[R] Extracing Unique Rows based on 2 Column

[Ragia Ibrahim]

Dear group,
kindly,  I have a data frame, as follows:


 Measure_id i j value      rank
1           1 2 3   2.0 1.0000000
2           1 5 1   2.0 1.0000000
3           1 2 1   1.5 0.7500000
4           1 5 2   1.5 0.7500000
5           1 7 3   1.5 1.0000000
6           1 2 4   1.0 0.5000000
7           1 7 5   1.0 0.6666667
8           2 5 2   2.5 1.0000000
9           2 2 1   2.0 1.0000000
10          2 2 4   2.0 1.0000000
..        ... . .   ...       ...

I want to select distinct rows based on two coulmn ( Measure_id  and i )

for example for Measure_id  = 1,2  the result would be....
1           1 2 3   2.0 1.0000000
2           1 5 1   2.0 1.0000000
5           1 7 3   1.5 1.0000000
8           2 5 2   2.5 1.0000000
9          2 2 1   2.0 1.0000000


kindly how I could do this?

example of the data frame are followed using dput.

dput(orderlist)

structure(list(Measure_id = c(1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 
2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 
5, 5, 5), i = c(2, 5, 2, 5, 7, 2, 7, 5, 2, 2, 7, 2, 5, 7, 2, 
2, 2, 5, 5, 7, 7, 2, 5, 2, 2, 5, 7, 7, 2, 2, 5, 2, 5, 7, 7), 
    j = c(3, 1, 1, 2, 3, 4, 5, 2, 1, 4, 5, 3, 1, 3, 1, 3, 4, 
    1, 2, 3, 5, 4, 2, 1, 3, 1, 3, 5, 1, 4, 2, 3, 1, 3, 5), value = c(2, 
    2, 1.5, 1.5, 1.5, 1, 1, 2.5, 2, 2, 2, 1.5, 1.5, 1, 1, 0, 
    0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 2, 2, 2, 1, 1, 1, 1), 
    rank = c(1, 1, 0.75, 0.75, 1, 0.5, 0.666666666666667, 1, 
    1, 1, 1, 0.75, 0.6, 0.5, 1, 0, 0, NaN, NaN, NaN, NaN, 1, 
    1, 0, 0, 0, NaN, NaN, 1, 1, 1, 0.5, 0.5, 1, 1)), class = c("grouped_df", 
"tbl_df", "tbl", "data.frame"), row.names = c(NA, -35L), .Names = c("Measure_id", 
"i", "j", "value", "rank"), vars = list(Measure_id), indices = list(
    0:6, 7:13, 14:20, 21:27, 28:34), group_sizes = c(7L, 7L, 
7L, 7L, 7L), biggest_group_size = 7L, labels = structure(list(
    Measure_id = c(1, 2, 3, 4, 5)), class = "data.frame", row.names = c(NA, 
-5L), .Names = "Measure_id", vars = list(Measure_id)))




thanks in advance
Ragia