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Find the identical rows in a matrix
How to find rows that have maximum value?How to do equality check of a large matrix and get the corresponding index position?List manipulation: Dropping first or last row or column of a matrixIs it possible to colour only one column of a matrix?Make a vector of sums of matrix rowsHow to operate on spans of rows in a matrix?Efficiently select the smallest magnitude element from each column of a matrixMatrix expansion and reorganisationMatrix using For LoopChanging the position of rows and columns in a matrix
$begingroup$
Suppose I have the following matrix:
M =
0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0,1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
TableForm[M, TableHeadings -> S1, S2, S3, S4, S5, S6, S7, S8]
In this case, it turns out that rows (S1, S8), (S2, S3, S4), (S5, S6, S7) have equal element values in identical column positions. I have a 1000 x 1000 matrix to examine and would appreciate any assistance in coding this problem.
list-manipulation matrix
edited 4 hours ago
m_goldberg
89.1k873200
asked 5 hours ago
PRGPRG
1028
$endgroup$
add a comment |
$begingroup$
Suppose I have the following matrix:
M =
0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0,1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
TableForm[M, TableHeadings -> S1, S2, S3, S4, S5, S6, S7, S8]
In this case, it turns out that rows (S1, S8), (S2, S3, S4), (S5, S6, S7) have equal element values in identical column positions. I have a 1000 x 1000 matrix to examine and would appreciate any assistance in coding this problem.
list-manipulation matrix
edited 4 hours ago
m_goldberg
89.1k873200
asked 5 hours ago
PRGPRG
1028
$endgroup$
2
$begingroup$
TryValues[PositionIndex[M]]
$endgroup$
– Coolwater
2 hours ago
$begingroup$
@Coolwater If there is a unique row, your method will fail. At least one needs to delete if list has length 1
$endgroup$
– Okkes Dulgerci
1 hour ago
add a comment |
$begingroup$
Suppose I have the following matrix:
M =
0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0,1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
TableForm[M, TableHeadings -> S1, S2, S3, S4, S5, S6, S7, S8]
In this case, it turns out that rows (S1, S8), (S2, S3, S4), (S5, S6, S7) have equal element values in identical column positions. I have a 1000 x 1000 matrix to examine and would appreciate any assistance in coding this problem.
list-manipulation matrix
edited 4 hours ago
m_goldberg
89.1k873200
asked 5 hours ago
PRGPRG
1028
$endgroup$
Suppose I have the following matrix:
M =
0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0,1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
TableForm[M, TableHeadings -> S1, S2, S3, S4, S5, S6, S7, S8]
In this case, it turns out that rows (S1, S8), (S2, S3, S4), (S5, S6, S7) have equal element values in identical column positions. I have a 1000 x 1000 matrix to examine and would appreciate any assistance in coding this problem.
list-manipulation matrix
list-manipulation matrix
edited 4 hours ago
m_goldberg
89.1k873200
asked 5 hours ago
PRGPRG
1028
edited 4 hours ago
m_goldberg
89.1k873200
asked 5 hours ago
PRGPRG
1028
edited 4 hours ago
m_goldberg
89.1k873200
edited 4 hours ago
m_goldberg
89.1k873200
edited 4 hours ago
m_goldberg
89.1k873200
89.1k873200
asked 5 hours ago
PRGPRG
1028
asked 5 hours ago
PRGPRG
1028
asked 5 hours ago
PRGPRG
1028
1028
2
$begingroup$
TryValues[PositionIndex[M]]
$endgroup$
– Coolwater
2 hours ago
$begingroup$
@Coolwater If there is a unique row, your method will fail. At least one needs to delete if list has length 1
$endgroup$
– Okkes Dulgerci
1 hour ago
add a comment |
2
$begingroup$
TryValues[PositionIndex[M]]
$endgroup$
– Coolwater
2 hours ago
$begingroup$
@Coolwater If there is a unique row, your method will fail. At least one needs to delete if list has length 1
$endgroup$
– Okkes Dulgerci
1 hour ago
2
2
$begingroup$
Try
Values[PositionIndex[M]]$endgroup$
– Coolwater
2 hours ago
$begingroup$
Try
Values[PositionIndex[M]]$endgroup$
– Coolwater
2 hours ago
$begingroup$
@Coolwater If there is a unique row, your method will fail. At least one needs to delete if list has length 1
$endgroup$
– Okkes Dulgerci
1 hour ago
$begingroup$
@Coolwater If there is a unique row, your method will fail. At least one needs to delete if list has length 1
$endgroup$
– Okkes Dulgerci
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
idx = DeleteDuplicates[Sort /@ Nearest[M -> Automatic, M, ∞, 0]]
1, 8, 2, 3, 4, 5, 6, 7
In order to obtain the labels of the rows, you may use the following:
labels = S1, S2, S3, S4, S5, S6, S7, S8;
Map[labels[[#]] &, idx, 2]
S1, S8, S2, S3, S4, S5, S6, S7
answered 5 hours ago
Henrik SchumacherHenrik Schumacher
61.2k585171
$endgroup$
$begingroup$
Henrik: Can I add the S in front of the result; e.g., (S1,S8),(S3,S4),(S5,S6,S7)?
$endgroup$
– PRG
4 hours ago
$begingroup$
MANY THANKS, HENRIK!
$endgroup$
– PRG
3 hours ago
$begingroup$
YOU'RE WELCOME, PRG! =D
$endgroup$
– Henrik Schumacher
3 hours ago
add a comment |
$begingroup$
Here is an alternative.
row, col = Dimensions@M;
Flatten /@ DeleteDuplicates[Position[#, ConstantArray[0, col]] & /@ Outer[Subtract, M, M, 1]]
1, 8, 2, 3, 4, 5, 6, 7
answered 1 hour ago
Okkes DulgerciOkkes Dulgerci
5,5511919
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
idx = DeleteDuplicates[Sort /@ Nearest[M -> Automatic, M, ∞, 0]]
1, 8, 2, 3, 4, 5, 6, 7
In order to obtain the labels of the rows, you may use the following:
labels = S1, S2, S3, S4, S5, S6, S7, S8;
Map[labels[[#]] &, idx, 2]
S1, S8, S2, S3, S4, S5, S6, S7
answered 5 hours ago
Henrik SchumacherHenrik Schumacher
61.2k585171
$endgroup$
$begingroup$
Henrik: Can I add the S in front of the result; e.g., (S1,S8),(S3,S4),(S5,S6,S7)?
$endgroup$
– PRG
4 hours ago
$begingroup$
MANY THANKS, HENRIK!
$endgroup$
– PRG
3 hours ago
$begingroup$
YOU'RE WELCOME, PRG! =D
$endgroup$
– Henrik Schumacher
3 hours ago
add a comment |
$begingroup$
idx = DeleteDuplicates[Sort /@ Nearest[M -> Automatic, M, ∞, 0]]
1, 8, 2, 3, 4, 5, 6, 7
In order to obtain the labels of the rows, you may use the following:
labels = S1, S2, S3, S4, S5, S6, S7, S8;
Map[labels[[#]] &, idx, 2]
S1, S8, S2, S3, S4, S5, S6, S7
answered 5 hours ago
Henrik SchumacherHenrik Schumacher
61.2k585171
$endgroup$
$begingroup$
Henrik: Can I add the S in front of the result; e.g., (S1,S8),(S3,S4),(S5,S6,S7)?
$endgroup$
– PRG
4 hours ago
$begingroup$
MANY THANKS, HENRIK!
$endgroup$
– PRG
3 hours ago
$begingroup$
YOU'RE WELCOME, PRG! =D
$endgroup$
– Henrik Schumacher
3 hours ago
add a comment |
$begingroup$
idx = DeleteDuplicates[Sort /@ Nearest[M -> Automatic, M, ∞, 0]]
1, 8, 2, 3, 4, 5, 6, 7
In order to obtain the labels of the rows, you may use the following:
labels = S1, S2, S3, S4, S5, S6, S7, S8;
Map[labels[[#]] &, idx, 2]
S1, S8, S2, S3, S4, S5, S6, S7
answered 5 hours ago
Henrik SchumacherHenrik Schumacher
61.2k585171
$endgroup$
idx = DeleteDuplicates[Sort /@ Nearest[M -> Automatic, M, ∞, 0]]
1, 8, 2, 3, 4, 5, 6, 7
In order to obtain the labels of the rows, you may use the following:
labels = S1, S2, S3, S4, S5, S6, S7, S8;
Map[labels[[#]] &, idx, 2]
S1, S8, S2, S3, S4, S5, S6, S7
answered 5 hours ago
Henrik SchumacherHenrik Schumacher
61.2k585171
edited 4 hours ago
answered 5 hours ago
Henrik SchumacherHenrik Schumacher
61.2k585171
answered 5 hours ago
Henrik SchumacherHenrik Schumacher
61.2k585171
answered 5 hours ago
Henrik SchumacherHenrik Schumacher
61.2k585171
61.2k585171
$begingroup$
Henrik: Can I add the S in front of the result; e.g., (S1,S8),(S3,S4),(S5,S6,S7)?
$endgroup$
– PRG
4 hours ago
$begingroup$
MANY THANKS, HENRIK!
$endgroup$
– PRG
3 hours ago
$begingroup$
YOU'RE WELCOME, PRG! =D
$endgroup$
– Henrik Schumacher
3 hours ago
add a comment |
$begingroup$
Henrik: Can I add the S in front of the result; e.g., (S1,S8),(S3,S4),(S5,S6,S7)?
$endgroup$
– PRG
4 hours ago
$begingroup$
MANY THANKS, HENRIK!
$endgroup$
– PRG
3 hours ago
$begingroup$
YOU'RE WELCOME, PRG! =D
$endgroup$
– Henrik Schumacher
3 hours ago
$begingroup$
Henrik: Can I add the S in front of the result; e.g., (S1,S8),(S3,S4),(S5,S6,S7)?
$endgroup$
– PRG
4 hours ago
$begingroup$
Henrik: Can I add the S in front of the result; e.g., (S1,S8),(S3,S4),(S5,S6,S7)?
$endgroup$
– PRG
4 hours ago
$begingroup$
MANY THANKS, HENRIK!
$endgroup$
– PRG
3 hours ago
$begingroup$
MANY THANKS, HENRIK!
$endgroup$
– PRG
3 hours ago
$begingroup$
YOU'RE WELCOME, PRG! =D
$endgroup$
– Henrik Schumacher
3 hours ago
$begingroup$
YOU'RE WELCOME, PRG! =D
$endgroup$
– Henrik Schumacher
3 hours ago
add a comment |
$begingroup$
Here is an alternative.
row, col = Dimensions@M;
Flatten /@ DeleteDuplicates[Position[#, ConstantArray[0, col]] & /@ Outer[Subtract, M, M, 1]]
1, 8, 2, 3, 4, 5, 6, 7
answered 1 hour ago
Okkes DulgerciOkkes Dulgerci
5,5511919
$endgroup$
add a comment |
$begingroup$
Here is an alternative.
row, col = Dimensions@M;
Flatten /@ DeleteDuplicates[Position[#, ConstantArray[0, col]] & /@ Outer[Subtract, M, M, 1]]
1, 8, 2, 3, 4, 5, 6, 7
answered 1 hour ago
Okkes DulgerciOkkes Dulgerci
5,5511919
$endgroup$
add a comment |
$begingroup$
Here is an alternative.
row, col = Dimensions@M;
Flatten /@ DeleteDuplicates[Position[#, ConstantArray[0, col]] & /@ Outer[Subtract, M, M, 1]]
1, 8, 2, 3, 4, 5, 6, 7
answered 1 hour ago
Okkes DulgerciOkkes Dulgerci
5,5511919
$endgroup$
Here is an alternative.
row, col = Dimensions@M;
Flatten /@ DeleteDuplicates[Position[#, ConstantArray[0, col]] & /@ Outer[Subtract, M, M, 1]]
1, 8, 2, 3, 4, 5, 6, 7
answered 1 hour ago
Okkes DulgerciOkkes Dulgerci
5,5511919
edited 1 hour ago
answered 1 hour ago
Okkes DulgerciOkkes Dulgerci
5,5511919
answered 1 hour ago
Okkes DulgerciOkkes Dulgerci
5,5511919
answered 1 hour ago
Okkes DulgerciOkkes Dulgerci
5,5511919
5,5511919
add a comment |
add a comment |
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$begingroup$
Try
Values[PositionIndex[M]]$endgroup$
– Coolwater
2 hours ago
$begingroup$
@Coolwater If there is a unique row, your method will fail. At least one needs to delete if list has length 1
$endgroup$
– Okkes Dulgerci
1 hour ago