Identify row indices corresponding to each distinct row of a matrix

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 at gmail.com> ?2018?11?8??? ??10:43???

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 at 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 at 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 at 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

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