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Create chunks from an array

Generating a list with duplicate entries

MapThread with List ManupulationCombine nested list entriesList all “duplicate” elements in TallyGenerating a List of Numbers with “Table” Causing DeviationDuplicate rows with multiple entriesHow to Floor specific entries of list within a matrix?Map entries of StructuredArrayGenerating list of labeled stringDeleting duplicate equations from a list of equationsFlattening a list desiring a specific format

3

$begingroup$

How can I get a function that gives the following output?

listX = {a, b, c, d}

numberX = 3

myDuplicatesList[listX, numberX] 

{{a,a,a},{b,b,b},{c,c,c},{d,d,d}}

list-manipulation

share|improve this question

edited 58 mins ago

MarcoB

37k556113

asked 1 hour ago

lisalisa

161

New contributor

lisa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Wolfram Challenges.
  $endgroup$
  – J. M. is computer-less♦
  29 mins ago
  

  
  
  
  

add a comment | 

3

$begingroup$

How can I get a function that gives the following output?

listX = {a, b, c, d}

numberX = 3

myDuplicatesList[listX, numberX] 

{{a,a,a},{b,b,b},{c,c,c},{d,d,d}}

list-manipulation

share|improve this question

edited 58 mins ago

MarcoB

37k556113

asked 1 hour ago

lisalisa

161

New contributor

lisa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Wolfram Challenges.
  $endgroup$
  – J. M. is computer-less♦
  29 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

How can I get a function that gives the following output?

listX = {a, b, c, d}

numberX = 3

myDuplicatesList[listX, numberX] 

{{a,a,a},{b,b,b},{c,c,c},{d,d,d}}

list-manipulation

share|improve this question

edited 58 mins ago

MarcoB

37k556113

asked 1 hour ago

lisalisa

161

New contributor

lisa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

listX = {a, b, c, d}

numberX = 3

myDuplicatesList[listX, numberX] 

share|improve this question

edited 58 mins ago

MarcoB

37k556113

asked 1 hour ago

lisalisa

161

New contributor

lisa 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 58 mins ago

MarcoB

37k556113

asked 1 hour ago

lisalisa

161

New contributor

lisa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 58 mins ago

MarcoB

37k556113

edited 58 mins ago

MarcoB

37k556113

asked 1 hour ago

lisalisa

161

New contributor

lisa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 1 hour ago

lisalisa

161

4 Answers
4

active

oldest

votes

2

$begingroup$

one way might be

listX    = {a, b, c, d}

numberX  = 3

Transpose[{listX}].{Table[1, {numberX}]}

share|improve this answer

answered 1 hour ago

NasserNasser

58.1k489206

$endgroup$

add a comment | 

1

$begingroup$

Transpose@ConstantArray[listX, numberX]

{{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}

share|improve this answer

answered 20 mins ago

RomanRoman

2,674717

$endgroup$

add a comment | 

1

$begingroup$

Alternatively:

listX = {a, b, c};

numberX = 3;

Table[ConstantArray[i, numberX], {i, listX}]

{{a, a, a}, { b, b, b}, {c, c, c}}

Or:

listX={a, b, c, d};

numberX=3;



Transpose@Table[i, {numberX}, {i, listX}]

share|improve this answer

edited 20 mins ago

answered 33 mins ago

MarcoBMarcoB

37k556113

$endgroup$

add a comment | 

0

$begingroup$

Transpose[{listX}[[ConstantArray[1, numberX]]]]

share|improve this answer

answered 31 mins ago

Henrik SchumacherHenrik Schumacher

55.7k576154

$endgroup$

add a comment | 

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2

$begingroup$

one way might be

listX    = {a, b, c, d}

numberX  = 3

Transpose[{listX}].{Table[1, {numberX}]}

share|improve this answer

answered 1 hour ago

NasserNasser

58.1k489206

$endgroup$

add a comment | 

2

$begingroup$

one way might be

listX    = {a, b, c, d}

numberX  = 3

Transpose[{listX}].{Table[1, {numberX}]}

share|improve this answer

answered 1 hour ago

NasserNasser

58.1k489206

$endgroup$

add a comment | 

$begingroup$

one way might be

listX    = {a, b, c, d}

numberX  = 3

Transpose[{listX}].{Table[1, {numberX}]}

share|improve this answer

answered 1 hour ago

NasserNasser

58.1k489206

$endgroup$

listX    = {a, b, c, d}

numberX  = 3

Transpose[{listX}].{Table[1, {numberX}]}

share|improve this answer

answered 1 hour ago

NasserNasser

58.1k489206

answered 1 hour ago

NasserNasser

58.1k489206

answered 1 hour ago

NasserNasser

58.1k489206

1

$begingroup$

Transpose@ConstantArray[listX, numberX]

{{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}

share|improve this answer

answered 20 mins ago

RomanRoman

2,674717

$endgroup$

add a comment | 

1

$begingroup$

Transpose@ConstantArray[listX, numberX]

{{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}

share|improve this answer

answered 20 mins ago

RomanRoman

2,674717

$endgroup$

add a comment | 

$begingroup$

Transpose@ConstantArray[listX, numberX]

{{a, a, a}, {b, b, b}, {c, c, c}, {d, d, d}}

share|improve this answer

answered 20 mins ago

RomanRoman

2,674717

$endgroup$

Transpose@ConstantArray[listX, numberX]

share|improve this answer

answered 20 mins ago

RomanRoman

2,674717

answered 20 mins ago

RomanRoman

2,674717

answered 20 mins ago

RomanRoman

2,674717

1

$begingroup$

Alternatively:

listX = {a, b, c};

numberX = 3;

Table[ConstantArray[i, numberX], {i, listX}]

{{a, a, a}, { b, b, b}, {c, c, c}}

Or:

listX={a, b, c, d};

numberX=3;



Transpose@Table[i, {numberX}, {i, listX}]

share|improve this answer

edited 20 mins ago

answered 33 mins ago

MarcoBMarcoB

37k556113

$endgroup$

add a comment | 

1

$begingroup$

Alternatively:

listX = {a, b, c};

numberX = 3;

Table[ConstantArray[i, numberX], {i, listX}]

{{a, a, a}, { b, b, b}, {c, c, c}}

Or:

listX={a, b, c, d};

numberX=3;



Transpose@Table[i, {numberX}, {i, listX}]

share|improve this answer

edited 20 mins ago

answered 33 mins ago

MarcoBMarcoB

37k556113

$endgroup$

add a comment | 

$begingroup$

Alternatively:

listX = {a, b, c};

numberX = 3;

Table[ConstantArray[i, numberX], {i, listX}]

{{a, a, a}, { b, b, b}, {c, c, c}}

Or:

listX={a, b, c, d};

numberX=3;



Transpose@Table[i, {numberX}, {i, listX}]

share|improve this answer

edited 20 mins ago

answered 33 mins ago

MarcoBMarcoB

37k556113

$endgroup$

listX = {a, b, c};

numberX = 3;

Table[ConstantArray[i, numberX], {i, listX}]

listX={a, b, c, d};

numberX=3;



Transpose@Table[i, {numberX}, {i, listX}]

share|improve this answer

edited 20 mins ago

answered 33 mins ago

MarcoBMarcoB

37k556113

answered 33 mins ago

MarcoBMarcoB

37k556113

answered 33 mins ago

MarcoBMarcoB

37k556113

0

$begingroup$

Transpose[{listX}[[ConstantArray[1, numberX]]]]

share|improve this answer

answered 31 mins ago

Henrik SchumacherHenrik Schumacher

55.7k576154

$endgroup$

add a comment | 

0

$begingroup$

Transpose[{listX}[[ConstantArray[1, numberX]]]]

share|improve this answer

answered 31 mins ago

Henrik SchumacherHenrik Schumacher

55.7k576154

$endgroup$

add a comment | 

$begingroup$

Transpose[{listX}[[ConstantArray[1, numberX]]]]

share|improve this answer

answered 31 mins ago

Henrik SchumacherHenrik Schumacher

55.7k576154

$endgroup$

Transpose[{listX}[[ConstantArray[1, numberX]]]]

share|improve this answer

answered 31 mins ago

Henrik SchumacherHenrik Schumacher

55.7k576154

answered 31 mins ago

Henrik SchumacherHenrik Schumacher

55.7k576154

answered 31 mins ago

Henrik SchumacherHenrik Schumacher

55.7k576154