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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 3 hours ago
m_goldberg
89.1k873200
asked 4 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 3 hours ago
m_goldberg
89.1k873200
asked 4 hours ago
PRGPRG
1028
$endgroup$
2
$begingroup$
TryValues[PositionIndex[M]]
$endgroup$
– Coolwater
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
24 mins 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 3 hours ago
m_goldberg
89.1k873200
asked 4 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 3 hours ago
m_goldberg
89.1k873200
asked 4 hours ago
PRGPRG
1028
edited 3 hours ago
m_goldberg
89.1k873200
asked 4 hours ago
PRGPRG
1028
edited 3 hours ago
m_goldberg
89.1k873200
edited 3 hours ago
m_goldberg
89.1k873200
edited 3 hours ago
m_goldberg
89.1k873200
89.1k873200
asked 4 hours ago
PRGPRG
1028
asked 4 hours ago
PRGPRG
1028
asked 4 hours ago
PRGPRG
1028
1028
2
$begingroup$
TryValues[PositionIndex[M]]
$endgroup$
– Coolwater
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
24 mins ago
add a comment |
2
$begingroup$
TryValues[PositionIndex[M]]
$endgroup$
– Coolwater
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
24 mins ago
2
2
$begingroup$
Try
Values[PositionIndex[M]]$endgroup$
– Coolwater
1 hour ago
$begingroup$
Try
Values[PositionIndex[M]]$endgroup$
– Coolwater
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
24 mins 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
24 mins 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 4 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
3 hours ago
$begingroup$
MANY THANKS, HENRIK!
$endgroup$
– PRG
2 hours ago
$begingroup$
YOU'RE WELCOME, PRG! =D
$endgroup$
– Henrik Schumacher
2 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 32 mins 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 4 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
3 hours ago
$begingroup$
MANY THANKS, HENRIK!
$endgroup$
– PRG
2 hours ago
$begingroup$
YOU'RE WELCOME, PRG! =D
$endgroup$
– Henrik Schumacher
2 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 4 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
3 hours ago
$begingroup$
MANY THANKS, HENRIK!
$endgroup$
– PRG
2 hours ago
$begingroup$
YOU'RE WELCOME, PRG! =D
$endgroup$
– Henrik Schumacher
2 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 4 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 4 hours ago
Henrik SchumacherHenrik Schumacher
61.2k585171
edited 3 hours ago
answered 4 hours ago
Henrik SchumacherHenrik Schumacher
61.2k585171
answered 4 hours ago
Henrik SchumacherHenrik Schumacher
61.2k585171
answered 4 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
3 hours ago
$begingroup$
MANY THANKS, HENRIK!
$endgroup$
– PRG
2 hours ago
$begingroup$
YOU'RE WELCOME, PRG! =D
$endgroup$
– Henrik Schumacher
2 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
3 hours ago
$begingroup$
MANY THANKS, HENRIK!
$endgroup$
– PRG
2 hours ago
$begingroup$
YOU'RE WELCOME, PRG! =D
$endgroup$
– Henrik Schumacher
2 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
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
3 hours ago
$begingroup$
MANY THANKS, HENRIK!
$endgroup$
– PRG
2 hours ago
$begingroup$
MANY THANKS, HENRIK!
$endgroup$
– PRG
2 hours ago
$begingroup$
YOU'RE WELCOME, PRG! =D
$endgroup$
– Henrik Schumacher
2 hours ago
$begingroup$
YOU'RE WELCOME, PRG! =D
$endgroup$
– Henrik Schumacher
2 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 32 mins 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 32 mins 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 32 mins 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 32 mins ago
Okkes DulgerciOkkes Dulgerci
5,5511919
edited 12 mins ago
answered 32 mins ago
Okkes DulgerciOkkes Dulgerci
5,5511919
answered 32 mins ago
Okkes DulgerciOkkes Dulgerci
5,5511919
answered 32 mins ago
Okkes DulgerciOkkes Dulgerci
5,5511919
5,5511919
add a comment |
add a comment |
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2
$begingroup$
Try
Values[PositionIndex[M]]$endgroup$
– Coolwater
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
24 mins ago