Ykhjuy

Another idea using dplyr,

library(dplyr)



df1 %>% 

 group_by(id) %>% 

 mutate(val = ifelse(row_number() == 1 & all(is.na(val)), 0, val)) %>% 

 na.omit()

which gives,

# A tibble: 5 x 2

# Groups:   id [2]

  id      val

  <fct> <dbl>

1 a         0

2 b         1

3 b         2

4 b         2

5 b         3

df1[is.na(df1)] <- 0

df1[!(duplicated(df1$id) & df1$val == 0), ]



  id val

1  a   0

5  b   1

6  b   2

7  b   2

8  b   3

We may do

df1 %>% group_by(id) %>% do(if(all(is.na(.$val))) replace(.[1, ], 2, 0) else na.omit(.))

# A tibble: 5 x 2

# Groups:   id [2]

#   id      val

#   <fct> <dbl>

# 1 a         0

# 2 b         1

# 3 b         2

# 4 b         2

# 5 b         3

After grouping by id, if everything in val is NA, then we leave only the first row with the second element replaced by 0, otherwise the same data is returned after applying na.omit.

In a more readable format that would be

df1 %>% group_by(id) %>% 

  do(if(all(is.na(.$val))) data.frame(id = .$id[1], val = 0) else na.omit(.))

(Here I presume that you indeed want to get rid of all NA values; otherwise there is no need for na.omit.)

Base R option is to find groups with all NAs and transform them by changing their val to 0 and select only unique rows so that there is only one row per group. We rbind this dataframe with the groups which are !all_NA.

all_NA <- with(df1, ave(is.na(val), id, FUN = all))

rbind(unique(transform(df1[all_NA, ], val = 0)), df1[!all_NA, ])



#  id val

#1  a   0

#5  b   1

#6  b   2

#7  b   2

#8  b   3

dplyr option looks ugly but one way is to make two groups of dataframes one with groups of all NA values and other with groups of all non-NA values. For groups with all NA values we add row with it's id and val as 0 and bind this to the other group.

library(dplyr)



bind_rows(df1 %>%

            group_by(id) %>%

            filter(all(!is.na(val))), 

          df1 %>%

             group_by(id) %>%

             filter(all(is.na(val))) %>%

             ungroup() %>%

             summarise(id = unique(id), 

                       val = 0)) %>%

arrange(id)





#   id      val

#  <fct> <dbl>

#1  a         0

#2  b         1

#3  b         2

#4  b         2

#5  b         3

Changed the df to make example more exhaustive -

df1 <- data.frame(id = rep(c("a", "b","c"), each = 4),

                  val = c(NA, NA, NA, NA, 1, 2, 2, 3,NA,2,NA,3))

library(dplyr)

df1 %>%

  group_by(id) %>%

  mutate(case=sum(is.na(val))==n(), row_num=row_number() ) %>%

  mutate(val=ifelse(is.na(val)&case,0,val)) %>%

  filter( !(case&row_num!=1) ) %>%

  select(id, val)

Output

  id      val

  <fct> <dbl>

1 a         0

2 b         1

3 b         2

4 b         2

5 b         3

6 c        NA

7 c         2

8 c        NA

9 c         3

Here is an option too:

df1 %>% 

  mutate_if(is.factor,as.character) %>% 

 mutate_all(funs(replace(.,is.na(.),0))) %>% 

  slice(4:nrow(.))

This gives:

 id val

1  a   0

2  b   1

3  b   2

4  b   2

5  b   3

Alternative:

df1 %>% 

  mutate_if(is.factor,as.character) %>% 

 mutate_all(funs(replace(.,is.na(.),0))) %>% 

  unique()

Here is a base R solution.

res <- lapply(split(df1, df1$id), function(DF){

  if(anyNA(DF$val)) {

    i <- is.na(DF$val)

    DF$val[i] <- 0

    DF <- rbind(DF[i & !duplicated(DF[i, ]), ], DF[!i, ])

  }

  DF

})

res <- do.call(rbind, res)

row.names(res) <- NULL

res

#  id val

#1  a   0

#2  b   1

#3  b   2

#4  b   2

#5  b   3

Edit.

A dplyr solution could be the following.
It was tested with the original dataset posted by the OP, with the dataset in Vivek Kalyanarangan's answer and with the dataset in markus' comment, renamed df2 and df3, respectively.

library(dplyr)



na2zero <- function(DF){

  DF %>%

    group_by(id) %>%

    mutate(val = ifelse(is.na(val), 0, val),

           crit = val == 0 & duplicated(val)) %>%

    filter(!crit) %>%

    select(-crit)

}



na2zero(df1)

na2zero(df2)

na2zero(df3)