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5

$begingroup$

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

list-manipulation matrix

share|improve this question

edited 8 hours ago

aleksander_si

asked 8 hours ago

aleksander_sialeksander_si

2844 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.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  5 mins ago
  
  

  
  
  
  

add a comment | 

5

$begingroup$

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

list-manipulation matrix

share|improve this question

edited 8 hours ago

aleksander_si

asked 8 hours ago

aleksander_sialeksander_si

2844 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.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  5 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

list-manipulation matrix

share|improve this question

edited 8 hours ago

aleksander_si

asked 8 hours ago

aleksander_sialeksander_si

2844 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.

$endgroup$

DiagonalMatrix[Hold/@a, b, c]//ReleaseHold

a, 0, 0, 0, b, c

share|improve this question

edited 8 hours ago

aleksander_si

asked 8 hours ago

aleksander_sialeksander_si

2844 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.

share|improve this question

edited 8 hours ago

aleksander_si

asked 8 hours ago

aleksander_sialeksander_si

2844 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

2844 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

2844 bronze badges

2 Answers
2

active

oldest

votes

1

$begingroup$

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

share|improve this answer

edited 3 mins ago

answered 12 mins ago

C. E.C. E.

52.9k33 gold badges102102 silver badges210210 bronze badges

$endgroup$

add a comment | 

6

$begingroup$

PadRight[Flatten /@ a, 0, 0, b, c]

a, 0, 0, 0, b, c

share|improve this answer

answered 7 hours ago

kglrkglr

203k1010 gold badges232232 silver badges463463 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very neat solution!! I just hope I stated the question in a form which makes the problem searchable.
  $endgroup$
  – aleksander_si
  10 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, in a situation where the diagonal matrix is as per the below example, then this approach should not be applied:
  $endgroup$
  – aleksander_si
  2 mins ago
  
  

  
  
  
  

add a comment | 

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1

$begingroup$

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

share|improve this answer

edited 3 mins ago

answered 12 mins ago

C. E.C. E.

52.9k33 gold badges102102 silver badges210210 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

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

share|improve this answer

edited 3 mins ago

answered 12 mins ago

C. E.C. E.

52.9k33 gold badges102102 silver badges210210 bronze badges

$endgroup$

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

share|improve this answer

edited 3 mins ago

answered 12 mins ago

C. E.C. E.

52.9k33 gold badges102102 silver badges210210 bronze badges

$endgroup$

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

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

share|improve this answer

edited 3 mins ago

answered 12 mins ago

C. E.C. E.

52.9k33 gold badges102102 silver badges210210 bronze badges

answered 12 mins ago

C. E.C. E.

52.9k33 gold badges102102 silver badges210210 bronze badges

answered 12 mins ago

C. E.C. E.

52.9k33 gold badges102102 silver badges210210 bronze badges

6

$begingroup$

PadRight[Flatten /@ a, 0, 0, b, c]

a, 0, 0, 0, b, c

share|improve this answer

answered 7 hours ago

kglrkglr

203k1010 gold badges232232 silver badges463463 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very neat solution!! I just hope I stated the question in a form which makes the problem searchable.
  $endgroup$
  – aleksander_si
  10 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, in a situation where the diagonal matrix is as per the below example, then this approach should not be applied:
  $endgroup$
  – aleksander_si
  2 mins ago
  
  

  
  
  
  

add a comment | 

6

$begingroup$

PadRight[Flatten /@ a, 0, 0, b, c]

a, 0, 0, 0, b, c

share|improve this answer

answered 7 hours ago

kglrkglr

203k1010 gold badges232232 silver badges463463 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very neat solution!! I just hope I stated the question in a form which makes the problem searchable.
  $endgroup$
  – aleksander_si
  10 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, in a situation where the diagonal matrix is as per the below example, then this approach should not be applied:
  $endgroup$
  – aleksander_si
  2 mins ago
  
  

  
  
  
  

add a comment | 

$begingroup$

PadRight[Flatten /@ a, 0, 0, b, c]

a, 0, 0, 0, b, c

share|improve this answer

answered 7 hours ago

kglrkglr

203k1010 gold badges232232 silver badges463463 bronze badges

$endgroup$

PadRight[Flatten /@ a, 0, 0, b, c]

share|improve this answer

answered 7 hours ago

kglrkglr

203k1010 gold badges232232 silver badges463463 bronze badges

answered 7 hours ago

kglrkglr

203k1010 gold badges232232 silver badges463463 bronze badges

answered 7 hours ago

kglrkglr

203k1010 gold badges232232 silver badges463463 bronze badges