[R] Identify row indices corresponding to each distinct row of a matrix

[Jeff Newmiller]

The duplicated function returns TRUE for rows that have already appeared... exactly one of the rows is not represented in the output of duplicated. For the intended purpose of removing duplicates this behavior is ideal. I have no idea what your intended purpose is, since every row has duplicates elsewhere in the matrix. If you really want every set identified this way then a loop/apply seems inevitable (most opportunities for optimization come about by not visiting every combination).

Cm <- as.matrix( C )
D <- which( !duplicated( Cm, MARGIN=1 ) )
nCm <- nrow( Cm )
F <- lapply( D, function(d) {
   idxrep <- rep( d, nCm )
   which( 0 == unname( rowSums( Cm[idxrep,] != Cm ) ) )
  } )


On November 8, 2018 1:42:40 PM PST, li li <hannah.hlx using gmail.com> wrote:
>Thanks to all the reply. I will try to use plain text in the future.
>One question regarding using "which( ! duplicated( m, MARGIN=1 ) )".
>This seems to return the fist row indices corresponding to the distinct
>rows but it does not give all the row indices
>corresponding to each of the distinct rows. For example, in the my
>example
>below, rows 1, 13 15 are all (1,9).
>Thanks.
>  Hanna
>> A <- matrix(c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16),8,2)
>> B <- rbind(A,A,A)
>> C <- as.data.frame(B[sample(nrow(B)),])
>> C
>   V1 V2
>1   1  9
>2   2 10
>3   3 11
>4   5 13
>5   7 15
>6   6 14
>7   4 12
>8   3 11
>9   8 16
>10  5 13
>11  7 15
>12  2 10
>13  1  9
>14  8 16
>15  1  9
>16  3 11
>17  7 15
>18  4 12
>19  2 10
>20  6 14
>21  4 12
>22  8 16
>23  5 13
>24  6 14
>> T <- unique(C)
>> T
>  V1 V2
>1  1  9
>2  2 10
>3  3 11
>4  5 13
>5  7 15
>6  6 14
>7  4 12
>9  8 16
>>
>> i <- 1
>> which(C[,1]==T[i,1]& C[,2]==T[i,2])
>[1]  1 13 15
>
>
>Bert Gunter <bgunter.4567 using gmail.com> 于2018年11月8日周四 上午10:43写道：
>
>> Yes -- much better than mine. I didn't know about the MARGIN argument
>of
>> duplicated().
>>
>> -- Bert
>>
>>
>> On Wed, Nov 7, 2018 at 10:32 PM Jeff Newmiller
><jdnewmil using dcn.davis.ca.us>
>> wrote:
>>
>>> Perhaps
>>>
>>> which( ! duplicated( m, MARGIN=1 ) )
>>>
>>> ? (untested)
>>>
>>> On November 7, 2018 9:20:57 PM PST, Bert Gunter
><bgunter.4567 using gmail.com>
>>> wrote:
>>> >A mess -- due to your continued use of html formatting.
>>> >
>>> >But something like this may do what you want (hard to tell with the
>>> >mess):
>>> >
>>> >> m <- matrix(1:16,nrow=8)[rep(1:8,2),]
>>> >> m
>>> >      [,1] [,2]
>>> > [1,]    1    9
>>> > [2,]    2   10
>>> > [3,]    3   11
>>> > [4,]    4   12
>>> > [5,]    5   13
>>> > [6,]    6   14
>>> > [7,]    7   15
>>> > [8,]    8   16
>>> > [9,]    1    9
>>> >[10,]    2   10
>>> >[11,]    3   11
>>> >[12,]    4   12
>>> >[13,]    5   13
>>> >[14,]    6   14
>>> >[15,]    7   15
>>> >[16,]    8   16
>>> >> vec <- apply(m,1,paste,collapse="-") ## converts rows into
>character
>>> >vector
>>> >> vec
>>> >[1] "1-9"  "2-10" "3-11" "4-12" "5-13" "6-14" "7-15" "8-16" "1-9"
>>> >"2-10"
>>> >"3-11" "4-12" "5-13" "6-14"
>>> >[15] "7-15" "8-16"
>>> >> ## Then maybe:
>>> >> tapply(seq_along(vec),vec, I)
>>> >$`1-9`
>>> >[1] 1 9
>>> >
>>> >$`2-10`
>>> >[1]  2 10
>>> >
>>> >$`3-11`
>>> >[1]  3 11
>>> >
>>> >$`4-12`
>>> >[1]  4 12
>>> >
>>> >$`5-13`
>>> >[1]  5 13
>>> >
>>> >$`6-14`
>>> >[1]  6 14
>>> >
>>> >$`7-15`
>>> >[1]  7 15
>>> >
>>> >$`8-16`
>>> >[1]  8 16
>>> >
>>> >> ## gives the row numbers for each unique row
>>> >
>>> >There may well be slicker ways to do this -- if this is actually
>what
>>> >you
>>> >want to do.
>>> >
>>> >-- Bert
>>> >
>>> >
>>> >
>>> >On Wed, Nov 7, 2018 at 7:56 PM li li <hannah.hlx using gmail.com> wrote:
>>> >
>>> >> Hi all,
>>> >>    I use the following example to illustrate my question. As you
>can
>>> >see,
>>> >> in matrix C some rows are repeated and I would like to find the
>>> >indices of
>>> >> the rows corresponding to each of the distinct rows.
>>> >>   For example, for the row c(1,9), I have used the "which"
>function
>>> >to
>>> >> identify the row indices corresponding to c(1,9). Using this
>>> >approach, in
>>> >> order to cover all distinct rows, I need to use a for loop.
>>> >>    I am wondering whether there is an easier way where a for loop
>can
>>> >be
>>> >> avoided?
>>> >>    Thanks very much!
>>> >>       Hanna
>>> >>
>>> >>
>>> >>
>>> >> > A <- matrix(c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16),8,2)> B
><-
>>> >> rbind(A,A,A)> C <- as.data.frame(B[sample(nrow(B)),])> C   V1 V2
>>> >> 1   1  9
>>> >> 2   2 10
>>> >> 3   3 11
>>> >> 4   5 13
>>> >> 5   7 15
>>> >> 6   6 14
>>> >> 7   4 12
>>> >> 8   3 11
>>> >> 9   8 16
>>> >> 10  5 13
>>> >> 11  7 15
>>> >> 12  2 10
>>> >> 13  1  9
>>> >> 14  8 16
>>> >> 15  1  9
>>> >> 16  3 11
>>> >> 17  7 15
>>> >> 18  4 12
>>> >> 19  2 10
>>> >> 20  6 14
>>> >> 21  4 12
>>> >> 22  8 16
>>> >> 23  5 13
>>> >> 24  6 14> T <- unique(C)> T  V1 V2
>>> >> 1  1  9
>>> >> 2  2 10
>>> >> 3  3 11
>>> >> 4  5 13
>>> >> 5  7 15
>>> >> 6  6 14
>>> >> 7  4 12
>>> >> 9  8 16> > i <- 1                    > which(C[,1]==T[i,1]&
>>> >> C[,2]==T[i,2])[1]  1 13 15
>>> >>
>>> >>         [[alternative HTML version deleted]]
>>> >>
>>> >> ______________________________________________
>>> >> R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
>>> >> https://stat.ethz.ch/mailman/listinfo/r-help
>>> >> PLEASE do read the posting guide
>>> >> http://www.R-project.org/posting-guide.html
>>> >> and provide commented, minimal, self-contained, reproducible
>code.
>>> >>
>>> >
>>> >       [[alternative HTML version deleted]]
>>> >
>>> >______________________________________________
>>> >R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
>>> >https://stat.ethz.ch/mailman/listinfo/r-help
>>> >PLEASE do read the posting guide
>>> >http://www.R-project.org/posting-guide.html
>>> >and provide commented, minimal, self-contained, reproducible code.
>>>
>>> --
>>> Sent from my phone. Please excuse my brevity.
>>>
>>

-- 
Sent from my phone. Please excuse my brevity.