选择行，使多列没有重复的值

Question

I have a data frame with these values(built in such a way):

id1 = (c(1,1,2,2))
id2 = (c(10,11,10,11))
value =c(50,50,50,50)
df = data.frame(id1,id2,value)

df : 
  value id1 id2
1    50   1  10
2    50   1  11
3    50   2  10
4    50   2  11

I would like to keep only rows where both id1 and id2 are unique(each value of id1 and id2 must appear only once),also there might be more then one duplicate of each id:

df_unique : 
value id1 id2
1    50   1  10
4    50   2  11

if I use the duplicated command on one of the columns and then the other,I would discard wanted rows.

A solution which will return (1,11) and (2,10) is also good,as long as each element in id1 and id2 are unique.

Another example with more rows:

id1 = (c(1,1,1,2,2,2,3,3,3))
id2 = (c(10,11,12,10,11,12,10,11,12))
value =rep(50,9)
df = data.frame(id1,id2,value)

df:
  id1 id2 value
1   1  10    50
2   1  11    50
3   1  12    50
4   2  10    50
5   2  11    50
6   2  12    50
7   3  10    50
8   3  11    50
9   3  12    50

Where a good answer is:(1,10),(2,11),(3,12), but also any other answer where both id1 and id2 appear once are good.

Thank you,

Jacob

Answer 1

If you know that the data are arranged as in your example, cycling through id2 for each value of id1 and in the same order, the solution is easy:

N <- 3 # Number of rows in the result
idx <- seq(1, N*N, by=N) + seq(0,to=N-1)
df[idx,]
##   id1 id2 value
## 1   1  10    50
## 5   2  11    50
## 9   3  12    50

I doubt that this is what you're asking. If the rows are in an unknown order or not all values are present in one column for each value in the other, you have to check each combination of N rows.

# Maximum number of result rows
N <- with(df, min(length(unique(id1)), length(unique(id2))))
N
## [1] 3

# Potential indices
index <- combn(seq(nrow(df)), N)

index is a matrix where each column represents three rows in df . Now to check for duplicated values:

good <- apply(index, 2, function(x) !any(duplicated(df[x,'id1']) | duplicated(df[x,'id2'])))

good has the value TRUE for a combination of rows that passes the test.

which(good)
## [1] 22 24 39 44 53 56
index[, good]
##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    1    1    2    2    3    3
## [2,]    5    6    4    6    4    5
## [3,]    9    8    9    7    8    7

Each column of the above matrix represents a combination of rows that passes the test.

This finds all the combinations. You might want to find just the first combination, so that you don't go on to test additional combinations after a hit is found. Then for is appropriate:

for (i in seq(ncol(index))) {
  x <- index[,i]
  if (!any(duplicated(df[x,'id1']) | duplicated(df[x,'id2']))) {
    rows <- x
    break
  }
}

df[rows,]
##   id1 id2 value
## 1   1  10    50
## 5   2  11    50
## 9   3  12    50

Note: Depending on the data, it is possible that with N=3 , you will get no rows that pass the test. In that case, repeat the procedure with N=2 , and so on. I leave that loop as an exercise for the reader.