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How to apply differences on part of a list and keep the rest

Is there a good way to map a function over a list to lists exclusively of a certain depth?Monitor the list for changesHow to apply a function of several arguments to a list?Split according to List and apply ruleSetting the value of a list itemCombining Map with DropHow to combine Nest and list PartDifferences applied to a list of matricesApplying Rest and Most to sublists of listMultiplying a list a vectors by the same matrix

4

$begingroup$

I have a list,

l1 = a, b, 3, c, e, f, 5, k, n, k, 12, m, s, t, 1, y

and want to apply differences on the third parts and keep the parts right of the numerals collected.

My result should be

l2 = 2, c, k, 7, k, m, -11, m, y

I tried Map and MapAt, but I could not get anywhere. I could work around split things up and connect again. But is there a better way to do it?

list-manipulation

share|improve this question

edited 1 hour ago

m_goldberg

89.4k873201

asked 9 hours ago

user57467user57467

563

$endgroup$

add a comment | 

4

$begingroup$

I have a list,

l1 = a, b, 3, c, e, f, 5, k, n, k, 12, m, s, t, 1, y

and want to apply differences on the third parts and keep the parts right of the numerals collected.

My result should be

l2 = 2, c, k, 7, k, m, -11, m, y

I tried Map and MapAt, but I could not get anywhere. I could work around split things up and connect again. But is there a better way to do it?

list-manipulation

share|improve this question

edited 1 hour ago

m_goldberg

89.4k873201

asked 9 hours ago

user57467user57467

563

$endgroup$

add a comment | 

$begingroup$

I have a list,

l1 = a, b, 3, c, e, f, 5, k, n, k, 12, m, s, t, 1, y

and want to apply differences on the third parts and keep the parts right of the numerals collected.

My result should be

l2 = 2, c, k, 7, k, m, -11, m, y

I tried Map and MapAt, but I could not get anywhere. I could work around split things up and connect again. But is there a better way to do it?

list-manipulation

share|improve this question

edited 1 hour ago

m_goldberg

89.4k873201

asked 9 hours ago

user57467user57467

563

$endgroup$

l1 = a, b, 3, c, e, f, 5, k, n, k, 12, m, s, t, 1, y

l2 = 2, c, k, 7, k, m, -11, m, y

share|improve this question

edited 1 hour ago

m_goldberg

89.4k873201

asked 9 hours ago

user57467user57467

563

share|improve this question

edited 1 hour ago

m_goldberg

89.4k873201

asked 9 hours ago

user57467user57467

563

edited 1 hour ago

m_goldberg

89.4k873201

edited 1 hour ago

m_goldberg

89.4k873201

asked 9 hours ago

user57467user57467

563

asked 9 hours ago

user57467user57467

563

4 Answers
4

active

oldest

votes

4

$begingroup$

Perhaps this?:

l1 = a, b, 3, c, e, f, 5, k, n, k, 12, m, s, t, 1, y;

l2 = Differences[l1[[All, 3 ;;]]] /. b_ - a_ :> Sequence[a, b]
(* 2, c, k, 7, k, m, -11, m, y *)

It assumes the letter symbols are simple and not complicated expressions.

This is more complicated, but more robust:

Flatten /@ 
 Transpose@
 MapAt[Differences, 
 Partition[Transpose@l1[[All, 3 ;;]], 1, 2, 1, 1], 1, All, 1]

share|improve this answer

edited 8 hours ago

answered 9 hours ago

Michael E2Michael E2

151k12204485

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you. I have to digest your answer.
  $endgroup$
  – user57467
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How would you do it if l1= z,a, b, 3, c, w,e, f, 5, k, q,n, k, 12, m, p,s, t, 1, y;
  $endgroup$
  – user57467
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user57467 I would change the third column to the fourth: l1[[All, 4 ;;]]. You can programmatically get the column with col = Position[First@l1, _?NumericQ][[1, 1]] or col = Position[First@l1, _Integer][[1, 1]]; then use l1[[All, col ;;]].
  $endgroup$
  – Michael E2
  7 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

You can also use BlockMap as follows:

BlockMap[#[[3]].-1, 1, ## & @@ Flatten@#[[4 ;;]] &@* Transpose, l1, 2, 1]

2, c, k, 7, k, m, -11, m, y

or

BlockMap[#[[1]].-1, 1, ## & @@ Flatten@ #[[2 ;;]] &@*Transpose, l1[[All, 3 ;;]], 2, 1]

2, c, k, 7, k, m, -11, m, y

share|improve this answer

edited 7 hours ago

answered 7 hours ago

kglrkglr

190k10209427

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am impressed!
  $endgroup$
  – user57467
  7 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

This is very similar to kglr's first solution but picks the relevant quantities a bit more explicitly:

l2 = BlockMap[#[[2, 3]] - #[[1, 3]], #[[1, 4]], #[[2, 4]] &, l1, 2, 1]

2, c, k, 7, k, m, -11, m, y

With a parameter to change the symbolic column quickly:

l2 = With[col = 3,
 BlockMap[#[[2,col]] - #[[1,col]], #[[1,col+1]], #[[2,col+1]] &, l1, 2, 1]]

2, c, k, 7, k, m, -11, m, y

share|improve this answer

edited 6 hours ago

answered 7 hours ago

RomanRoman

6,86011134

$endgroup$

add a comment | 

0

$begingroup$

A solution with MapThread on an offset Partition.

MapThread[Sequence @@ #@#2 &, Differences, Identity, Transpose@#] & /@ 
 Partition[l1[[All, 3 ;;]], 2, 1]

2, c, k, 7, k, m, -11, m, y

Differences is applied to the integers while Identity preserves the form of the symbols.

Hope this helps.

share|improve this answer

answered 2 hours ago

EdmundEdmund

26.9k330103

$endgroup$

add a comment | 

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4

$begingroup$

Perhaps this?:

l1 = a, b, 3, c, e, f, 5, k, n, k, 12, m, s, t, 1, y;

l2 = Differences[l1[[All, 3 ;;]]] /. b_ - a_ :> Sequence[a, b]
(* 2, c, k, 7, k, m, -11, m, y *)

It assumes the letter symbols are simple and not complicated expressions.

This is more complicated, but more robust:

Flatten /@ 
 Transpose@
 MapAt[Differences, 
 Partition[Transpose@l1[[All, 3 ;;]], 1, 2, 1, 1], 1, All, 1]

share|improve this answer

edited 8 hours ago

answered 9 hours ago

Michael E2Michael E2

151k12204485

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you. I have to digest your answer.
  $endgroup$
  – user57467
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How would you do it if l1= z,a, b, 3, c, w,e, f, 5, k, q,n, k, 12, m, p,s, t, 1, y;
  $endgroup$
  – user57467
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user57467 I would change the third column to the fourth: l1[[All, 4 ;;]]. You can programmatically get the column with col = Position[First@l1, _?NumericQ][[1, 1]] or col = Position[First@l1, _Integer][[1, 1]]; then use l1[[All, col ;;]].
  $endgroup$
  – Michael E2
  7 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

Perhaps this?:

l1 = a, b, 3, c, e, f, 5, k, n, k, 12, m, s, t, 1, y;

l2 = Differences[l1[[All, 3 ;;]]] /. b_ - a_ :> Sequence[a, b]
(* 2, c, k, 7, k, m, -11, m, y *)

It assumes the letter symbols are simple and not complicated expressions.

This is more complicated, but more robust:

Flatten /@ 
 Transpose@
 MapAt[Differences, 
 Partition[Transpose@l1[[All, 3 ;;]], 1, 2, 1, 1], 1, All, 1]

share|improve this answer

edited 8 hours ago

answered 9 hours ago

Michael E2Michael E2

151k12204485

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you. I have to digest your answer.
  $endgroup$
  – user57467
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How would you do it if l1= z,a, b, 3, c, w,e, f, 5, k, q,n, k, 12, m, p,s, t, 1, y;
  $endgroup$
  – user57467
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user57467 I would change the third column to the fourth: l1[[All, 4 ;;]]. You can programmatically get the column with col = Position[First@l1, _?NumericQ][[1, 1]] or col = Position[First@l1, _Integer][[1, 1]]; then use l1[[All, col ;;]].
  $endgroup$
  – Michael E2
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Perhaps this?:

l1 = a, b, 3, c, e, f, 5, k, n, k, 12, m, s, t, 1, y;

l2 = Differences[l1[[All, 3 ;;]]] /. b_ - a_ :> Sequence[a, b]
(* 2, c, k, 7, k, m, -11, m, y *)

It assumes the letter symbols are simple and not complicated expressions.

This is more complicated, but more robust:

Flatten /@ 
 Transpose@
 MapAt[Differences, 
 Partition[Transpose@l1[[All, 3 ;;]], 1, 2, 1, 1], 1, All, 1]

share|improve this answer

edited 8 hours ago

answered 9 hours ago

Michael E2Michael E2

151k12204485

$endgroup$

l1 = a, b, 3, c, e, f, 5, k, n, k, 12, m, s, t, 1, y;

l2 = Differences[l1[[All, 3 ;;]]] /. b_ - a_ :> Sequence[a, b]
(* 2, c, k, 7, k, m, -11, m, y *)

Flatten /@ 
 Transpose@
 MapAt[Differences, 
 Partition[Transpose@l1[[All, 3 ;;]], 1, 2, 1, 1], 1, All, 1]

share|improve this answer

edited 8 hours ago

answered 9 hours ago

Michael E2Michael E2

151k12204485

answered 9 hours ago

Michael E2Michael E2

151k12204485

answered 9 hours ago

Michael E2Michael E2

151k12204485

3

$begingroup$

You can also use BlockMap as follows:

BlockMap[#[[3]].-1, 1, ## & @@ Flatten@#[[4 ;;]] &@* Transpose, l1, 2, 1]

2, c, k, 7, k, m, -11, m, y

or

BlockMap[#[[1]].-1, 1, ## & @@ Flatten@ #[[2 ;;]] &@*Transpose, l1[[All, 3 ;;]], 2, 1]

2, c, k, 7, k, m, -11, m, y

share|improve this answer

edited 7 hours ago

answered 7 hours ago

kglrkglr

190k10209427

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am impressed!
  $endgroup$
  – user57467
  7 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

You can also use BlockMap as follows:

BlockMap[#[[3]].-1, 1, ## & @@ Flatten@#[[4 ;;]] &@* Transpose, l1, 2, 1]

2, c, k, 7, k, m, -11, m, y

or

BlockMap[#[[1]].-1, 1, ## & @@ Flatten@ #[[2 ;;]] &@*Transpose, l1[[All, 3 ;;]], 2, 1]

2, c, k, 7, k, m, -11, m, y

share|improve this answer

edited 7 hours ago

answered 7 hours ago

kglrkglr

190k10209427

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am impressed!
  $endgroup$
  – user57467
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

You can also use BlockMap as follows:

BlockMap[#[[3]].-1, 1, ## & @@ Flatten@#[[4 ;;]] &@* Transpose, l1, 2, 1]

2, c, k, 7, k, m, -11, m, y

or

BlockMap[#[[1]].-1, 1, ## & @@ Flatten@ #[[2 ;;]] &@*Transpose, l1[[All, 3 ;;]], 2, 1]

2, c, k, 7, k, m, -11, m, y

share|improve this answer

edited 7 hours ago

answered 7 hours ago

kglrkglr

190k10209427

$endgroup$

BlockMap[#[[3]].-1, 1, ## & @@ Flatten@#[[4 ;;]] &@* Transpose, l1, 2, 1]

BlockMap[#[[1]].-1, 1, ## & @@ Flatten@ #[[2 ;;]] &@*Transpose, l1[[All, 3 ;;]], 2, 1]

share|improve this answer

edited 7 hours ago

answered 7 hours ago

kglrkglr

190k10209427

answered 7 hours ago

kglrkglr

190k10209427

answered 7 hours ago

kglrkglr

190k10209427

2

$begingroup$

This is very similar to kglr's first solution but picks the relevant quantities a bit more explicitly:

l2 = BlockMap[#[[2, 3]] - #[[1, 3]], #[[1, 4]], #[[2, 4]] &, l1, 2, 1]

2, c, k, 7, k, m, -11, m, y

With a parameter to change the symbolic column quickly:

l2 = With[col = 3,
 BlockMap[#[[2,col]] - #[[1,col]], #[[1,col+1]], #[[2,col+1]] &, l1, 2, 1]]

2, c, k, 7, k, m, -11, m, y

share|improve this answer

edited 6 hours ago

answered 7 hours ago

RomanRoman

6,86011134

$endgroup$

add a comment | 

2

$begingroup$

This is very similar to kglr's first solution but picks the relevant quantities a bit more explicitly:

l2 = BlockMap[#[[2, 3]] - #[[1, 3]], #[[1, 4]], #[[2, 4]] &, l1, 2, 1]

2, c, k, 7, k, m, -11, m, y

With a parameter to change the symbolic column quickly:

l2 = With[col = 3,
 BlockMap[#[[2,col]] - #[[1,col]], #[[1,col+1]], #[[2,col+1]] &, l1, 2, 1]]

2, c, k, 7, k, m, -11, m, y

share|improve this answer

edited 6 hours ago

answered 7 hours ago

RomanRoman

6,86011134

$endgroup$

add a comment | 

$begingroup$

This is very similar to kglr's first solution but picks the relevant quantities a bit more explicitly:

l2 = BlockMap[#[[2, 3]] - #[[1, 3]], #[[1, 4]], #[[2, 4]] &, l1, 2, 1]

2, c, k, 7, k, m, -11, m, y

With a parameter to change the symbolic column quickly:

l2 = With[col = 3,
 BlockMap[#[[2,col]] - #[[1,col]], #[[1,col+1]], #[[2,col+1]] &, l1, 2, 1]]

2, c, k, 7, k, m, -11, m, y

share|improve this answer

edited 6 hours ago

answered 7 hours ago

RomanRoman

6,86011134

$endgroup$

l2 = BlockMap[#[[2, 3]] - #[[1, 3]], #[[1, 4]], #[[2, 4]] &, l1, 2, 1]

l2 = With[col = 3,
 BlockMap[#[[2,col]] - #[[1,col]], #[[1,col+1]], #[[2,col+1]] &, l1, 2, 1]]

share|improve this answer

edited 6 hours ago

answered 7 hours ago

RomanRoman

6,86011134

answered 7 hours ago

RomanRoman

6,86011134

answered 7 hours ago

RomanRoman

6,86011134

0

$begingroup$

A solution with MapThread on an offset Partition.

MapThread[Sequence @@ #@#2 &, Differences, Identity, Transpose@#] & /@ 
 Partition[l1[[All, 3 ;;]], 2, 1]

2, c, k, 7, k, m, -11, m, y

Differences is applied to the integers while Identity preserves the form of the symbols.

Hope this helps.

share|improve this answer

answered 2 hours ago

EdmundEdmund

26.9k330103

$endgroup$

add a comment | 

0

$begingroup$

A solution with MapThread on an offset Partition.

MapThread[Sequence @@ #@#2 &, Differences, Identity, Transpose@#] & /@ 
 Partition[l1[[All, 3 ;;]], 2, 1]

2, c, k, 7, k, m, -11, m, y

Differences is applied to the integers while Identity preserves the form of the symbols.

Hope this helps.

share|improve this answer

answered 2 hours ago

EdmundEdmund

26.9k330103

$endgroup$

add a comment | 

$begingroup$

A solution with MapThread on an offset Partition.

MapThread[Sequence @@ #@#2 &, Differences, Identity, Transpose@#] & /@ 
 Partition[l1[[All, 3 ;;]], 2, 1]

2, c, k, 7, k, m, -11, m, y

Differences is applied to the integers while Identity preserves the form of the symbols.

Hope this helps.

share|improve this answer

answered 2 hours ago

EdmundEdmund

26.9k330103

$endgroup$

MapThread[Sequence @@ #@#2 &, Differences, Identity, Transpose@#] & /@ 
 Partition[l1[[All, 3 ;;]], 2, 1]

share|improve this answer

answered 2 hours ago

EdmundEdmund

26.9k330103

answered 2 hours ago

EdmundEdmund

26.9k330103

answered 2 hours ago

EdmundEdmund

26.9k330103