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3

$begingroup$

I'm trying to repeat each element of a list x number of times, where x is the corresponding element of the same position in another list.

For example, I have list A = 1,2,3,4 and another list B = 3,1,4,2 and I'm trying to get C = 1,1,1,2,3,3,3,3,4,4.

How do I get C from A and B?

list-manipulation

share|improve this question

edited Sep 20 at 7:56

Alexey Popkov

48.1k44 gold badges118118 silver badges282282 bronze badges

asked Sep 18 at 16:12

reemodelsreemodels

21388 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What have you tried?
  $endgroup$
  – Edmund
  Sep 18 at 16:23
  

  
  
  
  

add a comment
 | 

3

$begingroup$

I'm trying to repeat each element of a list x number of times, where x is the corresponding element of the same position in another list.

For example, I have list A = 1,2,3,4 and another list B = 3,1,4,2 and I'm trying to get C = 1,1,1,2,3,3,3,3,4,4.

How do I get C from A and B?

list-manipulation

share|improve this question

edited Sep 20 at 7:56

Alexey Popkov

48.1k44 gold badges118118 silver badges282282 bronze badges

asked Sep 18 at 16:12

reemodelsreemodels

21388 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What have you tried?
  $endgroup$
  – Edmund
  Sep 18 at 16:23
  

  
  
  
  

add a comment
 | 

$begingroup$

I'm trying to repeat each element of a list x number of times, where x is the corresponding element of the same position in another list.

For example, I have list A = 1,2,3,4 and another list B = 3,1,4,2 and I'm trying to get C = 1,1,1,2,3,3,3,3,4,4.

How do I get C from A and B?

list-manipulation

share|improve this question

edited Sep 20 at 7:56

Alexey Popkov

48.1k44 gold badges118118 silver badges282282 bronze badges

asked Sep 18 at 16:12

reemodelsreemodels

21388 bronze badges

$endgroup$

share|improve this question

edited Sep 20 at 7:56

Alexey Popkov

48.1k44 gold badges118118 silver badges282282 bronze badges

asked Sep 18 at 16:12

reemodelsreemodels

21388 bronze badges

share|improve this question

edited Sep 20 at 7:56

Alexey Popkov

48.1k44 gold badges118118 silver badges282282 bronze badges

asked Sep 18 at 16:12

reemodelsreemodels

21388 bronze badges

edited Sep 20 at 7:56

Alexey Popkov

48.1k44 gold badges118118 silver badges282282 bronze badges

edited Sep 20 at 7:56

Alexey Popkov

48.1k44 gold badges118118 silver badges282282 bronze badges

asked Sep 18 at 16:12

reemodelsreemodels

21388 bronze badges

asked Sep 18 at 16:12

reemodelsreemodels

21388 bronze badges

5 Answers
5

active

oldest

votes

13

$begingroup$

Join @@ MapThread[Table, A,B]

1, 1, 1, 2, 3, 3, 3, 3, 4, 4

Join @@ Table @@@ Transpose @ A,B

1, 1, 1, 2, 3, 3, 3, 3, 4, 4

Join @@ MapThread[ConstantArray, A, B]

1, 1, 1, 2, 3, 3, 3, 3, 4, 4

Also

Internal`RepetitionFromMultiplicity @ Transpose[A, B]

1, 1, 1, 2, 3, 3, 3, 3, 4, 4

share|improve this answer

edited Sep 18 at 20:24

answered Sep 18 at 17:05

kglrkglr

227k1010 gold badges257257 silver badges518518 bronze badges

$endgroup$

add a comment
 | 

5

$begingroup$

kglr's proposals are nice, and the last (undocumented) one is very nice. As a variation, here is a solution using Inner[] + Flatten[]:

Flatten[Inner[ConstantArray, 1, 2, 3, 4, 3, 1, 4, 2, List]]
 1, 1, 1, 2, 3, 3, 3, 3, 4, 4

As kglr notes, a shorter version is

Inner[ConstantArray, 1, 2, 3, 4, 3, 1, 4, 2, Join]

or in version 10 and later,

Inner[Table, 1, 2, 3, 4, 3, 1, 4, 2, Join]

share|improve this answer

edited Sep 21 at 5:19

answered Sep 19 at 13:36

J. M. will be back soon♦J. M. will be back soon

104k1111 gold badges322322 silver badges484484 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  shorter Inner[ConstantArray, 1, 2, 3, 4, 3, 1, 4, 2, Join] (+1)
  $endgroup$
  – kglr
  Sep 19 at 14:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed, thank you!
  $endgroup$
  – J. M. will be back soon♦
  Sep 21 at 5:19
  

  
  
  
  

add a comment
 | 

3

$begingroup$

a = 1, 2, 3, 4;

b = 3, 1, 4, 2;

Using ConstantArray

c = Flatten[ConstantArray[#[[1]], #[[2]]] & /@
 Transpose[a, b]]

(* 1, 1, 1, 2, 3, 3, 3, 3, 4, 4 *)

or using Table

c = Flatten[Table[#[[1]], #[[2]]] & /@
 Transpose[a, b]]

(* 1, 1, 1, 2, 3, 3, 3, 3, 4, 4 *)

share|improve this answer

answered Sep 18 at 17:01

Bob HanlonBob Hanlon

69.6k33 gold badges3838 silver badges107107 bronze badges

$endgroup$

add a comment
 | 

3

$begingroup$

Another way of approaching this is to define a function that carries out the basic task. In this case, to repeat the x element y times.

f[x_, y_] := ConstantArray[x, y]; SetAttributes[f, Listable]

Making this function Listable allows very simple calling method:

f[a, b] // Flatten
1, 1, 1, 2, 3, 3, 3, 3, 4, 4

share|improve this answer

answered Sep 18 at 22:51

bill sbill s

57.9k33 gold badges8181 silver badges162162 bronze badges

$endgroup$

add a comment
 | 

0

$begingroup$

a = 1, 2, 3, 4;
b = 3, 1, 4, 2;

Flatten[Table[a[[n]], n, Length[a], b[[n]]]]
(* or *)
Flatten[Table[Table[a[[n]], b[[n]]], n, Length[a]]]

1, 1, 1, 2, 3, 3, 3, 3, 4, 4

Flatten[Table[ConstantArray[a[[n]], b[[n]]], n, Length[a]]]

1, 1, 1, 2, 3, 3, 3, 3, 4, 4

share|improve this answer

answered Sep 18 at 22:45

MelaGoMelaGo

3,83611 gold badge33 silver badges1212 bronze badges

$endgroup$

add a comment
 | 

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13

$begingroup$

Join @@ MapThread[Table, A,B]

1, 1, 1, 2, 3, 3, 3, 3, 4, 4

Join @@ Table @@@ Transpose @ A,B

1, 1, 1, 2, 3, 3, 3, 3, 4, 4

Join @@ MapThread[ConstantArray, A, B]

1, 1, 1, 2, 3, 3, 3, 3, 4, 4

Also

Internal`RepetitionFromMultiplicity @ Transpose[A, B]

1, 1, 1, 2, 3, 3, 3, 3, 4, 4

share|improve this answer

edited Sep 18 at 20:24

answered Sep 18 at 17:05

kglrkglr

227k1010 gold badges257257 silver badges518518 bronze badges

$endgroup$

add a comment
 | 

13

$begingroup$

Join @@ MapThread[Table, A,B]

1, 1, 1, 2, 3, 3, 3, 3, 4, 4

Join @@ Table @@@ Transpose @ A,B

1, 1, 1, 2, 3, 3, 3, 3, 4, 4

Join @@ MapThread[ConstantArray, A, B]

1, 1, 1, 2, 3, 3, 3, 3, 4, 4

Also

Internal`RepetitionFromMultiplicity @ Transpose[A, B]

1, 1, 1, 2, 3, 3, 3, 3, 4, 4

share|improve this answer

edited Sep 18 at 20:24

answered Sep 18 at 17:05

kglrkglr

227k1010 gold badges257257 silver badges518518 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Join @@ MapThread[Table, A,B]

1, 1, 1, 2, 3, 3, 3, 3, 4, 4

Join @@ Table @@@ Transpose @ A,B

1, 1, 1, 2, 3, 3, 3, 3, 4, 4

Join @@ MapThread[ConstantArray, A, B]

1, 1, 1, 2, 3, 3, 3, 3, 4, 4

Also

Internal`RepetitionFromMultiplicity @ Transpose[A, B]

1, 1, 1, 2, 3, 3, 3, 3, 4, 4

share|improve this answer

edited Sep 18 at 20:24

answered Sep 18 at 17:05

kglrkglr

227k1010 gold badges257257 silver badges518518 bronze badges

$endgroup$

Join @@ MapThread[Table, A,B]

Join @@ Table @@@ Transpose @ A,B

Join @@ MapThread[ConstantArray, A, B]

Internal`RepetitionFromMultiplicity @ Transpose[A, B]

share|improve this answer

edited Sep 18 at 20:24

answered Sep 18 at 17:05

kglrkglr

227k1010 gold badges257257 silver badges518518 bronze badges

answered Sep 18 at 17:05

kglrkglr

227k1010 gold badges257257 silver badges518518 bronze badges

answered Sep 18 at 17:05

kglrkglr

227k1010 gold badges257257 silver badges518518 bronze badges

5

$begingroup$

kglr's proposals are nice, and the last (undocumented) one is very nice. As a variation, here is a solution using Inner[] + Flatten[]:

Flatten[Inner[ConstantArray, 1, 2, 3, 4, 3, 1, 4, 2, List]]
 1, 1, 1, 2, 3, 3, 3, 3, 4, 4

As kglr notes, a shorter version is

Inner[ConstantArray, 1, 2, 3, 4, 3, 1, 4, 2, Join]

or in version 10 and later,

Inner[Table, 1, 2, 3, 4, 3, 1, 4, 2, Join]

share|improve this answer

edited Sep 21 at 5:19

answered Sep 19 at 13:36

J. M. will be back soon♦J. M. will be back soon

104k1111 gold badges322322 silver badges484484 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  shorter Inner[ConstantArray, 1, 2, 3, 4, 3, 1, 4, 2, Join] (+1)
  $endgroup$
  – kglr
  Sep 19 at 14:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed, thank you!
  $endgroup$
  – J. M. will be back soon♦
  Sep 21 at 5:19
  

  
  
  
  

add a comment
 | 

5

$begingroup$

kglr's proposals are nice, and the last (undocumented) one is very nice. As a variation, here is a solution using Inner[] + Flatten[]:

Flatten[Inner[ConstantArray, 1, 2, 3, 4, 3, 1, 4, 2, List]]
 1, 1, 1, 2, 3, 3, 3, 3, 4, 4

As kglr notes, a shorter version is

Inner[ConstantArray, 1, 2, 3, 4, 3, 1, 4, 2, Join]

or in version 10 and later,

Inner[Table, 1, 2, 3, 4, 3, 1, 4, 2, Join]

share|improve this answer

edited Sep 21 at 5:19

answered Sep 19 at 13:36

J. M. will be back soon♦J. M. will be back soon

104k1111 gold badges322322 silver badges484484 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  shorter Inner[ConstantArray, 1, 2, 3, 4, 3, 1, 4, 2, Join] (+1)
  $endgroup$
  – kglr
  Sep 19 at 14:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed, thank you!
  $endgroup$
  – J. M. will be back soon♦
  Sep 21 at 5:19
  

  
  
  
  

add a comment
 | 

$begingroup$

kglr's proposals are nice, and the last (undocumented) one is very nice. As a variation, here is a solution using Inner[] + Flatten[]:

Flatten[Inner[ConstantArray, 1, 2, 3, 4, 3, 1, 4, 2, List]]
 1, 1, 1, 2, 3, 3, 3, 3, 4, 4

As kglr notes, a shorter version is

Inner[ConstantArray, 1, 2, 3, 4, 3, 1, 4, 2, Join]

or in version 10 and later,

Inner[Table, 1, 2, 3, 4, 3, 1, 4, 2, Join]

share|improve this answer

edited Sep 21 at 5:19

answered Sep 19 at 13:36

J. M. will be back soon♦J. M. will be back soon

104k1111 gold badges322322 silver badges484484 bronze badges

$endgroup$

Flatten[Inner[ConstantArray, 1, 2, 3, 4, 3, 1, 4, 2, List]]
 1, 1, 1, 2, 3, 3, 3, 3, 4, 4

Inner[ConstantArray, 1, 2, 3, 4, 3, 1, 4, 2, Join]

Inner[Table, 1, 2, 3, 4, 3, 1, 4, 2, Join]

share|improve this answer

edited Sep 21 at 5:19

answered Sep 19 at 13:36

J. M. will be back soon♦J. M. will be back soon

104k1111 gold badges322322 silver badges484484 bronze badges

answered Sep 19 at 13:36

J. M. will be back soon♦J. M. will be back soon

104k1111 gold badges322322 silver badges484484 bronze badges

answered Sep 19 at 13:36

J. M. will be back soon♦J. M. will be back soon

104k1111 gold badges322322 silver badges484484 bronze badges

3

$begingroup$

a = 1, 2, 3, 4;

b = 3, 1, 4, 2;

Using ConstantArray

c = Flatten[ConstantArray[#[[1]], #[[2]]] & /@
 Transpose[a, b]]

(* 1, 1, 1, 2, 3, 3, 3, 3, 4, 4 *)

or using Table

c = Flatten[Table[#[[1]], #[[2]]] & /@
 Transpose[a, b]]

(* 1, 1, 1, 2, 3, 3, 3, 3, 4, 4 *)

share|improve this answer

answered Sep 18 at 17:01

Bob HanlonBob Hanlon

69.6k33 gold badges3838 silver badges107107 bronze badges

$endgroup$

add a comment
 | 

3

$begingroup$

a = 1, 2, 3, 4;

b = 3, 1, 4, 2;

Using ConstantArray

c = Flatten[ConstantArray[#[[1]], #[[2]]] & /@
 Transpose[a, b]]

(* 1, 1, 1, 2, 3, 3, 3, 3, 4, 4 *)

or using Table

c = Flatten[Table[#[[1]], #[[2]]] & /@
 Transpose[a, b]]

(* 1, 1, 1, 2, 3, 3, 3, 3, 4, 4 *)

share|improve this answer

answered Sep 18 at 17:01

Bob HanlonBob Hanlon

69.6k33 gold badges3838 silver badges107107 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

a = 1, 2, 3, 4;

b = 3, 1, 4, 2;

Using ConstantArray

c = Flatten[ConstantArray[#[[1]], #[[2]]] & /@
 Transpose[a, b]]

(* 1, 1, 1, 2, 3, 3, 3, 3, 4, 4 *)

or using Table

c = Flatten[Table[#[[1]], #[[2]]] & /@
 Transpose[a, b]]

(* 1, 1, 1, 2, 3, 3, 3, 3, 4, 4 *)

share|improve this answer

answered Sep 18 at 17:01

Bob HanlonBob Hanlon

69.6k33 gold badges3838 silver badges107107 bronze badges

$endgroup$

a = 1, 2, 3, 4;

b = 3, 1, 4, 2;

c = Flatten[ConstantArray[#[[1]], #[[2]]] & /@
 Transpose[a, b]]

(* 1, 1, 1, 2, 3, 3, 3, 3, 4, 4 *)

c = Flatten[Table[#[[1]], #[[2]]] & /@
 Transpose[a, b]]

(* 1, 1, 1, 2, 3, 3, 3, 3, 4, 4 *)

share|improve this answer

answered Sep 18 at 17:01

Bob HanlonBob Hanlon

69.6k33 gold badges3838 silver badges107107 bronze badges

answered Sep 18 at 17:01

Bob HanlonBob Hanlon

69.6k33 gold badges3838 silver badges107107 bronze badges

answered Sep 18 at 17:01

Bob HanlonBob Hanlon

69.6k33 gold badges3838 silver badges107107 bronze badges

3

$begingroup$

Another way of approaching this is to define a function that carries out the basic task. In this case, to repeat the x element y times.

f[x_, y_] := ConstantArray[x, y]; SetAttributes[f, Listable]

Making this function Listable allows very simple calling method:

f[a, b] // Flatten
1, 1, 1, 2, 3, 3, 3, 3, 4, 4

share|improve this answer

answered Sep 18 at 22:51

bill sbill s

57.9k33 gold badges8181 silver badges162162 bronze badges

$endgroup$

add a comment
 | 

3

$begingroup$

Another way of approaching this is to define a function that carries out the basic task. In this case, to repeat the x element y times.

f[x_, y_] := ConstantArray[x, y]; SetAttributes[f, Listable]

Making this function Listable allows very simple calling method:

f[a, b] // Flatten
1, 1, 1, 2, 3, 3, 3, 3, 4, 4

share|improve this answer

answered Sep 18 at 22:51

bill sbill s

57.9k33 gold badges8181 silver badges162162 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Another way of approaching this is to define a function that carries out the basic task. In this case, to repeat the x element y times.

f[x_, y_] := ConstantArray[x, y]; SetAttributes[f, Listable]

Making this function Listable allows very simple calling method:

f[a, b] // Flatten
1, 1, 1, 2, 3, 3, 3, 3, 4, 4

share|improve this answer

answered Sep 18 at 22:51

bill sbill s

57.9k33 gold badges8181 silver badges162162 bronze badges

$endgroup$

f[x_, y_] := ConstantArray[x, y]; SetAttributes[f, Listable]

f[a, b] // Flatten
1, 1, 1, 2, 3, 3, 3, 3, 4, 4

share|improve this answer

answered Sep 18 at 22:51

bill sbill s

57.9k33 gold badges8181 silver badges162162 bronze badges

answered Sep 18 at 22:51

bill sbill s

57.9k33 gold badges8181 silver badges162162 bronze badges

answered Sep 18 at 22:51

bill sbill s

57.9k33 gold badges8181 silver badges162162 bronze badges

0

$begingroup$

a = 1, 2, 3, 4;
b = 3, 1, 4, 2;

Flatten[Table[a[[n]], n, Length[a], b[[n]]]]
(* or *)
Flatten[Table[Table[a[[n]], b[[n]]], n, Length[a]]]

1, 1, 1, 2, 3, 3, 3, 3, 4, 4

Flatten[Table[ConstantArray[a[[n]], b[[n]]], n, Length[a]]]

1, 1, 1, 2, 3, 3, 3, 3, 4, 4

share|improve this answer

answered Sep 18 at 22:45

MelaGoMelaGo

3,83611 gold badge33 silver badges1212 bronze badges

$endgroup$

add a comment
 | 

0

$begingroup$

a = 1, 2, 3, 4;
b = 3, 1, 4, 2;

Flatten[Table[a[[n]], n, Length[a], b[[n]]]]
(* or *)
Flatten[Table[Table[a[[n]], b[[n]]], n, Length[a]]]

1, 1, 1, 2, 3, 3, 3, 3, 4, 4

Flatten[Table[ConstantArray[a[[n]], b[[n]]], n, Length[a]]]

1, 1, 1, 2, 3, 3, 3, 3, 4, 4

share|improve this answer

answered Sep 18 at 22:45

MelaGoMelaGo

3,83611 gold badge33 silver badges1212 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

a = 1, 2, 3, 4;
b = 3, 1, 4, 2;

Flatten[Table[a[[n]], n, Length[a], b[[n]]]]
(* or *)
Flatten[Table[Table[a[[n]], b[[n]]], n, Length[a]]]

1, 1, 1, 2, 3, 3, 3, 3, 4, 4

Flatten[Table[ConstantArray[a[[n]], b[[n]]], n, Length[a]]]

1, 1, 1, 2, 3, 3, 3, 3, 4, 4

share|improve this answer

answered Sep 18 at 22:45

MelaGoMelaGo

3,83611 gold badge33 silver badges1212 bronze badges

$endgroup$

a = 1, 2, 3, 4;
b = 3, 1, 4, 2;

Flatten[Table[a[[n]], n, Length[a], b[[n]]]]
(* or *)
Flatten[Table[Table[a[[n]], b[[n]]], n, Length[a]]]

Flatten[Table[ConstantArray[a[[n]], b[[n]]], n, Length[a]]]

share|improve this answer

answered Sep 18 at 22:45

MelaGoMelaGo

3,83611 gold badge33 silver badges1212 bronze badges

answered Sep 18 at 22:45

MelaGoMelaGo

3,83611 gold badge33 silver badges1212 bronze badges

answered Sep 18 at 22:45

MelaGoMelaGo

3,83611 gold badge33 silver badges1212 bronze badges