2009H1N1) %>%
group_by(X2009H1N1, type) %>%
mutate(aprox_MidLevelRange = imputeTS::na_interpolation(MidLevelRange, option = "stine"))
This gives you:
# A tibble: 12 x 6
# Groups: X2009H1N1, type [1]
X2009H1N1 type month MidLevelRange lowEst aprox_MidLevelRange
<fct> <fct> <date> <dbl> <dbl> <dbl>
1 0-17 years Cases 2009-04-12 0 0 0
2 0-17 years Cases 2009-05-12 NA NA 593718.
3 0-17 years Cases 2009-06-12 NA NA 1335612.
4 0-17 years Cases 2009-07-12 NA NA 2289061.
5 0-17 years Cases 2009-08-12 NA NA 3559604.
6 0-17 years Cases 2009-09-12 NA NA 5336975.
7 0-17 years Cases 2009-10-12 8000000 5000000 8000000
8 0-17 years Cases 2009-11-12 16000000 12000000 16000000
9 0-17 years Cases 2009-12-12 18000000 12000000 18000000
10 0-17 years Cases 2010-01-12 19000000 13000000 19000000
11 0-17 years Cases 2010-02-12 19000000 14000000 19000000
12 0-17 years Cases 2010-03-12 19000000 14000000 19000000
So just comparing interpolation functions I guess this could be the best option.
Just plot yourself the different interpolation options, to see the differences. In general this are the interpolation options:
imputeTS::na_interpolation(x, option ="linear")
imputeTS::na_interpolation(x, option ="spline")
imputeTS::na_interpolation(x, option ="stine")
linear / spline options from imputeTS are the same as zoo::approx()/ zoo::spline(). stine does not exist in zoo.