Fill NAs with either last or next non NA value in R

user_13311198 0
Viewed 667

I am trying to fill NA values in a column with other non-NA values within the same group in R. So my data looks something like this:

df
       id year pop
1  E1 2000  NA
2  E2 2000  NA
3  E2 2001  NA
4  E2 2003 120
5  E2 2005 125
6  E3 1999 115
7  E3 2001 300
8  E3 2003  NA
9  E4 2004  10
10 E4 2005  NA
11 E4 2008  NA
12 E4 2009   9
13 E5 2002  12
14 E5 2003  80

And I want NA values to have either the last non-NA or the next non-NA value of pop within the same group of id. To look something like this:

    df.desired
   id year pop
1  E1 2000  NA
2  E2 2000 120
3  E2 2001 120
4  E2 2003 120
5  E2 2005 125
6  E3 1999 115
7  E3 2001 300
8  E3 2003 300
9  E4 2004  10
10 E4 2005  10
11 E4 2008   9
12 E4 2009   9
13 E5 2002  12
14 E5 2003  80

I tried different things with both zoo::na.locf() and dplyr::fill() but I keep having two main issues: 1. I get errors because entire groups only have NA (like id == "E1" here) and 2. I can only choose either the last or the naxt non-NA value. These are some examples of what I've tried:

library(tidyverse)
library(zoo)
    df.desired <- df %>%
group_by(id) %>%
      arrange(year)%>%
      mutate(pop_imputated = pop)%>%
      fill(pop_imputated)%>%
      ungroup()


df.desired <- df %>%
  group_by(id) %>%
  arrange(year)%>%
  mutate(pop_imputated = zoo::na.locf(pop))%>%
  fill(pop_imputated)%>%
  ungroup()

Any ideas? Thanks a lot!

dplyr tidyverse na zoo r
6 Answers

Here is an answer that would match your expected output exactly: it will impute to the nearest non-missing value, either upward or downward.

Here is the code, using a spiced up version of your example:

library(tidyverse)
df = structure(list(id = c("E1", "E2", "E2", "E2", "E2", "E3", "E3", "E3", "E4", "E4", "E4", "E4", "E4", "E4", "E4", "E4", "E5", "E5"), 
                    year = c(2000L, 2000L, 2001L, 2003L, 2005L, 1999L, 2001L, 2003L, 2004L, 2005L, 2006L, 2007L, 2008L, 2009L, 2018L, 2019L, 2002L, 2003L), 
                    pop = c(NA, NA, NA, 120L, 125L, 115L, 300L, NA, 10L, NA, NA, NA, NA, 9L, NA, 8L, 12L, 80L), 
                    pop_exp = c(NA, 120L, 120L, 120L, 125L, 115L, 300L, 300L, 10L, 10L, 10L, 9L, 9L, 9L, 9L, 8L, 12L, 80L)), 
               class = "data.frame", row.names = c(NA, -18L))

fill_nearest = function(x){
  keys=which(!is.na(x))
  if(length(keys)==0) return(NA)
  b = map_dbl(seq.int(x), ~keys[which.min(abs(.x-keys))])
  x[b]
}

df %>% 
  group_by(id) %>% 
  arrange(id, year) %>%
  mutate(pop_imputated = fill_nearest(pop)) %>% 
  ungroup()
#> # A tibble: 18 x 5
#>    id     year   pop pop_exp pop_imputated
#>    <chr> <int> <int>   <int>         <int>
#>  1 E1     2000    NA      NA            NA
#>  2 E2     2000    NA     120           120
#>  3 E2     2001    NA     120           120
#>  4 E2     2003   120     120           120
#>  5 E2     2005   125     125           125
#>  6 E3     1999   115     115           115
#>  7 E3     2001   300     300           300
#>  8 E3     2003    NA     300           300
#>  9 E4     2004    10      10            10
#> 10 E4     2005    NA      10            10
#> 11 E4     2006    NA      10            10
#> 12 E4     2007    NA       9             9
#> 13 E4     2008    NA       9             9
#> 14 E4     2009     9       9             9
#> 15 E4     2018    NA       9             9
#> 16 E4     2019     8       8             8
#> 17 E5     2002    12      12            12
#> 18 E5     2003    80      80            80

Created on 2021-05-13 by the reprex package (v2.0.0)

As I had to use a purrr loop, it might get a bit slow in a huge dataset though.

EDIT: I suggested to add this option in tidyr::fill(): https://github.com/tidyverse/tidyr/issues/1119. The issue also contains a tweaked version of this function to use the year column as the reference to calculate the "distance" between the values. For instance, you would rather have row 15 as 8 than as 9 because the year is much closer.

user_38880000

Have you tried to change the .direction attribute of the tidyr::fill function? You can use "downup" (first downwards, then upwards), or viceversa "updown"

library(dplyr)
library(tidyr)

df %>%
  group_by(id) %>%
  mutate(pop_imputated = pop) %>%
  fill(pop_imputated, .direction = "downup") %>%
  ungroup()

# A tibble: 14 x 4
   id     year   pop pop_imputated
   <chr> <int> <int>         <int>
 1 E1     2000    NA            NA
 2 E2     2000    NA           120
 3 E2     2001    NA           120
 4 E2     2003   120           120
 5 E2     2005   125           125
 6 E3     1999   115           115
 7 E3     2001   300           300
 8 E3     2003    NA           300
 9 E4     2004    10            10
10 E4     2005    NA            10
11 E4     2008    NA            10
12 E4     2009     9             9
13 E5     2002    12            12
14 E5     2003    80            80

It looks similar to your desired output

user_74654620

library runner has a built in function fill_run which may also be used

df %>% 
  group_by(id) %>%
  mutate(pop = runner::fill_run(pop, run_for_first = T))
#> Warning in runner::fill_run(pop, run_for_first = T): All x values are NA
#> # A tibble: 14 x 3
#> # Groups:   id [5]
#>    id     year   pop
#>    <chr> <int> <int>
#>  1 E1     2000    NA
#>  2 E2     2000   120
#>  3 E2     2001   120
#>  4 E2     2003   120
#>  5 E2     2005   125
#>  6 E3     1999   115
#>  7 E3     2001   300
#>  8 E3     2003   300
#>  9 E4     2004    10
#> 10 E4     2005    10
#> 11 E4     2008    10
#> 12 E4     2009     9
#> 13 E5     2002    12
#> 14 E5     2003    80

Created on 2021-05-13 by the reprex package (v2.0.0)

user_28848590

I hope this is what you were looking for. I filled all NA values with the last non NA values in a group.

library(dplyr)

df %>%
  group_by(id) %>%
  mutate(across(pop, ~ coalesce(.x, last(.x[!is.na(.x)]))))


# A tibble: 14 x 3
# Groups:   id [5]
   id     year   pop
   <chr> <int> <int>
 1 E1     2000    NA
 2 E2     2000   125
 3 E2     2001   125
 4 E2     2003   120
 5 E2     2005   125
 6 E3     1999   115
 7 E3     2001   300
 8 E3     2003   300
 9 E4     2004    10
10 E4     2005     9
11 E4     2008     9
12 E4     2009     9
13 E5     2002    12
14 E5     2003    80
user_143145200

Another solution uses nalocf(NA last observation carried forward); as it operates top-down, we first need to re-arrange the dataframe so the first popvalue is non-NA:

library(zoo)
df %>%
  arrange(desc(id)) %>%
  mutate(pop = na.locf(pop))
   id year pop
13 E5 2002  12
14 E5 2003  80
9  E4 2004  10
10 E4 2005  10
11 E4 2008  10
12 E4 2009   9
6  E3 1999 115
7  E3 2001 300
8  E3 2003 300
2  E2 2000 300
3  E2 2001 300
4  E2 2003 120
5  E2 2005 125
1  E1 2000 125

We can of course reinstitute the original order:

library(zoo)
df %>%
  arrange(desc(id)) %>%
  mutate(pop = na.locf(pop)) %>%
  arrange(id)
user_80399780

Since na.approx accepts approx arguments (see ?approx and ?na.approx) we can use na.approx with method = "constant" and rule = 2. Also sort the data back to the original order if you want exactly the output shown in the question.

library(dplyr)
library(zoo)

df %>%
  group_by(id) %>%
  arrange(year)%>%
  mutate(pop_imputated = na.approx(pop, method = "const", rule = 2, na.rm = FALSE)) %>%
  ungroup() %>%
  arrange(id, year)

giving:

# A tibble: 14 x 4
   id     year   pop pop_imputated
   <chr> <int> <int>         <dbl>
 1 E1     2000    NA            NA
 2 E2     2000    NA           120
 3 E2     2001    NA           120
 4 E2     2003   120           120
 5 E2     2005   125           125
 6 E3     1999   115           115
 7 E3     2001   300           300
 8 E3     2003    NA           300
 9 E4     2004    10            10
10 E4     2005    NA            10
11 E4     2008    NA            10
12 E4     2009     9             9
13 E5     2002    12            12
14 E5     2003    80            80

Note

Lines <- "       id year pop
1  E1 2000  NA
2  E2 2000  NA
3  E2 2001  NA
4  E2 2003 120
5  E2 2005 125
6  E3 1999 115
7  E3 2001 300
8  E3 2003  NA
9  E4 2004  10
10 E4 2005  NA
11 E4 2008  NA
12 E4 2009   9
13 E5 2002  12
14 E5 2003  80"
df <- read.table(text = Lines)
user_5165480
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