pair-wise duplicate removal from dataframe [duplic

问题:

This question already has an answer here:

• Select equivalent rows [A-B & B-A] [duplicate] 1 answer

This seems like a simple problem but I can\'t seem to figure it out. I\'d like to remove duplicates from a dataframe (df) if two columns have the same values, even if those values are in the reverse order. What I mean is, say you have the following data frame:

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)

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

If I now remove duplicates, I get the following data frame:

df[duplicated(df),]

  a b
3 A B
6 B A
8 C B

However, I would also like to remove the row 6 in this data frame, since \"A\", \"B\" is the same as \"B\", \"A\". How can I do this automatically?

Ideally I could specify which two columns to compare since the data frames could have varying columns and can be quite large.

Thanks!

回答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.

回答2:

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

回答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 

回答4:

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),]