# NOT RUN {
sample_data1 %>%
head
## on level 1, each column is tested as a unique identifier. the VAL columns have no
## duplicates and hence qualify, even though they normally would be considered as IDs
## on level 3, combinations of 3 columns are tested. implying that ID_COL 1,2,3 form a unique key
## level 2 does not appear, implying that combinations of any 2 ID_COLs do not form a unique key
sample_data1 %>%
determine_distinct(listviewer = FALSE)
# }
Run the code above in your browser using DataLab