combine rows in data frame containing NA to make complete row

I know this is a duplicate Q but I can't seem to find the post again

Using the following data

df <- data.frame(A=c(1,1,2,2),B=c(NA,2,NA,4),C=c(3,NA,NA,5),D=c(NA,2,3,NA),E=c(5,NA,NA,4))

  A  B  C  D  E
  1 NA  3 NA  5
  1  2 NA  2 NA
  2 NA NA  3 NA
  2  4  5 NA  4

Grouping by A, I'd like the following output using a tidyverse solution

  A  B  C  D  E
  1  2  3  2  5
  2  4  5  3  4

I have many groups in A. I think I saw an answer using coalesce but am unsure how to get it work. I'd like a solution that works with characters as well. Thanks!

Solution 1:

I haven't figured out how to put the coalesce_by_column function inside the dplyr pipeline, but this works:

coalesce_by_column <- function(df) {
  return(coalesce(df[1], df[2]))
}

df %>%
  group_by(A) %>%
  summarise_all(coalesce_by_column)

##       A     B     C     D     E
##   <dbl> <dbl> <dbl> <dbl> <dbl>
## 1     1     2     3     2     5
## 2     2     4     5     3     4

Edit: include @Jon Harmon's solution for more than 2 members of a group

# Supply lists by splicing them into dots:
coalesce_by_column <- function(df) {
  return(dplyr::coalesce(!!! as.list(df)))
}

df %>%
  group_by(A) %>%
  summarise_all(coalesce_by_column)

#> # A tibble: 2 x 5
#>       A     B     C     D     E
#>   <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1     1     2     3     2     5
#> 2     2     4     5     3     4

Solution 2:

We can use fill to fill all the missing values. And then filter just one row for each group.

library(dplyr)
library(tidyr)

df2 <- df %>%
  group_by(A) %>%
  fill(everything(), .direction = "down") %>%
  fill(everything(), .direction = "up") %>%
  slice(1)

And thanks to @Roger-123, the above code can be further simplified as follows.

df2 <- df %>%
  group_by(A) %>%
  fill(everything(), .direction = "downup") %>%
  slice(1)

Solution 3:

Not tidyverse but here's one base R solution

df <- data.frame(A=c(1,1),B=c(NA,2),C=c(3,NA),D=c(NA,2),E=c(5,NA))
sapply(df, function(x) x[!is.na(x)][1])
#A B C D E 
#1 2 3 2 5 

With updated data

do.call(rbind, lapply(split(df, df$A), function(a) sapply(a, function(x) x[!is.na(x)][1])))
#  A B C D E
#1 1 2 3 2 5
#2 2 4 5 3 4

Solution 4:

Here is an even more general solution (using unique, na.omit to sort of create coalesce), which can handle more than two rows with overlapping information. Super simply and forward.

> df <- data.frame(A=c(1,1,2,2,2),B=c(NA,2,NA,4,4),C=c(3,NA,NA,5,NA),D=c(NA,2,3,NA,NA),E=c(5,NA,NA,4,4))

> df
  A  B  C  D  E
1 1 NA  3 NA  5
2 1  2 NA  2 NA
3 2 NA NA  3 NA
4 2  4  5 NA  4
5 2  4 NA NA  4

> df %>% group_by(A) %>% summarise_all(funs( na.omit(unique(.)) ))
# A tibble: 2 x 5
      A     B     C     D     E
  <dbl> <dbl> <dbl> <dbl> <dbl>
1     1     2     3     2     5
2     2     4     5     3     4