I would like to add all missing dates between min and max date in a data.frame
and linear interpolate all missing values, like
df <- data.fra
Another nice and short solution (using imputeTS):
library(imputeTS)
x <- zoo(df$value,df$date)
x <- na.interpolation(x, option = "linear")
print(x)
Here are a few solutions.
1) zoo Convert the data frame to a zoo series and use na.approx
with an xout=
of sequential dates to get the final series
library(zoo)
z <- read.zoo(mydf)
zz <- na.approx(z, xout = seq(start(z), end(z), "day"))
giving:
> zz
2015-10-05 2015-10-06 2015-10-07 2015-10-08 2015-10-09 2015-10-10 2015-10-11
8.000000 6.333333 4.666667 3.000000 9.000000 8.200000 7.400000
2015-10-12 2015-10-13 2015-10-14
6.600000 5.800000 5.000000
It may be more convenient to leave it in zoo form so you can use all the facilities of zoo but if you need it in data frame form just use
DF <- fortify.zoo(zz)
1a) zoo/magrittr The above could alternately be expressed as a magrittr pipeline:
library(magrittr)
df %>% read.zoo %>% na.approx(xout = seq(start(.), end(.), "day")) %>% fortify.zoo
(or omit the fortify.zoo
part if you want zoo output).
2) base R We can essentially do the same thing without packages like this:
n <- nrow(mydf)
with(mydf, data.frame(approx(date, value, xout = seq(date[1], date[n], "day"))))
Here is one way. I created a data frame with a sequence of date using the first and last date. Using full_join()
in the dplyr
package, I merged the data frame and mydf
. I then used na.approx()
in the zoo package to handle the interpolation in the mutate()
part.
mydf <- data.frame(date = as.Date(c("2015-10-05","2015-10-08","2015-10-09",
"2015-10-12","2015-10-14")),
value = c(8,3,9,NA,5))
library(dplyr)
library(zoo)
data.frame(date = seq(mydf$date[1], mydf$date[nrow(mydf)], by = 1)) %>%
full_join(mydf, by = "date") %>%
mutate(approx = na.approx(value))
# date value approx
#1 2015-10-05 8 8.000000
#2 2015-10-06 NA 6.333333
#3 2015-10-07 NA 4.666667
#4 2015-10-08 3 3.000000
#5 2015-10-09 9 9.000000
#6 2015-10-10 NA 8.200000
#7 2015-10-11 NA 7.400000
#8 2015-10-12 NA 6.600000
#9 2015-10-13 NA 5.800000
#10 2015-10-14 5 5.000000
I think your code would look much clear and simple if you use Forecast package.
library(forecast)
x <- zoo(df$value,df$date)
x <- as.ts(x)
x <- na.interp(x)
print(x)