Alternating Total

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up vote
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down vote

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Any cleaner solutions than what I came up with? (Maybe by multiplying every other entry by -1)

altTotal[x_] := Total@x[[;; ;; 2]] - Total@x[[2 ;; ;; 2]]

list-manipulation

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asked Aug 21 at 21:43

qwr

1878

add a comment | 

up vote
3
down vote

favorite

1

Any cleaner solutions than what I came up with? (Maybe by multiplying every other entry by -1)

altTotal[x_] := Total@x[[;; ;; 2]] - Total@x[[2 ;; ;; 2]]

list-manipulation

share|improve this question

asked Aug 21 at 21:43

qwr

1878

add a comment | 

up vote
3
down vote

favorite

1

up vote
3
down vote

favorite

1

1

Any cleaner solutions than what I came up with? (Maybe by multiplying every other entry by -1)

altTotal[x_] := Total@x[[;; ;; 2]] - Total@x[[2 ;; ;; 2]]

list-manipulation

share|improve this question

asked Aug 21 at 21:43

qwr

1878

Any cleaner solutions than what I came up with? (Maybe by multiplying every other entry by -1)

altTotal[x_] := Total@x[[;; ;; 2]] - Total@x[[2 ;; ;; 2]]

list-manipulation

share|improve this question

asked Aug 21 at 21:43

qwr

1878

share|improve this question

share|improve this question

asked Aug 21 at 21:43

qwr

1878

asked Aug 21 at 21:43

qwr

1878

asked Aug 21 at 21:43

qwr

1878

1878

add a comment | 

add a comment | 

5 Answers
5

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oldest

votes

up vote
3
down vote

Maybe

altTotal2[x_] := Subtract @@ Total[Partition[x, 2]]

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answered Aug 21 at 22:01

Henrik Schumacher

36.2k249102

add a comment | 

up vote
3
down vote

altTotal3[x_] := Total@x - 2 Total@x[[2 ;; ;; 2]]

is a bit faster.

share|improve this answer

answered Aug 21 at 22:10

JimB

15k12456

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is a very nice example for floating point arithmetic being significantly faster than memory operations.
  – Henrik Schumacher
  Aug 22 at 9:47
  

  
  
  
  

add a comment | 

up vote
2
down vote

Here's a way to take + and - each successive member using Riffle:

altTotal5[x_]:=Total[x Riffle[ConstantArray[1, Length[x]], -1][[1 ;; Length[x]]]]

share|improve this answer

answered Aug 22 at 3:48

bill s

50.6k373142

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up vote
2
down vote

altTotal4[x_] := Total@Developer`PartitionMap[Subtract @@ # &, x, 2]

Edit: altTotal4 is quite surprisingly slow. The following Fold is much better but still not competititve.

altTotal5[x_] := First@Fold[With[s = #1[[2]], s*#2 + #1[[1]], -s] &, 0, 1, -1, x]

share|improve this answer

edited Aug 22 at 12:38

answered Aug 22 at 0:15

Alan

5,7621022

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's slower by more than two orders of magnitude than altTotal3.
  – Henrik Schumacher
  Aug 22 at 9:44
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

"Maybe" goes as below:

altTotal4 = SparseArray[i_?EvenQ -> -1, Length[#], 1].# &

Or

altTotal5[x_] := Total@MapIndexed[(-1)^(1 + First[#2]) # &, x]

share|improve this answer

edited Aug 22 at 3:27

answered Aug 22 at 3:21

Αλέξανδρος Ζεγγ

1,419720

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote

Maybe

altTotal2[x_] := Subtract @@ Total[Partition[x, 2]]

share|improve this answer

answered Aug 21 at 22:01

Henrik Schumacher

36.2k249102

add a comment | 

up vote
3
down vote

Maybe

altTotal2[x_] := Subtract @@ Total[Partition[x, 2]]

share|improve this answer

answered Aug 21 at 22:01

Henrik Schumacher

36.2k249102

add a comment | 

up vote
3
down vote

up vote
3
down vote

Maybe

altTotal2[x_] := Subtract @@ Total[Partition[x, 2]]

share|improve this answer

answered Aug 21 at 22:01

Henrik Schumacher

36.2k249102

Maybe

altTotal2[x_] := Subtract @@ Total[Partition[x, 2]]

share|improve this answer

answered Aug 21 at 22:01

Henrik Schumacher

36.2k249102

share|improve this answer

share|improve this answer

answered Aug 21 at 22:01

Henrik Schumacher

36.2k249102

answered Aug 21 at 22:01

Henrik Schumacher

36.2k249102

answered Aug 21 at 22:01

Henrik Schumacher

36.2k249102

36.2k249102

add a comment | 

add a comment | 

up vote
3
down vote

altTotal3[x_] := Total@x - 2 Total@x[[2 ;; ;; 2]]

is a bit faster.

share|improve this answer

answered Aug 21 at 22:10

JimB

15k12456

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is a very nice example for floating point arithmetic being significantly faster than memory operations.
  – Henrik Schumacher
  Aug 22 at 9:47
  

  
  
  
  

add a comment | 

up vote
3
down vote

altTotal3[x_] := Total@x - 2 Total@x[[2 ;; ;; 2]]

is a bit faster.

share|improve this answer

answered Aug 21 at 22:10

JimB

15k12456

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is a very nice example for floating point arithmetic being significantly faster than memory operations.
  – Henrik Schumacher
  Aug 22 at 9:47
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

altTotal3[x_] := Total@x - 2 Total@x[[2 ;; ;; 2]]

is a bit faster.

share|improve this answer

answered Aug 21 at 22:10

JimB

15k12456

altTotal3[x_] := Total@x - 2 Total@x[[2 ;; ;; 2]]

is a bit faster.

share|improve this answer

answered Aug 21 at 22:10

JimB

15k12456

share|improve this answer

share|improve this answer

answered Aug 21 at 22:10

JimB

15k12456

answered Aug 21 at 22:10

JimB

15k12456

answered Aug 21 at 22:10

JimB

15k12456

15k12456

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is a very nice example for floating point arithmetic being significantly faster than memory operations.
  – Henrik Schumacher
  Aug 22 at 9:47
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is a very nice example for floating point arithmetic being significantly faster than memory operations.
  – Henrik Schumacher
  Aug 22 at 9:47
  

  
  
  
  

That is a very nice example for floating point arithmetic being significantly faster than memory operations.
– Henrik Schumacher
Aug 22 at 9:47

That is a very nice example for floating point arithmetic being significantly faster than memory operations.
– Henrik Schumacher
Aug 22 at 9:47

add a comment | 

up vote
2
down vote

Here's a way to take + and - each successive member using Riffle:

altTotal5[x_]:=Total[x Riffle[ConstantArray[1, Length[x]], -1][[1 ;; Length[x]]]]

share|improve this answer

answered Aug 22 at 3:48

bill s

50.6k373142

add a comment | 

up vote
2
down vote

Here's a way to take + and - each successive member using Riffle:

altTotal5[x_]:=Total[x Riffle[ConstantArray[1, Length[x]], -1][[1 ;; Length[x]]]]

share|improve this answer

answered Aug 22 at 3:48

bill s

50.6k373142

add a comment | 

up vote
2
down vote

up vote
2
down vote

Here's a way to take + and - each successive member using Riffle:

altTotal5[x_]:=Total[x Riffle[ConstantArray[1, Length[x]], -1][[1 ;; Length[x]]]]

share|improve this answer

answered Aug 22 at 3:48

bill s

50.6k373142

Here's a way to take + and - each successive member using Riffle:

altTotal5[x_]:=Total[x Riffle[ConstantArray[1, Length[x]], -1][[1 ;; Length[x]]]]

share|improve this answer

answered Aug 22 at 3:48

bill s

50.6k373142

share|improve this answer

share|improve this answer

answered Aug 22 at 3:48

bill s

50.6k373142

answered Aug 22 at 3:48

bill s

50.6k373142

answered Aug 22 at 3:48

bill s

50.6k373142

50.6k373142

add a comment | 

add a comment | 

up vote
2
down vote

altTotal4[x_] := Total@Developer`PartitionMap[Subtract @@ # &, x, 2]

Edit: altTotal4 is quite surprisingly slow. The following Fold is much better but still not competititve.

altTotal5[x_] := First@Fold[With[s = #1[[2]], s*#2 + #1[[1]], -s] &, 0, 1, -1, x]

share|improve this answer

edited Aug 22 at 12:38

answered Aug 22 at 0:15

Alan

5,7621022

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's slower by more than two orders of magnitude than altTotal3.
  – Henrik Schumacher
  Aug 22 at 9:44
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

altTotal4[x_] := Total@Developer`PartitionMap[Subtract @@ # &, x, 2]

Edit: altTotal4 is quite surprisingly slow. The following Fold is much better but still not competititve.

altTotal5[x_] := First@Fold[With[s = #1[[2]], s*#2 + #1[[1]], -s] &, 0, 1, -1, x]

share|improve this answer

edited Aug 22 at 12:38

answered Aug 22 at 0:15

Alan

5,7621022

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's slower by more than two orders of magnitude than altTotal3.
  – Henrik Schumacher
  Aug 22 at 9:44
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

altTotal4[x_] := Total@Developer`PartitionMap[Subtract @@ # &, x, 2]

Edit: altTotal4 is quite surprisingly slow. The following Fold is much better but still not competititve.

altTotal5[x_] := First@Fold[With[s = #1[[2]], s*#2 + #1[[1]], -s] &, 0, 1, -1, x]

share|improve this answer

edited Aug 22 at 12:38

answered Aug 22 at 0:15

Alan

5,7621022

altTotal4[x_] := Total@Developer`PartitionMap[Subtract @@ # &, x, 2]

Edit: altTotal4 is quite surprisingly slow. The following Fold is much better but still not competititve.

altTotal5[x_] := First@Fold[With[s = #1[[2]], s*#2 + #1[[1]], -s] &, 0, 1, -1, x]

share|improve this answer

edited Aug 22 at 12:38

answered Aug 22 at 0:15

Alan

5,7621022

share|improve this answer

share|improve this answer

edited Aug 22 at 12:38

edited Aug 22 at 12:38

edited Aug 22 at 12:38

answered Aug 22 at 0:15

Alan

5,7621022

answered Aug 22 at 0:15

Alan

5,7621022

answered Aug 22 at 0:15

Alan

5,7621022

5,7621022

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's slower by more than two orders of magnitude than altTotal3.
  – Henrik Schumacher
  Aug 22 at 9:44
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's slower by more than two orders of magnitude than altTotal3.
  – Henrik Schumacher
  Aug 22 at 9:44
  
  

  
  
  
  

That's slower by more than two orders of magnitude than altTotal3.
– Henrik Schumacher
Aug 22 at 9:44

That's slower by more than two orders of magnitude than altTotal3.
– Henrik Schumacher
Aug 22 at 9:44

add a comment | 

up vote
0
down vote

"Maybe" goes as below:

altTotal4 = SparseArray[i_?EvenQ -> -1, Length[#], 1].# &

Or

altTotal5[x_] := Total@MapIndexed[(-1)^(1 + First[#2]) # &, x]

share|improve this answer

edited Aug 22 at 3:27

answered Aug 22 at 3:21

Αλέξανδρος Ζεγγ

1,419720

add a comment | 

up vote
0
down vote

"Maybe" goes as below:

altTotal4 = SparseArray[i_?EvenQ -> -1, Length[#], 1].# &

Or

altTotal5[x_] := Total@MapIndexed[(-1)^(1 + First[#2]) # &, x]

share|improve this answer

edited Aug 22 at 3:27

answered Aug 22 at 3:21

Αλέξανδρος Ζεγγ

1,419720

add a comment | 

up vote
0
down vote

up vote
0
down vote

"Maybe" goes as below:

altTotal4 = SparseArray[i_?EvenQ -> -1, Length[#], 1].# &

Or

altTotal5[x_] := Total@MapIndexed[(-1)^(1 + First[#2]) # &, x]

share|improve this answer

edited Aug 22 at 3:27

answered Aug 22 at 3:21

Αλέξανδρος Ζεγγ

1,419720

"Maybe" goes as below:

altTotal4 = SparseArray[i_?EvenQ -> -1, Length[#], 1].# &

Or

altTotal5[x_] := Total@MapIndexed[(-1)^(1 + First[#2]) # &, x]

share|improve this answer

edited Aug 22 at 3:27

answered Aug 22 at 3:21

Αλέξανδρος Ζεγγ

1,419720

share|improve this answer

share|improve this answer

edited Aug 22 at 3:27

edited Aug 22 at 3:27

edited Aug 22 at 3:27

answered Aug 22 at 3:21

Αλέξανδρος Ζεγγ

1,419720

answered Aug 22 at 3:21

Αλέξανδρος Ζεγγ

1,419720

answered Aug 22 at 3:21

Αλέξανδρος Ζεγγ

1,419720

1,419720

add a comment | 

add a comment | 

 

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