How to fill NAs with LOCF by factors in data frame, split by country

Question

I have the following data frame (simplified) with the country variable as a factor and the value variable has missing values:

country value
AUT     NA
AUT            


        

Answer 1

I'm a little late to this conversation, but here is a data.table way, which will be much faster for larger data sets:

library(zoo)
library(data.table)

# Convert to data table
setDT(data)

data[, value := na.locf(value, na.rm = FALSE), by = country]

data
   country  value
1:     AUT     NA
2:     AUT      5
3:     AUT      5
4:     AUT      5
5:     GER     NA
6:     GER     NA
7:     GER      7
8:     GER      7
9:     GER      7

# And if you want to convert "data" back to a data frame...
setDF(data)

Answer 2

A combination of the packages dplyr and imputeTS can do the job.

library(dplyr)
library(imputeTS)
data %>% group_by(country) %>% 
mutate(value = na.locf(value, na.remaining="keep"))   

With the na.remaining parameter of the na.locf function of imputeTS you have additionally the option to choose, what to do with the trailing NAs.

These are the options:

• "keep" - return the series with NAs
• "rm" - remove remaining NAs
• "mean" - replace remaining NAs by overall mean
• "rev" - perform nocb / locf from the reverse direction

By choosing "mean" you would for example get a result with 7 for every GER in the specific example.

Answer 3

You simply need to split by country, then a do either a zoo::na.locf() or na.fill, filling to the right. Here is an example explicitly showing the three-component arg syntax of na.fill:

library(plyr)
library(zoo)

data <- data.frame(country=c("AUT", "AUT", "AUT", "AUT", "GER", "GER", "GER", "GER", "GER"), value=c(NA, 5, NA, NA, NA, NA, 7, NA, NA))

# The following is equivalent to na.locf
na.fill.right <- function(...) { na.fill(..., list(left=NA,interior=NA,right="extend")) }

ddply(data, .(country), na.fill.right)

  country value
1     AUT  <NA>
2     AUT     5
3     AUT     5
4     AUT     5
5     GER  <NA>
6     GER  <NA>
7     GER     7
8     GER     7
9     GER     7

Answer 4

If speed is a consideration then this unstack/stack solution is about 4 to 6 times faster than the others on my system although it does entail a slightly longer line of code:

stack(lapply(unstack(data, value ~ country), na.locf, na.rm = FALSE))

Another approach is:

transform(data, value = ave(value, country, FUN = na.locf0))

Answer 5

Here's a ddply solution. Try this

library(plyr)
ddply(DF, .(country), na.locf)
  country value
1     AUT  <NA>
2     AUT     5
3     AUT     5
4     AUT     5
5     GER  <NA>
6     GER  <NA>
7     GER     7
8     GER     7
9     GER     7

Edit From ddply help you can find that

.variables:  variables to split data frame by, 
as quoted variables, a formula or character vector.

so another alternatives to get what you want are:

ddply(DF, "country", na.locf)
ddply(DF, ~country, na.locf)

note that replacing .variables with DF$variable is not allowed, that's why you got an error when doing this.

DF is your data.frame

Answer 6

Split the data.frame with by and use na.locf on the subsets:

do.call(rbind,by(data,data$country,na.locf))

If you would like to remove the row names:

do.call(rbind,unname(by(data,data$country,na.locf)))