R help - Nov 2018 - 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
>> >>
>> >>         [[alternative HTML version deleted]]
>> >>
>> >> ______________________________________________
>> >> R-help at 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 at 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.
>>
>
	[[alternative HTML version deleted]]

One way, rather clumsy, is to convert your data.frame in a character vector
or list. via an invertible tranformation, and use match on it.  E.g.,
> tmp <- do.call(paste, c(list(sep="\001"), unname(C))) #
convert to
character
> # or tmp <- split(C, seq_len(nrow(C))) # convert to list of its rows
> g <- match(tmp, unique(tmp))
> g
 [1] 1 2 3 4 5 6 7 3 8 4 5 2 1 8 1 3 5 7 2 6 7 8 4 6
> split(seq_along(g), g)
$`1`
[1]  1 13 15

$`2`
[1]  2 12 19
...
$`8`
[1]  9 14 22

Both ways work for nice enough inputs, but converting to text can cause
problems if the 'sep' is in any of input text and match() on lists of
lists
used
to have problems when the inner lists were big.


Bill Dunlap
TIBCO Software
wdunlap tibco.com

On Thu, Nov 8, 2018 at 1:42 PM, li li <hannah.hlx at 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 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
> >> >>
> >> >>         [[alternative HTML version deleted]]
> >> >>
> >> >> ______________________________________________
> >> >> R-help at 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 at 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.
> >>
> >
>
>         [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at 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]]

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 at 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 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
>>> >>
>>> >>         [[alternative HTML version deleted]]
>>> >>
>>> >> ______________________________________________
>>> >> R-help at 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 at 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.

Thanks. It makes sense.

Jeff Newmiller <jdnewmil at dcn.davis.ca.us> ?2018?11?8??? ??8:05???
> 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 at 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 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
> >>> >>
> >>> >>         [[alternative HTML version deleted]]
> >>> >>
> >>> >> ______________________________________________
> >>> >> R-help at 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 at 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.
>
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