How to convert diagonal matrix to rectangular matrixIs there a built in function to obtain the back diagonal of a matrix?How to select all elements above the main diagonal of matrix?Diagonal times dense matrix, high precisionHow to extract the list of all matrices from a Block Diagonal Matrix?Get rid of infinity in matrix elements (by separate definition of diagonal and off-diagonal elements)Summation over diagonal blocksHow to transform symmetric matrix to diagonal?Sum of elements in a diagonalReplace diagonal elements in sparse matrixConstructing a block diagonal matrix
How to convert diagonal matrix to rectangular matrix
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How to convert diagonal matrix to rectangular matrix
Is there a built in function to obtain the back diagonal of a matrix?How to select all elements above the main diagonal of matrix?Diagonal times dense matrix, high precisionHow to extract the list of all matrices from a Block Diagonal Matrix?Get rid of infinity in matrix elements (by separate definition of diagonal and off-diagonal elements)Summation over diagonal blocksHow to transform symmetric matrix to diagonal?Sum of elements in a diagonalReplace diagonal elements in sparse matrixConstructing a block diagonal matrix
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Suppose you have the following diagonal matrix:
DiagonalMatrix[Hold/@a, b, c]//ReleaseHold
a, 0, 0, b, c
How can the above matrix be converted to the following rectangular one?
a, 0, 0, 0, b, c
Or another example:
b, c, 0, 0, a} -> b, c, 0, 0, 0, a
list-manipulation matrix
asked 8 hours ago
aleksander_sialeksander_si
284 bronze badges
New contributor
aleksander_si is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Suppose you have the following diagonal matrix:
DiagonalMatrix[Hold/@a, b, c]//ReleaseHold
a, 0, 0, b, c
How can the above matrix be converted to the following rectangular one?
a, 0, 0, 0, b, c
Or another example:
b, c, 0, 0, a} -> b, c, 0, 0, 0, a
list-manipulation matrix
asked 8 hours ago
aleksander_sialeksander_si
284 bronze badges
New contributor
aleksander_si is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Could you clarify what should happen for e.g. the matrixa, d, e, 0, 0, b, cas in my answer? Whether my answer is correct or not hinges on this, because it gives a different answer than kglr.
$endgroup$
– C. E.
20 mins ago
add a comment |
$begingroup$
Suppose you have the following diagonal matrix:
DiagonalMatrix[Hold/@a, b, c]//ReleaseHold
a, 0, 0, b, c
How can the above matrix be converted to the following rectangular one?
a, 0, 0, 0, b, c
Or another example:
b, c, 0, 0, a} -> b, c, 0, 0, 0, a
list-manipulation matrix
asked 8 hours ago
aleksander_sialeksander_si
284 bronze badges
New contributor
aleksander_si is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Suppose you have the following diagonal matrix:
DiagonalMatrix[Hold/@a, b, c]//ReleaseHold
a, 0, 0, b, c
How can the above matrix be converted to the following rectangular one?
a, 0, 0, 0, b, c
Or another example:
b, c, 0, 0, a} -> b, c, 0, 0, 0, a
list-manipulation matrix
list-manipulation matrix
asked 8 hours ago
aleksander_sialeksander_si
284 bronze badges
New contributor
aleksander_si is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
aleksander_sialeksander_si
284 bronze badges
New contributor
aleksander_si is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 mins ago
aleksander_si
asked 8 hours ago
aleksander_sialeksander_si
284 bronze badges
New contributor
aleksander_si is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
aleksander_sialeksander_si
284 bronze badges
asked 8 hours ago
aleksander_sialeksander_si
284 bronze badges
284 bronze badges
New contributor
aleksander_si is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
aleksander_si is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Could you clarify what should happen for e.g. the matrixa, d, e, 0, 0, b, cas in my answer? Whether my answer is correct or not hinges on this, because it gives a different answer than kglr.
$endgroup$
– C. E.
20 mins ago
add a comment |
1
$begingroup$
Could you clarify what should happen for e.g. the matrixa, d, e, 0, 0, b, cas in my answer? Whether my answer is correct or not hinges on this, because it gives a different answer than kglr.
$endgroup$
– C. E.
20 mins ago
1
1
$begingroup$
Could you clarify what should happen for e.g. the matrix
a, d, e, 0, 0, b, c as in my answer? Whether my answer is correct or not hinges on this, because it gives a different answer than kglr.$endgroup$
– C. E.
20 mins ago
$begingroup$
Could you clarify what should happen for e.g. the matrix
a, d, e, 0, 0, b, c as in my answer? Whether my answer is correct or not hinges on this, because it gives a different answer than kglr.$endgroup$
– C. E.
20 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
kglr's solution fails if the number of elements on the diagonal in any of the preceding rows is larger than the number of elements on the diagonal of the last row.
For example:
PadRight[Flatten /@ a, d, e, 0, 0, b, c] // MatrixForm
m = a, 0, 0, 0, b, c, d, e, f, 0, 0, 0, b, c
PadRight[Flatten /@ m] // MatrixForm
Here is a different solution:
diag = Flatten[#] & /@ Diagonal[m];
ncols = Length@Flatten[diag];
offsets = Most@Prepend[Accumulate[Length /@ diag], 0];
row[values_, offset_, ncols_] := PadRight[ArrayPad[values, offset, 0], ncols]
matrix[diag_, offsets_, ncols_] := MapThread[row[#, #2, ncols] &, diag, offsets]
m = a, 0, 0, 0, b, c, d, e, f, 0, 0, 0, b, c
matrix[diag, offsets, ncols] // MatrixForm
answered 27 mins ago
C. E.C. E.
52.9k3 gold badges102 silver badges210 bronze badges
$endgroup$
$begingroup$
Thank you very much for the effort, as per my test, this solution seems to fully solve the problem!
$endgroup$
– aleksander_si
5 mins ago
add a comment |
$begingroup$
PadRight[Flatten /@ a, 0, 0, b, c]
a, 0, 0, 0, b, c
answered 7 hours ago
kglrkglr
203k10 gold badges232 silver badges463 bronze badges
$endgroup$
$begingroup$
Yes, in a situation where the diagonal matrix is as per the below example, then this approach should not be applied: a,b,0,0,c}
$endgroup$
– aleksander_si
17 mins ago
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
$begingroup$
kglr's solution fails if the number of elements on the diagonal in any of the preceding rows is larger than the number of elements on the diagonal of the last row.
For example:
PadRight[Flatten /@ a, d, e, 0, 0, b, c] // MatrixForm
m = a, 0, 0, 0, b, c, d, e, f, 0, 0, 0, b, c
PadRight[Flatten /@ m] // MatrixForm
Here is a different solution:
diag = Flatten[#] & /@ Diagonal[m];
ncols = Length@Flatten[diag];
offsets = Most@Prepend[Accumulate[Length /@ diag], 0];
row[values_, offset_, ncols_] := PadRight[ArrayPad[values, offset, 0], ncols]
matrix[diag_, offsets_, ncols_] := MapThread[row[#, #2, ncols] &, diag, offsets]
m = a, 0, 0, 0, b, c, d, e, f, 0, 0, 0, b, c
matrix[diag, offsets, ncols] // MatrixForm
answered 27 mins ago
C. E.C. E.
52.9k3 gold badges102 silver badges210 bronze badges
$endgroup$
$begingroup$
Thank you very much for the effort, as per my test, this solution seems to fully solve the problem!
$endgroup$
– aleksander_si
5 mins ago
add a comment |
$begingroup$
kglr's solution fails if the number of elements on the diagonal in any of the preceding rows is larger than the number of elements on the diagonal of the last row.
For example:
PadRight[Flatten /@ a, d, e, 0, 0, b, c] // MatrixForm
m = a, 0, 0, 0, b, c, d, e, f, 0, 0, 0, b, c
PadRight[Flatten /@ m] // MatrixForm
Here is a different solution:
diag = Flatten[#] & /@ Diagonal[m];
ncols = Length@Flatten[diag];
offsets = Most@Prepend[Accumulate[Length /@ diag], 0];
row[values_, offset_, ncols_] := PadRight[ArrayPad[values, offset, 0], ncols]
matrix[diag_, offsets_, ncols_] := MapThread[row[#, #2, ncols] &, diag, offsets]
m = a, 0, 0, 0, b, c, d, e, f, 0, 0, 0, b, c
matrix[diag, offsets, ncols] // MatrixForm
answered 27 mins ago
C. E.C. E.
52.9k3 gold badges102 silver badges210 bronze badges
$endgroup$
$begingroup$
Thank you very much for the effort, as per my test, this solution seems to fully solve the problem!
$endgroup$
– aleksander_si
5 mins ago
add a comment |
$begingroup$
kglr's solution fails if the number of elements on the diagonal in any of the preceding rows is larger than the number of elements on the diagonal of the last row.
For example:
PadRight[Flatten /@ a, d, e, 0, 0, b, c] // MatrixForm
m = a, 0, 0, 0, b, c, d, e, f, 0, 0, 0, b, c
PadRight[Flatten /@ m] // MatrixForm
Here is a different solution:
diag = Flatten[#] & /@ Diagonal[m];
ncols = Length@Flatten[diag];
offsets = Most@Prepend[Accumulate[Length /@ diag], 0];
row[values_, offset_, ncols_] := PadRight[ArrayPad[values, offset, 0], ncols]
matrix[diag_, offsets_, ncols_] := MapThread[row[#, #2, ncols] &, diag, offsets]
m = a, 0, 0, 0, b, c, d, e, f, 0, 0, 0, b, c
matrix[diag, offsets, ncols] // MatrixForm
answered 27 mins ago
C. E.C. E.
52.9k3 gold badges102 silver badges210 bronze badges
$endgroup$
kglr's solution fails if the number of elements on the diagonal in any of the preceding rows is larger than the number of elements on the diagonal of the last row.
For example:
PadRight[Flatten /@ a, d, e, 0, 0, b, c] // MatrixForm
m = a, 0, 0, 0, b, c, d, e, f, 0, 0, 0, b, c
PadRight[Flatten /@ m] // MatrixForm
Here is a different solution:
diag = Flatten[#] & /@ Diagonal[m];
ncols = Length@Flatten[diag];
offsets = Most@Prepend[Accumulate[Length /@ diag], 0];
row[values_, offset_, ncols_] := PadRight[ArrayPad[values, offset, 0], ncols]
matrix[diag_, offsets_, ncols_] := MapThread[row[#, #2, ncols] &, diag, offsets]
m = a, 0, 0, 0, b, c, d, e, f, 0, 0, 0, b, c
matrix[diag, offsets, ncols] // MatrixForm
answered 27 mins ago
C. E.C. E.
52.9k3 gold badges102 silver badges210 bronze badges
edited 19 mins ago
answered 27 mins ago
C. E.C. E.
52.9k3 gold badges102 silver badges210 bronze badges
answered 27 mins ago
C. E.C. E.
52.9k3 gold badges102 silver badges210 bronze badges
answered 27 mins ago
C. E.C. E.
52.9k3 gold badges102 silver badges210 bronze badges
52.9k3 gold badges102 silver badges210 bronze badges
$begingroup$
Thank you very much for the effort, as per my test, this solution seems to fully solve the problem!
$endgroup$
– aleksander_si
5 mins ago
add a comment |
$begingroup$
Thank you very much for the effort, as per my test, this solution seems to fully solve the problem!
$endgroup$
– aleksander_si
5 mins ago
$begingroup$
Thank you very much for the effort, as per my test, this solution seems to fully solve the problem!
$endgroup$
– aleksander_si
5 mins ago
$begingroup$
Thank you very much for the effort, as per my test, this solution seems to fully solve the problem!
$endgroup$
– aleksander_si
5 mins ago
add a comment |
$begingroup$
PadRight[Flatten /@ a, 0, 0, b, c]
a, 0, 0, 0, b, c
answered 7 hours ago
kglrkglr
203k10 gold badges232 silver badges463 bronze badges
$endgroup$
$begingroup$
Yes, in a situation where the diagonal matrix is as per the below example, then this approach should not be applied: a,b,0,0,c}
$endgroup$
– aleksander_si
17 mins ago
add a comment |
$begingroup$
PadRight[Flatten /@ a, 0, 0, b, c]
a, 0, 0, 0, b, c
answered 7 hours ago
kglrkglr
203k10 gold badges232 silver badges463 bronze badges
$endgroup$
$begingroup$
Yes, in a situation where the diagonal matrix is as per the below example, then this approach should not be applied: a,b,0,0,c}
$endgroup$
– aleksander_si
17 mins ago
add a comment |
$begingroup$
PadRight[Flatten /@ a, 0, 0, b, c]
a, 0, 0, 0, b, c
answered 7 hours ago
kglrkglr
203k10 gold badges232 silver badges463 bronze badges
$endgroup$
PadRight[Flatten /@ a, 0, 0, b, c]
a, 0, 0, 0, b, c
answered 7 hours ago
kglrkglr
203k10 gold badges232 silver badges463 bronze badges
answered 7 hours ago
kglrkglr
203k10 gold badges232 silver badges463 bronze badges
answered 7 hours ago
kglrkglr
203k10 gold badges232 silver badges463 bronze badges
answered 7 hours ago
kglrkglr
203k10 gold badges232 silver badges463 bronze badges
203k10 gold badges232 silver badges463 bronze badges
$begingroup$
Yes, in a situation where the diagonal matrix is as per the below example, then this approach should not be applied: a,b,0,0,c}
$endgroup$
– aleksander_si
17 mins ago
add a comment |
$begingroup$
Yes, in a situation where the diagonal matrix is as per the below example, then this approach should not be applied: a,b,0,0,c}
$endgroup$
– aleksander_si
17 mins ago
$begingroup$
Yes, in a situation where the diagonal matrix is as per the below example, then this approach should not be applied: a,b,0,0,c}
$endgroup$
– aleksander_si
17 mins ago
$begingroup$
Yes, in a situation where the diagonal matrix is as per the below example, then this approach should not be applied: a,b,0,0,c}
$endgroup$
– aleksander_si
17 mins ago
add a comment |
aleksander_si is a new contributor. Be nice, and check out our Code of Conduct.
aleksander_si is a new contributor. Be nice, and check out our Code of Conduct.
aleksander_si is a new contributor. Be nice, and check out our Code of Conduct.
aleksander_si is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
Could you clarify what should happen for e.g. the matrix
a, d, e, 0, 0, b, cas in my answer? Whether my answer is correct or not hinges on this, because it gives a different answer than kglr.$endgroup$
– C. E.
20 mins ago