# Identify row indices corresponding to each distinct row of a matrix

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## Identify row indices corresponding to each distinct row of a matrix

 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]] ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: Identify row indices corresponding to each distinct row of a matrix

 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 <[hidden email]> 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]] > > ______________________________________________ > [hidden email] 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]] ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: Identify row indices corresponding to each distinct row of a matrix

 Perhaps which( ! duplicated( m, MARGIN=1 ) ) ? (untested) On November 7, 2018 9:20:57 PM PST, Bert Gunter <[hidden email]> 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 <[hidden email]> 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]] >> >> ______________________________________________ >> [hidden email] 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]] > >______________________________________________ >[hidden email] 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. ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: Identify row indices corresponding to each distinct row of a matrix

 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 <[hidden email]> wrote: > Perhaps > > which( ! duplicated( m, MARGIN=1 ) ) > > ? (untested) > > On November 7, 2018 9:20:57 PM PST, Bert Gunter <[hidden email]> > 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 <[hidden email]> 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]] > >> > >> ______________________________________________ > >> [hidden email] 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]] > > > >______________________________________________ > >[hidden email] 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]] ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: 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 <[hidden email]> 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 <[hidden email]> > wrote: > >> Perhaps >> >> which( ! duplicated( m, MARGIN=1 ) ) >> >> ? (untested) >> >> On November 7, 2018 9:20:57 PM PST, Bert Gunter <[hidden email]> >> 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 <[hidden email]> 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]] >> >> >> >> ______________________________________________ >> >> [hidden email] 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]] >> > >> >______________________________________________ >> >[hidden email] 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]] ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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