pair-wise duplicate removal from dataframe [duplicate]

Answer 1

One solution is to first sort each row of df:

for (i in 1:nrow(df))
{
    df[i, ] = sort(df[i, ])
}
df

a b
1 A A
2 A B
3 A B
4 B C
5 A B
6 A B
7 B C
8 B C

At that point it's just a matter of removing the duplicated elements:

df = df[!duplicated(df),]
df
  a b 
1 A A
2 A B
4 B C

As thelatemail mentioned in the comments, your code actualy keeps the duplicates. You need to use !duplicated to remove them.

Answer 2

Using apply will be a better option than loops.

newDf <- data.frame(t(apply(df,1,sort)))

All you need to do now is remove duplicates.

newDf <- newDf[!duplicated(newDf),]

Answer 3

The other answers use a for loop to assign a value for each and every row. While this is not an issue if you have 100 rows, or even a thousand, you're going to be waiting a while if you have large data of the order of 1M rows.

Stealing from the other linked answer using data.table, you could try something like:

df[!duplicated(data.frame(list(do.call(pmin,df),do.call(pmax,df)))),]

A comparison benchmark with a larger dataset (df2):

df2 <- df[sample(1:nrow(df),50000,replace=TRUE),]

system.time(
  df2[!duplicated(data.frame(list(do.call(pmin,df2),do.call(pmax,df2)))),]
)
# user  system elapsed 
# 0.07    0.00    0.06 

system.time({
  for (i in 1:nrow(df2))
  {
      df2[i, ] = sort(df2[i, ])
  }
  df2[!duplicated(df2),]
}
)
#   user  system elapsed 
#  42.07    0.02   42.09 

Answer 4

Extending Ari's answer, to specify columns to check if other columns are also there:

a <- c(rep("A", 3), rep("B", 3), rep("C",2))
b <- c('A','B','B','C','A','A','B','B')
df <-data.frame(a,b)

df$c = sample(1:10,8)
df$d = sample(LETTERS,8)
df
  a b  c d
1 A A 10 B
2 A B  8 S
3 A B  7 J
4 B C  3 Q
5 B A  2 I
6 B A  6 U
7 C B  4 L
8 C B  5 V

cols = c(1,2)
newdf = df[,cols]
for (i in 1:nrow(df)){
    newdf[i, ] = sort(df[i,cols])
}

df[!duplicated(newdf),]
  a b c d
1 A A 8 X
2 A B 7 L
4 B C 2 P