How to replace NA with set of values

Question

I have the following data frame:

library(dplyr)
library(tibble)


df <- tibble(
  source = c("a", "b", "c", "d", "e"),
  score = c(10, 5, NA, 3, NA ) ) 


df

It looks like this:

# A tibble: 5 x 2
  source score
  <chr>  <dbl>
1 a         10 . # current max value
2 b          5
3 c         NA
4 d          3
5 e         NA

What I want to do is to replace NA in score column with values ranging for existing max + n onwards. Where n range from 1 to total number of rows of the df

Resulting in this (hand-coded) :

  source score
  a         10
  b          5
  c         11 # obtained from 10 + 1
  d          3
  e         12 #  obtained from 10 + 2

How can I achieve that?

score 10 · Answer 1 · answered Feb 14 '20 at 07:53

Another option :

transform(df, score = pmin(max(score, na.rm = TRUE) + 
                      cumsum(is.na(score)), score, na.rm = TRUE))

#  source score
#1      a    10
#2      b     5
#3      c    11
#4      d     3
#5      e    12

If you want to do this in dplyr

library(dplyr)
df %>% mutate(score = pmin(max(score, na.rm = TRUE) + 
                      cumsum(is.na(score)), score, na.rm = TRUE))

score 6 · Answer 2 · answered Feb 14 '20 at 07:40

6

A base R solution

df$score[is.na(df$score)] <- seq(which(is.na(df$score))) + max(df$score,na.rm = TRUE)

such that

> df
# A tibble: 5 x 2
  source score
  <chr>  <dbl>
1 a         10
2 b          5
3 c         11
4 d          3
5 e         12

answered Feb 14 '20 at 07:40

ThomasIsCoding

96,636
9
24
81

Already the most concise but could shorten `seq(which(is.na(df$score)))` to `1:sum(is.na(df$score))` – s_baldur Feb 14 '20 at 09:17
@sindri_baldur thanks. that one has been offered by https://stackoverflow.com/a/60222864/12158757 – ThomasIsCoding Feb 14 '20 at 09:23

Sotos · Answer 3 · 2020-02-14T08:55:57.960

6

Here is a dplyr approach,

df %>% 
 mutate(score = replace(score, 
                       is.na(score), 
                       (max(score, na.rm = TRUE) + (cumsum(is.na(score))))[is.na(score)])
                       )

which gives,

# A tibble: 5 x 2
  source score
  <chr>  <dbl>
1 a         10
2 b          5
3 c         11
4 d          3
5 e         12

edited Feb 14 '20 at 08:55

answered Feb 14 '20 at 07:46

Sotos

51,121
6
32
66

score 4 · Answer 4 · answered Feb 14 '20 at 07:48

4

With dplyr:

library(dplyr)

df %>%
  mutate_at("score", ~ ifelse(is.na(.), max(., na.rm = TRUE) + cumsum(is.na(.)), .))

Result:

# A tibble: 5 x 2
  source score
  <chr>  <dbl>
1 a         10
2 b          5
3 c         11
4 d          3
5 e         12

answered Feb 14 '20 at 07:48

Aron Strandberg

3,040
9
15

score 3 · Answer 5 · answered Feb 14 '20 at 07:45

A dplyr solution.

df %>%
  mutate(na_count = cumsum(is.na(score)),
         score = ifelse(is.na(score), max(score, na.rm = TRUE) + na_count, score)) %>%
  select(-na_count)
## A tibble: 5 x 2
#  source score
#  <chr>  <dbl>
#1 a         10
#2 b          5
#3 c         11
#4 d          3
#5 e         12

score 2 · Answer 6 · answered Feb 14 '20 at 08:50

Another one, quite similar to ThomasIsCoding's solution:

> df$score[is.na(df$score)]<-max(df$score, na.rm=T)+(1:sum(is.na(df$score)))
> df
# A tibble: 5 x 2
  source score
  <chr>  <dbl>
1 a         10
2 b          5
3 c         11
4 d          3
5 e         12

score 2 · Answer 7 · answered Feb 14 '20 at 09:49

Not quite elegant as compared to the base R solutions, but still possible:

library(data.table)
setDT(df)

max.score = df[, max(score, na.rm = TRUE)]
df[is.na(score), score :=(1:.N) + max.score]

Or in one line but a bit slower:

df[is.na(score), score := (1:.N) + df[, max(score, na.rm = TRUE)]]
df
   source score
1:      a    10
2:      b     5
3:      c    11
4:      d     3
5:      e    12

How to replace NA with set of values

7 Answers7