Help with sum specific elements of a matrix in mathematica

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I am trying to sum over specific indices in a matrix.

For example, if I have the matrix

a = 1, 2, 3, 3, 4, 5, 6, 6, 7, 8, 9, 9, 1, 5, 9, 7

i need to sum the index 2, 1 and 4, 3 , i.e 4 + 9 = 13 automatically.

I try the code

Sum[a[[ii]][[jj]], ii, 2, 4, 2, jj, 1, 4 , 2]

and this is equals to 20. (WTF) (i think this shows the result of 4+6+1+9) Why?

In a general form I need to compute the sum of indices 2, 1, 4, 3, 6, 5, 8, 7, 10, 9, 12, 11, 14,13, 16,15 for a 16x16 square matrix.

Thanks for the help

Regards

matrix

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edited 37 mins ago

kglr

165k8188388

asked 1 hour ago

Santiago Marín Agudelo

migrated from stats.stackexchange.com 50 mins ago

This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.

add a comment | 

up vote
1
down vote

favorite

I am trying to sum over specific indices in a matrix.

For example, if I have the matrix

a = 1, 2, 3, 3, 4, 5, 6, 6, 7, 8, 9, 9, 1, 5, 9, 7

i need to sum the index 2, 1 and 4, 3 , i.e 4 + 9 = 13 automatically.

I try the code

Sum[a[[ii]][[jj]], ii, 2, 4, 2, jj, 1, 4 , 2]

and this is equals to 20. (WTF) (i think this shows the result of 4+6+1+9) Why?

In a general form I need to compute the sum of indices 2, 1, 4, 3, 6, 5, 8, 7, 10, 9, 12, 11, 14,13, 16,15 for a 16x16 square matrix.

Thanks for the help

Regards

matrix

share|improve this question

edited 37 mins ago

kglr

165k8188388

asked 1 hour ago

Santiago Marín Agudelo

migrated from stats.stackexchange.com 50 mins ago

This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I am trying to sum over specific indices in a matrix.

For example, if I have the matrix

a = 1, 2, 3, 3, 4, 5, 6, 6, 7, 8, 9, 9, 1, 5, 9, 7

i need to sum the index 2, 1 and 4, 3 , i.e 4 + 9 = 13 automatically.

I try the code

Sum[a[[ii]][[jj]], ii, 2, 4, 2, jj, 1, 4 , 2]

and this is equals to 20. (WTF) (i think this shows the result of 4+6+1+9) Why?

In a general form I need to compute the sum of indices 2, 1, 4, 3, 6, 5, 8, 7, 10, 9, 12, 11, 14,13, 16,15 for a 16x16 square matrix.

Thanks for the help

Regards

matrix

share|improve this question

edited 37 mins ago

kglr

165k8188388

asked 1 hour ago

Santiago Marín Agudelo

I am trying to sum over specific indices in a matrix.

For example, if I have the matrix

a = 1, 2, 3, 3, 4, 5, 6, 6, 7, 8, 9, 9, 1, 5, 9, 7

i need to sum the index 2, 1 and 4, 3 , i.e 4 + 9 = 13 automatically.

I try the code

Sum[a[[ii]][[jj]], ii, 2, 4, 2, jj, 1, 4 , 2]

and this is equals to 20. (WTF) (i think this shows the result of 4+6+1+9) Why?

In a general form I need to compute the sum of indices 2, 1, 4, 3, 6, 5, 8, 7, 10, 9, 12, 11, 14,13, 16,15 for a 16x16 square matrix.

Thanks for the help

Regards

matrix

matrix

share|improve this question

edited 37 mins ago

kglr

165k8188388

asked 1 hour ago

Santiago Marín Agudelo

share|improve this question

edited 37 mins ago

kglr

165k8188388

asked 1 hour ago

Santiago Marín Agudelo

share|improve this question

share|improve this question

edited 37 mins ago

kglr

165k8188388

edited 37 mins ago

kglr

165k8188388

edited 37 mins ago

kglr

165k8188388

165k8188388

asked 1 hour ago

Santiago Marín Agudelo

asked 1 hour ago

Santiago Marín Agudelo

asked 1 hour ago

Santiago Marín Agudelo

migrated from stats.stackexchange.com 50 mins ago

This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.

migrated from stats.stackexchange.com 50 mins ago

This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
4
down vote

a = 1, 2, 3, 3, 4, 5, 6, 6, 7, 8, 9, 9, 1, 5, 9, 7;
positions = 2, 1, 4, 3;
Total@Extract[a, positions]

share|improve this answer

answered 1 hour ago

Alan

5,8691023

add a comment | 

up vote
2
down vote

To see why it doesn't work as you expected, look at the indices that you generated.

Table[ii, jj, ii, 2, 4, 2, jj, 1, 4, 2]

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

You generated four indices so the sum was over four entries. Use

a = 1, 2, 3, 3, 4, 5, 6, 6, 7, 8, 9, 9, 1, 5, 9, 7;

Sum[a[[Sequence @@ i]], i, 2, 1, 4, 3]

(* 13 *)

share|improve this answer

answered 21 mins ago

Bob Hanlon

56.2k23590

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up vote
1
down vote

I would have used Extract if Alan had not posted it first.

So, here is a way to use Sum:

Sum[a[[## & @@ i]], i, 2, 1, 4, 3]

13

But that is also posted a minute earlier by Bob Hanlon.

That leaves

☺ = ♯, ♯♯ [Function] +## & @@ (♯[[##]] & @@@ ♯♯);
☺[a, 2, 1, 4, 3]

13

and

☺☺ = +## & @@ (♯ [Function] #[[## & @@ ♯]]) /@ #2 &;
☺☺[a, 2, 1, 4, 3]

13

share|improve this answer

edited 9 mins ago

answered 20 mins ago

kglr

165k8188388

• 
  
  
  

  
  

  
  
  
  
  
  

  
  "A monad is just a monoid in the category of endofunctors, what's the problem?"
  – kglr
  8 mins ago
  

  
  
  
  

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote

a = 1, 2, 3, 3, 4, 5, 6, 6, 7, 8, 9, 9, 1, 5, 9, 7;
positions = 2, 1, 4, 3;
Total@Extract[a, positions]

share|improve this answer

answered 1 hour ago

Alan

5,8691023

add a comment | 

up vote
4
down vote

a = 1, 2, 3, 3, 4, 5, 6, 6, 7, 8, 9, 9, 1, 5, 9, 7;
positions = 2, 1, 4, 3;
Total@Extract[a, positions]

share|improve this answer

answered 1 hour ago

Alan

5,8691023

add a comment | 

up vote
4
down vote

up vote
4
down vote

a = 1, 2, 3, 3, 4, 5, 6, 6, 7, 8, 9, 9, 1, 5, 9, 7;
positions = 2, 1, 4, 3;
Total@Extract[a, positions]

share|improve this answer

answered 1 hour ago

Alan

5,8691023

a = 1, 2, 3, 3, 4, 5, 6, 6, 7, 8, 9, 9, 1, 5, 9, 7;
positions = 2, 1, 4, 3;
Total@Extract[a, positions]

share|improve this answer

answered 1 hour ago

Alan

5,8691023

share|improve this answer

share|improve this answer

answered 1 hour ago

Alan

5,8691023

answered 1 hour ago

Alan

5,8691023

answered 1 hour ago

Alan

5,8691023

5,8691023

add a comment | 

add a comment | 

up vote
2
down vote

To see why it doesn't work as you expected, look at the indices that you generated.

Table[ii, jj, ii, 2, 4, 2, jj, 1, 4, 2]

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

You generated four indices so the sum was over four entries. Use

a = 1, 2, 3, 3, 4, 5, 6, 6, 7, 8, 9, 9, 1, 5, 9, 7;

Sum[a[[Sequence @@ i]], i, 2, 1, 4, 3]

(* 13 *)

share|improve this answer

answered 21 mins ago

Bob Hanlon

56.2k23590

add a comment | 

up vote
2
down vote

To see why it doesn't work as you expected, look at the indices that you generated.

Table[ii, jj, ii, 2, 4, 2, jj, 1, 4, 2]

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

You generated four indices so the sum was over four entries. Use

a = 1, 2, 3, 3, 4, 5, 6, 6, 7, 8, 9, 9, 1, 5, 9, 7;

Sum[a[[Sequence @@ i]], i, 2, 1, 4, 3]

(* 13 *)

share|improve this answer

answered 21 mins ago

Bob Hanlon

56.2k23590

add a comment | 

up vote
2
down vote

up vote
2
down vote

To see why it doesn't work as you expected, look at the indices that you generated.

Table[ii, jj, ii, 2, 4, 2, jj, 1, 4, 2]

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

You generated four indices so the sum was over four entries. Use

a = 1, 2, 3, 3, 4, 5, 6, 6, 7, 8, 9, 9, 1, 5, 9, 7;

Sum[a[[Sequence @@ i]], i, 2, 1, 4, 3]

(* 13 *)

share|improve this answer

answered 21 mins ago

Bob Hanlon

56.2k23590

To see why it doesn't work as you expected, look at the indices that you generated.

Table[ii, jj, ii, 2, 4, 2, jj, 1, 4, 2]

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

You generated four indices so the sum was over four entries. Use

a = 1, 2, 3, 3, 4, 5, 6, 6, 7, 8, 9, 9, 1, 5, 9, 7;

Sum[a[[Sequence @@ i]], i, 2, 1, 4, 3]

(* 13 *)

share|improve this answer

answered 21 mins ago

Bob Hanlon

56.2k23590

share|improve this answer

share|improve this answer

answered 21 mins ago

Bob Hanlon

56.2k23590

answered 21 mins ago

Bob Hanlon

56.2k23590

answered 21 mins ago

Bob Hanlon

56.2k23590

56.2k23590

add a comment | 

add a comment | 

up vote
1
down vote

I would have used Extract if Alan had not posted it first.

So, here is a way to use Sum:

Sum[a[[## & @@ i]], i, 2, 1, 4, 3]

13

But that is also posted a minute earlier by Bob Hanlon.

That leaves

☺ = ♯, ♯♯ [Function] +## & @@ (♯[[##]] & @@@ ♯♯);
☺[a, 2, 1, 4, 3]

13

and

☺☺ = +## & @@ (♯ [Function] #[[## & @@ ♯]]) /@ #2 &;
☺☺[a, 2, 1, 4, 3]

13

share|improve this answer

edited 9 mins ago

answered 20 mins ago

kglr

165k8188388

• 
  
  
  

  
  

  
  
  
  
  
  

  
  "A monad is just a monoid in the category of endofunctors, what's the problem?"
  – kglr
  8 mins ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

I would have used Extract if Alan had not posted it first.

So, here is a way to use Sum:

Sum[a[[## & @@ i]], i, 2, 1, 4, 3]

13

But that is also posted a minute earlier by Bob Hanlon.

That leaves

☺ = ♯, ♯♯ [Function] +## & @@ (♯[[##]] & @@@ ♯♯);
☺[a, 2, 1, 4, 3]

13

and

☺☺ = +## & @@ (♯ [Function] #[[## & @@ ♯]]) /@ #2 &;
☺☺[a, 2, 1, 4, 3]

13

share|improve this answer

edited 9 mins ago

answered 20 mins ago

kglr

165k8188388

• 
  
  
  

  
  

  
  
  
  
  
  

  
  "A monad is just a monoid in the category of endofunctors, what's the problem?"
  – kglr
  8 mins ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

I would have used Extract if Alan had not posted it first.

So, here is a way to use Sum:

Sum[a[[## & @@ i]], i, 2, 1, 4, 3]

13

But that is also posted a minute earlier by Bob Hanlon.

That leaves

☺ = ♯, ♯♯ [Function] +## & @@ (♯[[##]] & @@@ ♯♯);
☺[a, 2, 1, 4, 3]

13

and

☺☺ = +## & @@ (♯ [Function] #[[## & @@ ♯]]) /@ #2 &;
☺☺[a, 2, 1, 4, 3]

13

share|improve this answer

edited 9 mins ago

answered 20 mins ago

kglr

165k8188388

I would have used Extract if Alan had not posted it first.

So, here is a way to use Sum:

Sum[a[[## & @@ i]], i, 2, 1, 4, 3]

13

But that is also posted a minute earlier by Bob Hanlon.

That leaves

☺ = ♯, ♯♯ [Function] +## & @@ (♯[[##]] & @@@ ♯♯);
☺[a, 2, 1, 4, 3]

13

and

☺☺ = +## & @@ (♯ [Function] #[[## & @@ ♯]]) /@ #2 &;
☺☺[a, 2, 1, 4, 3]

13

share|improve this answer

edited 9 mins ago

answered 20 mins ago

kglr

165k8188388

share|improve this answer

share|improve this answer

edited 9 mins ago

edited 9 mins ago

edited 9 mins ago

answered 20 mins ago

kglr

165k8188388

answered 20 mins ago

kglr

165k8188388

answered 20 mins ago

kglr

165k8188388

165k8188388

• 
  
  
  

  
  

  
  
  
  
  
  

  
  "A monad is just a monoid in the category of endofunctors, what's the problem?"
  – kglr
  8 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  "A monad is just a monoid in the category of endofunctors, what's the problem?"
  – kglr
  8 mins ago
  

  
  
  
  

"A monad is just a monoid in the category of endofunctors, what's the problem?"
– kglr
8 mins ago

"A monad is just a monoid in the category of endofunctors, what's the problem?"
– kglr
8 mins ago

add a comment | 

 

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