CODEWITHSUNDEEP
rif-statementdplyrreplacetibble
littleworth
littleworth

Reputation: 5169

How to replace NA with set of values

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?

Upvotes: 17

Views: 747

Answers (7)

Serhii
Serhii

Reputation: 422

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

Upvotes: 2

Sotos
Sotos

Reputation: 51592

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

Upvotes: 6

Łukasz Deryło
Łukasz Deryło

Reputation: 1870

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

Upvotes: 2

Ronak Shah
Ronak Shah

Reputation: 389265

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))

Upvotes: 10

Aron Strandberg
Aron Strandberg

Reputation: 3090

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

Upvotes: 4

Rui Barradas
Rui Barradas

Reputation: 76651

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

Upvotes: 3

ThomasIsCoding
ThomasIsCoding

Reputation: 102680

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

Upvotes: 6

Related Questions