How to replace NA with most recent non-NA by group? [duplicate]

Answer 1

Depending upon what you are doing next, you may prefer the data in a nested form.

(nested <- df %>% 
  group_by(name) %>% 
  summarize(
    age = na.omit(age)[1], 
    birthplace = na.omit(birthplace)[1], 
    value = list(value)
  )
)
## # A tibble: 4 x 4
##     name   age birthplace     value
##   <fctr> <dbl>     <fctr>    <list>
## 1      A    28      city1 <int [2]>
## 2      B    NA      city2 <int [3]>
## 3      C    NA         NA <int [1]>
## 4      D    53         NA <int [2]>

If you need to compute on individual values, you can always unnest it later.

nested %>% tidyr::unnest()
## # A tibble: 8 x 4
##     name   age birthplace value
##   <fctr> <dbl>     <fctr> <int>
## 1      A    28      city1   100
## 2      A    28      city1   101
## 3      B    NA      city2   102
## 4      B    NA      city2   103
## 5      B    NA      city2   104
## 6      C    NA         NA   105
## 7      D    53         NA   106
## 8      D    53         NA   107

Answer 2

You can wrap the na.locf in do

df %>% group_by(name) %>% do(na.locf(., na.rm = FALSE))

Answer 3

Here is a base R solution. The fill function invokes ave using na.omit(x)[1] as in Richie Cotton's solution.

fill <- function(...) ave(..., FUN = function(x) na.omit(x)[1])
transform(df, birthplace = fill(birthplace, name), age = fill(age, name))

Note: This also works with na.locf. Replace fill with:

library(zoo)
fill <- function(...) ave(..., FUN = function(x) na.locf(x, na.rm = FALSE))

Answer 4

As another base R solution, here is a poor man's na.locf

fill_down <- function(v) {
    if (length(v) > 1) {
        keep <- c(TRUE, !is.na(v[-1]))
        v[keep][cumsum(keep)]
    } else v
}

To fill down by group, the approach is to use tapply() to split and apply to each group, and split<- to combine groups to the original geometry, as

fill_down_by_group <- function(v, grp) {
    ## original 'by hand':
    ##     split(v, grp) <- tapply(v, grp, fill_down)
    ##     v
    ## done by built-in function `ave()`
    ave(v, grp, FUN=fill_down)
}

To process multiple columns, one might

elts <- c("age", "birthplace")
df[elts] <- lapply(df[elts], fill_down_by_group, df$name)

Notes

1. I would be interested in seeing how a dplyr solution handles many columns, without hard-coding each? Answering my own question, I guess this is

   library(dplyr); library(tidyr)
   df %>% group_by(name) %>% fill_(elts)
   

2. A more efficient base solution when the groups are already 'grouped' (e.g., identical(grp, sort(grp))) is

   fill_down_by_grouped <- function(v, grp) {
       if (length(v) > 1) {
           keep <- !(duplicated(v) & is.na(v))
           v[keep][cumsum(keep)]
       } else v
   }
   

3. For me, fill_down() on a vector with about 10M elements takes ~225ms; fill_down_by_grouped() takes ~300ms independent of the number of groups; fill_down_by_group() scales with the number of groups; for 10000 groups ~2s, 10M groups about 36s

Answer 5

Consider also a nested apply base solution running a rolling head() for each column:

df <- setNames(data.frame(lapply(names(df), function(d)
               sapply(1:nrow(df), function(i)
                      head(df[df[1:i, c("name")] == df$name[i], c(d)], 1))
        )), names(df))

Answer 6

You could this through a merge too. Just do a join on name column. Then do a group by on value.

library(sqldf)
sqldf('select t1.name, t2.age, t2.birthplace,t1.value from df t1 inner join df t2 on t1.name=t2.name group by t1.value')