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4

$begingroup$

Say I have two lists of pairs

a = {{1, 2}, {3, 4}};

b = {{5, 6}, {7, 8}};

and I want to add the 2nd element of every pair in b to the 2nd element of every pair in a, so the expected result would be {{1, 8}, {3, 12}}.

How can I do this?

I am a total beginner with Mathematica, so sorry if this is to basic :)

list-manipulation

share|improve this question

edited 4 hours ago

m_goldberg

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asked 8 hours ago

JimJim

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New contributor

Jim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

add a comment
 | 

4

$begingroup$

Say I have two lists of pairs

a = {{1, 2}, {3, 4}};

b = {{5, 6}, {7, 8}};

and I want to add the 2nd element of every pair in b to the 2nd element of every pair in a, so the expected result would be {{1, 8}, {3, 12}}.

How can I do this?

I am a total beginner with Mathematica, so sorry if this is to basic :)

list-manipulation

share|improve this question

edited 4 hours ago

m_goldberg

91.9k88 gold badges7575 silver badges210210 bronze badges

asked 8 hours ago

JimJim

2133 bronze badges

New contributor

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

$endgroup$

add a comment
 | 

$begingroup$

Say I have two lists of pairs

a = {{1, 2}, {3, 4}};

b = {{5, 6}, {7, 8}};

and I want to add the 2nd element of every pair in b to the 2nd element of every pair in a, so the expected result would be {{1, 8}, {3, 12}}.

How can I do this?

I am a total beginner with Mathematica, so sorry if this is to basic :)

list-manipulation

share|improve this question

edited 4 hours ago

m_goldberg

91.9k88 gold badges7575 silver badges210210 bronze badges

asked 8 hours ago

JimJim

2133 bronze badges

New contributor

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

$endgroup$

a = {{1, 2}, {3, 4}};

b = {{5, 6}, {7, 8}};

share|improve this question

edited 4 hours ago

m_goldberg

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asked 8 hours ago

JimJim

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Jim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|improve this question

edited 4 hours ago

m_goldberg

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asked 8 hours ago

JimJim

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Jim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 4 hours ago

m_goldberg

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edited 4 hours ago

m_goldberg

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asked 8 hours ago

JimJim

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asked 8 hours ago

JimJim

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6 Answers
6

active

oldest

votes

1

$begingroup$

lst1 = {{1, 2}, {2, 4}};



lst1a = lst1;

lst1a[[All, 2]] *= 2;

lst1a

{{1, 4}, {2, 8}}

lst2 = {{3, 5}, {10, 2}};



lst1b = lst1;

lst1b[[All, 2]] = lst1[[All, 2]] + lst2[[All, 2]];

lst1b

{{1, 7}, {2, 6}}

share|improve this answer

answered 8 hours ago

kglrkglr

220k1010 gold badges250250 silver badges504504 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

Try this:

{{1, 2}, {2, 4}} + ({{1, 2}, {2, 4}} /. {x_,y _} -> {0, y})



(* {{2, 4}, {4, 8}}  *)

or this

(# + (# /. {x_, y_} -> {0, y})) &[{{1, 2}, {2, 4}}]



(* {{1, 2}, {4, 8}}  *)

Have fun!

share|improve this answer

answered 7 hours ago

Alexei BoulbitchAlexei Boulbitch

23k2727 silver badges7878 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

list = {{1, 2}, {2, 4}};



MapAt[2 # &, {All, 2}] @ list

{{1, 4}, {2, 8}}

share|improve this answer

answered 7 hours ago

eldoeldo

30.4k22 gold badges2828 silver badges8585 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi @eldo! It's been a while... :)
  $endgroup$
  – Michael E2
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, nice to see you again :}
  $endgroup$
  – eldo
  7 hours ago
  

  
  
  
  

add a comment
 | 

1

$begingroup$

lst = {{1, 2}, {2, 4}};

MapThread[{First[#1], Last[#1] + Last[#2]} &, {lst, lst}]

{{1, 4}, {2, 8}}

share|improve this answer

answered 7 hours ago

Suba ThomasSuba Thomas

4,1261111 silver badges2020 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

I like to do this kind of thing by writing a custom function to perform the the desired operation on two given pairs. I then map this function over the two lists with MapThread. In most cases, by using Mathematica's argument-pattern destructuring, the custom function can be expressed extremely simply. This is one of those cases.

The custom function is

f[{x_, y_}, {_, z_}] := {x, y + z}

Note that the righthand side is a literal expression of the desired result.

Now let's contrive some test data. I write the generator so that two lists of pairs can easily be given any length.

With[{n = 3}, {a, b} = Partition[Partition[Range[4 n], 2], n]];

a

b

{{1, 2}, {3, 4}, {5, 6}}

 {{7, 8}, {9, 10}, {11, 12}}

Now we MapThread the function f over a and b.

MapThread[f, {a, b}]

{1, 10}, {3, 14}, {5, 18}}

share|improve this answer

edited 34 mins ago

answered 3 hours ago

m_goldbergm_goldberg

91.9k88 gold badges7575 silver badges210210 bronze badges

$endgroup$

add a comment
 | 

0

$begingroup$

list1 = {{1, 2}, {2, 4}};

Adding the second elements from each pair of the same list:

{#[[1]], 2 #[[2]]} & /@ list1

{{1, 4}, {2, 8}}

Adding the second element from a different list:

list2 = {{a, b}, {c, d}};

{#[[1, 1]], #[[1, 2]] + #[[2, 2]]} & /@ Transpose[{list1, list2}]

{{1, 2 + b}, {2, 4 + d}}

share|improve this answer

answered 6 hours ago

MelaGoMelaGo

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$endgroup$

add a comment
 | 

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1

$begingroup$

lst1 = {{1, 2}, {2, 4}};



lst1a = lst1;

lst1a[[All, 2]] *= 2;

lst1a

{{1, 4}, {2, 8}}

lst2 = {{3, 5}, {10, 2}};



lst1b = lst1;

lst1b[[All, 2]] = lst1[[All, 2]] + lst2[[All, 2]];

lst1b

{{1, 7}, {2, 6}}

share|improve this answer

answered 8 hours ago

kglrkglr

220k1010 gold badges250250 silver badges504504 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

lst1 = {{1, 2}, {2, 4}};



lst1a = lst1;

lst1a[[All, 2]] *= 2;

lst1a

{{1, 4}, {2, 8}}

lst2 = {{3, 5}, {10, 2}};



lst1b = lst1;

lst1b[[All, 2]] = lst1[[All, 2]] + lst2[[All, 2]];

lst1b

{{1, 7}, {2, 6}}

share|improve this answer

answered 8 hours ago

kglrkglr

220k1010 gold badges250250 silver badges504504 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

lst1 = {{1, 2}, {2, 4}};



lst1a = lst1;

lst1a[[All, 2]] *= 2;

lst1a

{{1, 4}, {2, 8}}

lst2 = {{3, 5}, {10, 2}};



lst1b = lst1;

lst1b[[All, 2]] = lst1[[All, 2]] + lst2[[All, 2]];

lst1b

{{1, 7}, {2, 6}}

share|improve this answer

answered 8 hours ago

kglrkglr

220k1010 gold badges250250 silver badges504504 bronze badges

$endgroup$

lst1 = {{1, 2}, {2, 4}};



lst1a = lst1;

lst1a[[All, 2]] *= 2;

lst1a

lst2 = {{3, 5}, {10, 2}};



lst1b = lst1;

lst1b[[All, 2]] = lst1[[All, 2]] + lst2[[All, 2]];

lst1b

share|improve this answer

answered 8 hours ago

kglrkglr

220k1010 gold badges250250 silver badges504504 bronze badges

answered 8 hours ago

kglrkglr

220k1010 gold badges250250 silver badges504504 bronze badges

answered 8 hours ago

kglrkglr

220k1010 gold badges250250 silver badges504504 bronze badges

1

$begingroup$

Try this:

{{1, 2}, {2, 4}} + ({{1, 2}, {2, 4}} /. {x_,y _} -> {0, y})



(* {{2, 4}, {4, 8}}  *)

or this

(# + (# /. {x_, y_} -> {0, y})) &[{{1, 2}, {2, 4}}]



(* {{1, 2}, {4, 8}}  *)

Have fun!

share|improve this answer

answered 7 hours ago

Alexei BoulbitchAlexei Boulbitch

23k2727 silver badges7878 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

Try this:

{{1, 2}, {2, 4}} + ({{1, 2}, {2, 4}} /. {x_,y _} -> {0, y})



(* {{2, 4}, {4, 8}}  *)

or this

(# + (# /. {x_, y_} -> {0, y})) &[{{1, 2}, {2, 4}}]



(* {{1, 2}, {4, 8}}  *)

Have fun!

share|improve this answer

answered 7 hours ago

Alexei BoulbitchAlexei Boulbitch

23k2727 silver badges7878 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Try this:

{{1, 2}, {2, 4}} + ({{1, 2}, {2, 4}} /. {x_,y _} -> {0, y})



(* {{2, 4}, {4, 8}}  *)

or this

(# + (# /. {x_, y_} -> {0, y})) &[{{1, 2}, {2, 4}}]



(* {{1, 2}, {4, 8}}  *)

Have fun!

share|improve this answer

answered 7 hours ago

Alexei BoulbitchAlexei Boulbitch

23k2727 silver badges7878 bronze badges

$endgroup$

{{1, 2}, {2, 4}} + ({{1, 2}, {2, 4}} /. {x_,y _} -> {0, y})



(* {{2, 4}, {4, 8}}  *)

(# + (# /. {x_, y_} -> {0, y})) &[{{1, 2}, {2, 4}}]



(* {{1, 2}, {4, 8}}  *)

share|improve this answer

answered 7 hours ago

Alexei BoulbitchAlexei Boulbitch

23k2727 silver badges7878 bronze badges

answered 7 hours ago

Alexei BoulbitchAlexei Boulbitch

23k2727 silver badges7878 bronze badges

answered 7 hours ago

Alexei BoulbitchAlexei Boulbitch

23k2727 silver badges7878 bronze badges

1

$begingroup$

list = {{1, 2}, {2, 4}};



MapAt[2 # &, {All, 2}] @ list

{{1, 4}, {2, 8}}

share|improve this answer

answered 7 hours ago

eldoeldo

30.4k22 gold badges2828 silver badges8585 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi @eldo! It's been a while... :)
  $endgroup$
  – Michael E2
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, nice to see you again :}
  $endgroup$
  – eldo
  7 hours ago
  

  
  
  
  

add a comment
 | 

1

$begingroup$

list = {{1, 2}, {2, 4}};



MapAt[2 # &, {All, 2}] @ list

{{1, 4}, {2, 8}}

share|improve this answer

answered 7 hours ago

eldoeldo

30.4k22 gold badges2828 silver badges8585 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi @eldo! It's been a while... :)
  $endgroup$
  – Michael E2
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, nice to see you again :}
  $endgroup$
  – eldo
  7 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

list = {{1, 2}, {2, 4}};



MapAt[2 # &, {All, 2}] @ list

{{1, 4}, {2, 8}}

share|improve this answer

answered 7 hours ago

eldoeldo

30.4k22 gold badges2828 silver badges8585 bronze badges

$endgroup$

list = {{1, 2}, {2, 4}};



MapAt[2 # &, {All, 2}] @ list

share|improve this answer

answered 7 hours ago

eldoeldo

30.4k22 gold badges2828 silver badges8585 bronze badges

answered 7 hours ago

eldoeldo

30.4k22 gold badges2828 silver badges8585 bronze badges

answered 7 hours ago

eldoeldo

30.4k22 gold badges2828 silver badges8585 bronze badges

1

$begingroup$

lst = {{1, 2}, {2, 4}};

MapThread[{First[#1], Last[#1] + Last[#2]} &, {lst, lst}]

{{1, 4}, {2, 8}}

share|improve this answer

answered 7 hours ago

Suba ThomasSuba Thomas

4,1261111 silver badges2020 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

lst = {{1, 2}, {2, 4}};

MapThread[{First[#1], Last[#1] + Last[#2]} &, {lst, lst}]

{{1, 4}, {2, 8}}

share|improve this answer

answered 7 hours ago

Suba ThomasSuba Thomas

4,1261111 silver badges2020 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

lst = {{1, 2}, {2, 4}};

MapThread[{First[#1], Last[#1] + Last[#2]} &, {lst, lst}]

{{1, 4}, {2, 8}}

share|improve this answer

answered 7 hours ago

Suba ThomasSuba Thomas

4,1261111 silver badges2020 bronze badges

$endgroup$

lst = {{1, 2}, {2, 4}};

MapThread[{First[#1], Last[#1] + Last[#2]} &, {lst, lst}]

share|improve this answer

answered 7 hours ago

Suba ThomasSuba Thomas

4,1261111 silver badges2020 bronze badges

answered 7 hours ago

Suba ThomasSuba Thomas

4,1261111 silver badges2020 bronze badges

answered 7 hours ago

Suba ThomasSuba Thomas

4,1261111 silver badges2020 bronze badges

1

$begingroup$

I like to do this kind of thing by writing a custom function to perform the the desired operation on two given pairs. I then map this function over the two lists with MapThread. In most cases, by using Mathematica's argument-pattern destructuring, the custom function can be expressed extremely simply. This is one of those cases.

The custom function is

f[{x_, y_}, {_, z_}] := {x, y + z}

Note that the righthand side is a literal expression of the desired result.

Now let's contrive some test data. I write the generator so that two lists of pairs can easily be given any length.

With[{n = 3}, {a, b} = Partition[Partition[Range[4 n], 2], n]];

a

b

{{1, 2}, {3, 4}, {5, 6}}

 {{7, 8}, {9, 10}, {11, 12}}

Now we MapThread the function f over a and b.

MapThread[f, {a, b}]

{1, 10}, {3, 14}, {5, 18}}

share|improve this answer

edited 34 mins ago

answered 3 hours ago

m_goldbergm_goldberg

91.9k88 gold badges7575 silver badges210210 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

I like to do this kind of thing by writing a custom function to perform the the desired operation on two given pairs. I then map this function over the two lists with MapThread. In most cases, by using Mathematica's argument-pattern destructuring, the custom function can be expressed extremely simply. This is one of those cases.

The custom function is

f[{x_, y_}, {_, z_}] := {x, y + z}

Note that the righthand side is a literal expression of the desired result.

Now let's contrive some test data. I write the generator so that two lists of pairs can easily be given any length.

With[{n = 3}, {a, b} = Partition[Partition[Range[4 n], 2], n]];

a

b

{{1, 2}, {3, 4}, {5, 6}}

 {{7, 8}, {9, 10}, {11, 12}}

Now we MapThread the function f over a and b.

MapThread[f, {a, b}]

{1, 10}, {3, 14}, {5, 18}}

share|improve this answer

edited 34 mins ago

answered 3 hours ago

m_goldbergm_goldberg

91.9k88 gold badges7575 silver badges210210 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

I like to do this kind of thing by writing a custom function to perform the the desired operation on two given pairs. I then map this function over the two lists with MapThread. In most cases, by using Mathematica's argument-pattern destructuring, the custom function can be expressed extremely simply. This is one of those cases.

The custom function is

f[{x_, y_}, {_, z_}] := {x, y + z}

Note that the righthand side is a literal expression of the desired result.

Now let's contrive some test data. I write the generator so that two lists of pairs can easily be given any length.

With[{n = 3}, {a, b} = Partition[Partition[Range[4 n], 2], n]];

a

b

{{1, 2}, {3, 4}, {5, 6}}

 {{7, 8}, {9, 10}, {11, 12}}

Now we MapThread the function f over a and b.

MapThread[f, {a, b}]

{1, 10}, {3, 14}, {5, 18}}

share|improve this answer

edited 34 mins ago

answered 3 hours ago

m_goldbergm_goldberg

91.9k88 gold badges7575 silver badges210210 bronze badges

$endgroup$

f[{x_, y_}, {_, z_}] := {x, y + z}

With[{n = 3}, {a, b} = Partition[Partition[Range[4 n], 2], n]];

a

b

MapThread[f, {a, b}]

share|improve this answer

edited 34 mins ago

answered 3 hours ago

m_goldbergm_goldberg

91.9k88 gold badges7575 silver badges210210 bronze badges

answered 3 hours ago

m_goldbergm_goldberg

91.9k88 gold badges7575 silver badges210210 bronze badges

answered 3 hours ago

m_goldbergm_goldberg

91.9k88 gold badges7575 silver badges210210 bronze badges

0

$begingroup$

list1 = {{1, 2}, {2, 4}};

Adding the second elements from each pair of the same list:

{#[[1]], 2 #[[2]]} & /@ list1

{{1, 4}, {2, 8}}

Adding the second element from a different list:

list2 = {{a, b}, {c, d}};

{#[[1, 1]], #[[1, 2]] + #[[2, 2]]} & /@ Transpose[{list1, list2}]

{{1, 2 + b}, {2, 4 + d}}

share|improve this answer

answered 6 hours ago

MelaGoMelaGo

3,14611 gold badge22 silver badges1010 bronze badges

$endgroup$

add a comment
 | 

0

$begingroup$

list1 = {{1, 2}, {2, 4}};

Adding the second elements from each pair of the same list:

{#[[1]], 2 #[[2]]} & /@ list1

{{1, 4}, {2, 8}}

Adding the second element from a different list:

list2 = {{a, b}, {c, d}};

{#[[1, 1]], #[[1, 2]] + #[[2, 2]]} & /@ Transpose[{list1, list2}]

{{1, 2 + b}, {2, 4 + d}}

share|improve this answer

answered 6 hours ago

MelaGoMelaGo

3,14611 gold badge22 silver badges1010 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

list1 = {{1, 2}, {2, 4}};

Adding the second elements from each pair of the same list:

{#[[1]], 2 #[[2]]} & /@ list1

{{1, 4}, {2, 8}}

Adding the second element from a different list:

list2 = {{a, b}, {c, d}};

{#[[1, 1]], #[[1, 2]] + #[[2, 2]]} & /@ Transpose[{list1, list2}]

{{1, 2 + b}, {2, 4 + d}}

share|improve this answer

answered 6 hours ago

MelaGoMelaGo

3,14611 gold badge22 silver badges1010 bronze badges

$endgroup$

list1 = {{1, 2}, {2, 4}};

{#[[1]], 2 #[[2]]} & /@ list1

list2 = {{a, b}, {c, d}};

{#[[1, 1]], #[[1, 2]] + #[[2, 2]]} & /@ Transpose[{list1, list2}]

share|improve this answer

answered 6 hours ago

MelaGoMelaGo

3,14611 gold badge22 silver badges1010 bronze badges

answered 6 hours ago

MelaGoMelaGo

3,14611 gold badge22 silver badges1010 bronze badges

answered 6 hours ago

MelaGoMelaGo

3,14611 gold badge22 silver badges1010 bronze badges