Using good method to produce a regular matrix












2












$begingroup$


The matrixform is as follow, and how can I use good method to produce it?enter image description here



H = {{1, 1, 0}, {2, 2, 0}, {0, 1, 1},

     {0, 2, 2}, {1, 0, 1}, {2, 0, 2}}





list-manipulation matrix array







share|improve this question











$endgroup$












  • $begingroup$
    If the picture can't show,then here is the matrixform:H={{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}
    $endgroup$
    – KarryMa
    Apr 6 at 9:22












  • $begingroup$
    KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, {ConstantArray[1, {3, 3}], RotateRight[Range[3]]}], {{1}, {2}} ]
    $endgroup$
    – Henrik Schumacher
    Apr 6 at 9:25












  • $begingroup$
    Great,thank you very much!
    $endgroup$
    – KarryMa
    Apr 6 at 9:26










  • $begingroup$
    Or Normal@KroneckerProduct[ SparseArray[{Band[{1, 1}] -> 1, Band[{1, 2}] -> 1, Band[{3, 1}] -> 1}, {3, 3}], {{1}, {2}} ].
    $endgroup$
    – Henrik Schumacher
    Apr 6 at 9:27






  • 3




    $begingroup$
    Transpose@KroneckerProduct[Permutations[{1, 0, 1}], {1, 2}]
    $endgroup$
    – LouisB
    Apr 6 at 9:45
















2












$begingroup$


The matrixform is as follow, and how can I use good method to produce it?enter image description here



H = {{1, 1, 0}, {2, 2, 0}, {0, 1, 1},

     {0, 2, 2}, {1, 0, 1}, {2, 0, 2}}





list-manipulation matrix array







share|improve this question











$endgroup$












  • $begingroup$
    If the picture can't show,then here is the matrixform:H={{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}
    $endgroup$
    – KarryMa
    Apr 6 at 9:22












  • $begingroup$
    KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, {ConstantArray[1, {3, 3}], RotateRight[Range[3]]}], {{1}, {2}} ]
    $endgroup$
    – Henrik Schumacher
    Apr 6 at 9:25












  • $begingroup$
    Great,thank you very much!
    $endgroup$
    – KarryMa
    Apr 6 at 9:26










  • $begingroup$
    Or Normal@KroneckerProduct[ SparseArray[{Band[{1, 1}] -> 1, Band[{1, 2}] -> 1, Band[{3, 1}] -> 1}, {3, 3}], {{1}, {2}} ].
    $endgroup$
    – Henrik Schumacher
    Apr 6 at 9:27






  • 3




    $begingroup$
    Transpose@KroneckerProduct[Permutations[{1, 0, 1}], {1, 2}]
    $endgroup$
    – LouisB
    Apr 6 at 9:45














2












2








2


0



$begingroup$


The matrixform is as follow, and how can I use good method to produce it?enter image description here



H = {{1, 1, 0}, {2, 2, 0}, {0, 1, 1},

     {0, 2, 2}, {1, 0, 1}, {2, 0, 2}}





list-manipulation matrix array







share|improve this question











$endgroup$




The matrixform is as follow, and how can I use good method to produce it?enter image description here



H = {{1, 1, 0}, {2, 2, 0}, {0, 1, 1},

     {0, 2, 2}, {1, 0, 1}, {2, 0, 2}}





list-manipulation matrix array




list-manipulation matrix array






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Apr 7 at 13:06









MarcoB

38.8k558116




38.8k558116










asked Apr 6 at 9:18









KarryMaKarryMa

163




163












  • $begingroup$
    If the picture can't show,then here is the matrixform:H={{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}
    $endgroup$
    – KarryMa
    Apr 6 at 9:22












  • $begingroup$
    KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, {ConstantArray[1, {3, 3}], RotateRight[Range[3]]}], {{1}, {2}} ]
    $endgroup$
    – Henrik Schumacher
    Apr 6 at 9:25












  • $begingroup$
    Great,thank you very much!
    $endgroup$
    – KarryMa
    Apr 6 at 9:26










  • $begingroup$
    Or Normal@KroneckerProduct[ SparseArray[{Band[{1, 1}] -> 1, Band[{1, 2}] -> 1, Band[{3, 1}] -> 1}, {3, 3}], {{1}, {2}} ].
    $endgroup$
    – Henrik Schumacher
    Apr 6 at 9:27






  • 3




    $begingroup$
    Transpose@KroneckerProduct[Permutations[{1, 0, 1}], {1, 2}]
    $endgroup$
    – LouisB
    Apr 6 at 9:45


















  • $begingroup$
    If the picture can't show,then here is the matrixform:H={{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}
    $endgroup$
    – KarryMa
    Apr 6 at 9:22












  • $begingroup$
    KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, {ConstantArray[1, {3, 3}], RotateRight[Range[3]]}], {{1}, {2}} ]
    $endgroup$
    – Henrik Schumacher
    Apr 6 at 9:25












  • $begingroup$
    Great,thank you very much!
    $endgroup$
    – KarryMa
    Apr 6 at 9:26










  • $begingroup$
    Or Normal@KroneckerProduct[ SparseArray[{Band[{1, 1}] -> 1, Band[{1, 2}] -> 1, Band[{3, 1}] -> 1}, {3, 3}], {{1}, {2}} ].
    $endgroup$
    – Henrik Schumacher
    Apr 6 at 9:27






  • 3




    $begingroup$
    Transpose@KroneckerProduct[Permutations[{1, 0, 1}], {1, 2}]
    $endgroup$
    – LouisB
    Apr 6 at 9:45
















$begingroup$
If the picture can't show,then here is the matrixform:H={{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}
$endgroup$
– KarryMa
Apr 6 at 9:22






$begingroup$
If the picture can't show,then here is the matrixform:H={{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}
$endgroup$
– KarryMa
Apr 6 at 9:22














$begingroup$
KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, {ConstantArray[1, {3, 3}], RotateRight[Range[3]]}], {{1}, {2}} ]
$endgroup$
– Henrik Schumacher
Apr 6 at 9:25






$begingroup$
KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, {ConstantArray[1, {3, 3}], RotateRight[Range[3]]}], {{1}, {2}} ]
$endgroup$
– Henrik Schumacher
Apr 6 at 9:25














$begingroup$
Great,thank you very much!
$endgroup$
– KarryMa
Apr 6 at 9:26




$begingroup$
Great,thank you very much!
$endgroup$
– KarryMa
Apr 6 at 9:26












$begingroup$
Or Normal@KroneckerProduct[ SparseArray[{Band[{1, 1}] -> 1, Band[{1, 2}] -> 1, Band[{3, 1}] -> 1}, {3, 3}], {{1}, {2}} ].
$endgroup$
– Henrik Schumacher
Apr 6 at 9:27




$begingroup$
Or Normal@KroneckerProduct[ SparseArray[{Band[{1, 1}] -> 1, Band[{1, 2}] -> 1, Band[{3, 1}] -> 1}, {3, 3}], {{1}, {2}} ].
$endgroup$
– Henrik Schumacher
Apr 6 at 9:27




3




3




$begingroup$
Transpose@KroneckerProduct[Permutations[{1, 0, 1}], {1, 2}]
$endgroup$
– LouisB
Apr 6 at 9:45




$begingroup$
Transpose@KroneckerProduct[Permutations[{1, 0, 1}], {1, 2}]
$endgroup$
– LouisB
Apr 6 at 9:45










2 Answers
2






active

oldest

votes


















2












$begingroup$

IntegerDigits[{12,24,4,8,10,20},3,3]



{{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}




also.. 



s = Transpose[Permutations /@ {{1, 1, 0}, {2, 2, 0}}];

Flatten[{{s[[1]]},Reverse@Rest@s},2]



{{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}







share|improve this answer











$endgroup$









  • 2




    $begingroup$
    I like the answer,thanks!
    $endgroup$
    – KarryMa
    Apr 6 at 9:55










  • $begingroup$
    @KarryMa Alternatively to PadLeft you can add 3 as 3rd argument IntegerDigits[{12, 24, 4, 8, 10, 20}, 3, 3].
    $endgroup$
    – Coolwater
    Apr 6 at 14:49










  • $begingroup$
    yes, you are so right!
    $endgroup$
    – J42161217
    Apr 6 at 14:51



















1












$begingroup$

A nice tool for this job is ArrayFlatten[ ]



a = {{1}, {2}}


$left(
begin{array}{c}
1 \
2 \
end{array}
right)$



Not sure why your rows are ordered the way they are. Are you trying to have a non-zero diagonal? 



{{a,a,0},{0,a,a},{a,0,a}}// ArrayFlatten


$left(
begin{array}{ccc}
1 & 1 & 0 \
2 & 2 & 0 \
0 & 1 & 1 \
0 & 2 & 2 \
1 & 0 & 1 \
2 & 0 & 2 \
end{array}
right)$






share|improve this answer











$endgroup$













  • $begingroup$
    This is not the exact result the OP is asking. The order of the elements is different. Also the "permutation" solution is already posted in my answer
    $endgroup$
    – J42161217
    Apr 6 at 14:58










  • $begingroup$
    My purpose was to introduce ArrayFlatten. Not enough info to algorithmically determine the order of perms.
    $endgroup$
    – MikeY
    Apr 6 at 15:27










  • $begingroup$
    Actually,this is the answer I originally wanted,then I will use ''PauliMatrix'' and ''Map'' to get what I want.Thank you for your help!
    $endgroup$
    – KarryMa
    Apr 7 at 2:35












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

IntegerDigits[{12,24,4,8,10,20},3,3]



{{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}




also.. 



s = Transpose[Permutations /@ {{1, 1, 0}, {2, 2, 0}}];

Flatten[{{s[[1]]},Reverse@Rest@s},2]



{{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}







share|improve this answer











$endgroup$









  • 2




    $begingroup$
    I like the answer,thanks!
    $endgroup$
    – KarryMa
    Apr 6 at 9:55










  • $begingroup$
    @KarryMa Alternatively to PadLeft you can add 3 as 3rd argument IntegerDigits[{12, 24, 4, 8, 10, 20}, 3, 3].
    $endgroup$
    – Coolwater
    Apr 6 at 14:49










  • $begingroup$
    yes, you are so right!
    $endgroup$
    – J42161217
    Apr 6 at 14:51
















2












$begingroup$

IntegerDigits[{12,24,4,8,10,20},3,3]



{{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}




also.. 



s = Transpose[Permutations /@ {{1, 1, 0}, {2, 2, 0}}];

Flatten[{{s[[1]]},Reverse@Rest@s},2]



{{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}







share|improve this answer











$endgroup$









  • 2




    $begingroup$
    I like the answer,thanks!
    $endgroup$
    – KarryMa
    Apr 6 at 9:55










  • $begingroup$
    @KarryMa Alternatively to PadLeft you can add 3 as 3rd argument IntegerDigits[{12, 24, 4, 8, 10, 20}, 3, 3].
    $endgroup$
    – Coolwater
    Apr 6 at 14:49










  • $begingroup$
    yes, you are so right!
    $endgroup$
    – J42161217
    Apr 6 at 14:51














2












2








2





$begingroup$

IntegerDigits[{12,24,4,8,10,20},3,3]



{{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}




also.. 



s = Transpose[Permutations /@ {{1, 1, 0}, {2, 2, 0}}];

Flatten[{{s[[1]]},Reverse@Rest@s},2]



{{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}







share|improve this answer











$endgroup$



IntegerDigits[{12,24,4,8,10,20},3,3]



{{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}




also.. 



s = Transpose[Permutations /@ {{1, 1, 0}, {2, 2, 0}}];

Flatten[{{s[[1]]},Reverse@Rest@s},2]



{{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}








share|improve this answer














share|improve this answer



share|improve this answer








edited Apr 6 at 14:52

























answered Apr 6 at 9:29









J42161217J42161217

4,688324




4,688324








  • 2




    $begingroup$
    I like the answer,thanks!
    $endgroup$
    – KarryMa
    Apr 6 at 9:55










  • $begingroup$
    @KarryMa Alternatively to PadLeft you can add 3 as 3rd argument IntegerDigits[{12, 24, 4, 8, 10, 20}, 3, 3].
    $endgroup$
    – Coolwater
    Apr 6 at 14:49










  • $begingroup$
    yes, you are so right!
    $endgroup$
    – J42161217
    Apr 6 at 14:51














  • 2




    $begingroup$
    I like the answer,thanks!
    $endgroup$
    – KarryMa
    Apr 6 at 9:55










  • $begingroup$
    @KarryMa Alternatively to PadLeft you can add 3 as 3rd argument IntegerDigits[{12, 24, 4, 8, 10, 20}, 3, 3].
    $endgroup$
    – Coolwater
    Apr 6 at 14:49










  • $begingroup$
    yes, you are so right!
    $endgroup$
    – J42161217
    Apr 6 at 14:51








2




2




$begingroup$
I like the answer,thanks!
$endgroup$
– KarryMa
Apr 6 at 9:55




$begingroup$
I like the answer,thanks!
$endgroup$
– KarryMa
Apr 6 at 9:55












$begingroup$
@KarryMa Alternatively to PadLeft you can add 3 as 3rd argument IntegerDigits[{12, 24, 4, 8, 10, 20}, 3, 3].
$endgroup$
– Coolwater
Apr 6 at 14:49




$begingroup$
@KarryMa Alternatively to PadLeft you can add 3 as 3rd argument IntegerDigits[{12, 24, 4, 8, 10, 20}, 3, 3].
$endgroup$
– Coolwater
Apr 6 at 14:49












$begingroup$
yes, you are so right!
$endgroup$
– J42161217
Apr 6 at 14:51




$begingroup$
yes, you are so right!
$endgroup$
– J42161217
Apr 6 at 14:51











1












$begingroup$

A nice tool for this job is ArrayFlatten[ ]



a = {{1}, {2}}


$left(
begin{array}{c}
1 \
2 \
end{array}
right)$



Not sure why your rows are ordered the way they are. Are you trying to have a non-zero diagonal? 



{{a,a,0},{0,a,a},{a,0,a}}// ArrayFlatten


$left(
begin{array}{ccc}
1 & 1 & 0 \
2 & 2 & 0 \
0 & 1 & 1 \
0 & 2 & 2 \
1 & 0 & 1 \
2 & 0 & 2 \
end{array}
right)$






share|improve this answer











$endgroup$













  • $begingroup$
    This is not the exact result the OP is asking. The order of the elements is different. Also the "permutation" solution is already posted in my answer
    $endgroup$
    – J42161217
    Apr 6 at 14:58










  • $begingroup$
    My purpose was to introduce ArrayFlatten. Not enough info to algorithmically determine the order of perms.
    $endgroup$
    – MikeY
    Apr 6 at 15:27










  • $begingroup$
    Actually,this is the answer I originally wanted,then I will use ''PauliMatrix'' and ''Map'' to get what I want.Thank you for your help!
    $endgroup$
    – KarryMa
    Apr 7 at 2:35
















1












$begingroup$

A nice tool for this job is ArrayFlatten[ ]



a = {{1}, {2}}


$left(
begin{array}{c}
1 \
2 \
end{array}
right)$



Not sure why your rows are ordered the way they are. Are you trying to have a non-zero diagonal? 



{{a,a,0},{0,a,a},{a,0,a}}// ArrayFlatten


$left(
begin{array}{ccc}
1 & 1 & 0 \
2 & 2 & 0 \
0 & 1 & 1 \
0 & 2 & 2 \
1 & 0 & 1 \
2 & 0 & 2 \
end{array}
right)$






share|improve this answer











$endgroup$













  • $begingroup$
    This is not the exact result the OP is asking. The order of the elements is different. Also the "permutation" solution is already posted in my answer
    $endgroup$
    – J42161217
    Apr 6 at 14:58










  • $begingroup$
    My purpose was to introduce ArrayFlatten. Not enough info to algorithmically determine the order of perms.
    $endgroup$
    – MikeY
    Apr 6 at 15:27










  • $begingroup$
    Actually,this is the answer I originally wanted,then I will use ''PauliMatrix'' and ''Map'' to get what I want.Thank you for your help!
    $endgroup$
    – KarryMa
    Apr 7 at 2:35














1












1








1





$begingroup$

A nice tool for this job is ArrayFlatten[ ]



a = {{1}, {2}}


$left(
begin{array}{c}
1 \
2 \
end{array}
right)$



Not sure why your rows are ordered the way they are. Are you trying to have a non-zero diagonal? 



{{a,a,0},{0,a,a},{a,0,a}}// ArrayFlatten


$left(
begin{array}{ccc}
1 & 1 & 0 \
2 & 2 & 0 \
0 & 1 & 1 \
0 & 2 & 2 \
1 & 0 & 1 \
2 & 0 & 2 \
end{array}
right)$






share|improve this answer











$endgroup$



A nice tool for this job is ArrayFlatten[ ]



a = {{1}, {2}}


$left(
begin{array}{c}
1 \
2 \
end{array}
right)$



Not sure why your rows are ordered the way they are. Are you trying to have a non-zero diagonal? 



{{a,a,0},{0,a,a},{a,0,a}}// ArrayFlatten


$left(
begin{array}{ccc}
1 & 1 & 0 \
2 & 2 & 0 \
0 & 1 & 1 \
0 & 2 & 2 \
1 & 0 & 1 \
2 & 0 & 2 \
end{array}
right)$







share|improve this answer














share|improve this answer



share|improve this answer








edited Apr 6 at 15:25

























answered Apr 6 at 13:21









MikeYMikeY

3,828916




3,828916












  • $begingroup$
    This is not the exact result the OP is asking. The order of the elements is different. Also the "permutation" solution is already posted in my answer
    $endgroup$
    – J42161217
    Apr 6 at 14:58










  • $begingroup$
    My purpose was to introduce ArrayFlatten. Not enough info to algorithmically determine the order of perms.
    $endgroup$
    – MikeY
    Apr 6 at 15:27










  • $begingroup$
    Actually,this is the answer I originally wanted,then I will use ''PauliMatrix'' and ''Map'' to get what I want.Thank you for your help!
    $endgroup$
    – KarryMa
    Apr 7 at 2:35


















  • $begingroup$
    This is not the exact result the OP is asking. The order of the elements is different. Also the "permutation" solution is already posted in my answer
    $endgroup$
    – J42161217
    Apr 6 at 14:58










  • $begingroup$
    My purpose was to introduce ArrayFlatten. Not enough info to algorithmically determine the order of perms.
    $endgroup$
    – MikeY
    Apr 6 at 15:27










  • $begingroup$
    Actually,this is the answer I originally wanted,then I will use ''PauliMatrix'' and ''Map'' to get what I want.Thank you for your help!
    $endgroup$
    – KarryMa
    Apr 7 at 2:35
















$begingroup$
This is not the exact result the OP is asking. The order of the elements is different. Also the "permutation" solution is already posted in my answer
$endgroup$
– J42161217
Apr 6 at 14:58




$begingroup$
This is not the exact result the OP is asking. The order of the elements is different. Also the "permutation" solution is already posted in my answer
$endgroup$
– J42161217
Apr 6 at 14:58












$begingroup$
My purpose was to introduce ArrayFlatten. Not enough info to algorithmically determine the order of perms.
$endgroup$
– MikeY
Apr 6 at 15:27




$begingroup$
My purpose was to introduce ArrayFlatten. Not enough info to algorithmically determine the order of perms.
$endgroup$
– MikeY
Apr 6 at 15:27












$begingroup$
Actually,this is the answer I originally wanted,then I will use ''PauliMatrix'' and ''Map'' to get what I want.Thank you for your help!
$endgroup$
– KarryMa
Apr 7 at 2:35




$begingroup$
Actually,this is the answer I originally wanted,then I will use ''PauliMatrix'' and ''Map'' to get what I want.Thank you for your help!
$endgroup$
– KarryMa
Apr 7 at 2:35


















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Paul Cézanne

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Pour l’article ayant un titre homophone, voir Sézanne. Paul Cézanne.mw-parser-output .entete.artiste{background-image:url("//upload.wikimedia.org/wikipedia/commons/f/f3/Picto_infobox_artiste.png")} Autoportrait vers 1861 Naissance 19 janvier 1839 Aix-en-Provence (France) Décès 22 octobre 1906 (à 67 ans) Aix-en-Provence (France) Nationalité Française Activité Peintre Formation Académie de Charles Suisse Maître Joseph Gilbert, Antoine Guillemet, Camille Pissarro Mouvement Impressionnisme postimpressionnisme Influencé par Eugène Delacroix, Gustave Courbet, Véronèse A influencé Braque, Picasso, Matisse Conjoint Hortense Fiquet (depuis 1886) Enfant Paul Cézanne  ( d ) Œuvres principales La Montagne Sainte-Victoire, Portrait de madame Cézanne signature modifier - modifier le code - modifier Wikidata Paul Cézanne , né le 19 janvier 1839 à Aix-en-Provence et mort le 22 octobre 1906 dans la mê...
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UIScrollView CustomStickyHeader Resize height generates problems when scroll is too fast

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0 I have been trying for a while to create a UIViewController that has a CustomStickyHeader and a UITableView. The idea I have been following is use the UITableView scrolling for changing the header view with the value of the contentOffset. The layout for the ViewController (AlbumsViewController) looks like: The UItableView Scrolling functions from the delegate are implemented like that: func scrollViewDidScroll(_ scrollView: UIScrollView) { let endReached = contentTableView.contentOffset.y >= (contentTableView.contentSize.height - contentTableView.frame.size.height) let isUP = lastContentOffset > scrollView.contentOffset.y if scrollView.contentOffset.y < 0 { headerHeightConstraint.constant += abs(scrollView.contentOffset.y) } else{ if !endReached{ ...
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Angular material date-picker (MatDatepicker) auto completes the date on focus out

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3 3 I am using angular material/mat-datepicker in my project and whenever the user enters one digit inside of it, and goes out of focus - the date picker is auto completing with a random date. Example Just enter 1 and go out of focus, or enter 5/5/20 The date picker is adding date automatically Update : Also I am using reactive forms and I'm validating the input with that approach Can I disable this ? angular datepicker angular-material angular6 angular-material-datetimepicker share | improve this question edited Nov 20 '18 at 13:21 ...
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