如何在数据框中组合两个观察结果并用相互矛盾的条目填充 NA

Question

I'd like to combine like observations such that NA s in observation A are filled with entries in observation B. If observation A and observation B have contradicting entries, eg, two different values in the same field, I'd like the resulting data frame to return an NA in that field.

Example.

Consider the following data frame

df1 <- data.frame(APPLIANT = c("tom", "tom"), 
                  PERMIT = c(31, 31), 
                  ISSUED_YR = c("2018", NA), 
                  TRANSFERED = c("Y", "N"))

It looks like

  APPLIANT PERMIT ISSUED_YR TRANSFERED
1      tom     31      2018          Y
2      tom     31      <NA>          N

I'd like my final data frame to look like

  APPLIANT PERMIT ISSUED_YR TRANSFERED
1      tom     31      2018         NA

I was thinking of using an apply function. maybe something like

apply(df1, 2, FUN = function(one_col){
if(length(unique(one_col)) == 1){one_col}else{ one_col[!is.na(one_col)]}
})

But im not sure how to handle the 'contradicting' data points in an elegant way.... I also do not feel like my solution is that elegant to begin with. If there is something simpler that would be ideal!

Answer 1

Maybe this could help if there is only two observations involved:

library(dplyr)

df1 %>%
  mutate(across(everything(), ~ case_when(
    length(unique(.x)) > 1 & !any(is.na(.x)) ~ NA_character_,
    TRUE ~ as.character(coalesce(.x, .x[!is.na(.x)]))
  ))) %>%
  distinct()

  APPLIANT PERMIT ISSUED_YR TRANSFERED
1      tom     31      2018       <NA>

Answer 2

If there are more than 1 unique value in a column return NA else return the non-NA value.

library(dplyr)

df1 %>%
  group_by(APPLIANT) %>%
  summarise(across(.fns = ~if(n_distinct(., na.rm = TRUE) > 1) NA else na.omit(.)[1]))

#APPLIANT PERMIT ISSUED_YR TRANSFERED
#  <chr>     <dbl> <chr>     <lgl>     
#1 tom          31 2018      NA        

Answer 3

for some reason the above suggestions worked on my example data but not on the actual data. In my real data set, I have some columns that are date objects, maybe this posed a problem

What seemed to work for me but is not as 'pretty' was the following

df %>%
      mutate_all(funs(if(length(unique(.)) == 1){ 
        unique(.)
      }else{
        if(any(is.na(.))){
          (.)[!is.na(.)]
        }else{
          NA
          }
        })) %>% 
      distinct()