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Apply MapThread to all but one variable

How do you efficiently return all of a List but one element?All values for a function with two arguments without OuterEfficiently finding the maximum value of a column in a matrixnested use of Apply/Map/MapThread in pure functionsMapThread AlternativesFinding neighbors from listMapThread problemapply binary operation to all adjacent pairsFlip sign of one variable in listFind numbers from Mean, Variance and Correlation coefficient

1

$begingroup$

I would like to know what is the most efficient to implement the following computation. Given three lists

    a = {a_1,a_2, a_3, …, a_n}

    b = {b_1,b_2, b_3, …, b_n}

    c = {c_1,c_2, c_3, …, c_n} 

and a function $f(x_1,x_2,x_3)$, obtain

     f(a_1,b_1,c_1)   f(a_1,b_1,c_2)   .....   f(a_1,b_1,c_n)  

     f(a_2,b_2,c_1)   f(a_2,b_2,c_2)   .....   f(a_2,b_2,c_n) 

     .....            .....            .....   .....

     f(a_n,b_n,c_1)   f(a_n,b_n,c_2)   .....   f(a_n,b_n,c_n) 

I cannot find a solution not using For.

list-manipulation

share|improve this question

edited 1 hour ago

corey979

20.9k64382

asked 2 hours ago

SmerdjakovSmerdjakov

1255

$endgroup$

add a comment | 

1

$begingroup$

I would like to know what is the most efficient to implement the following computation. Given three lists

    a = {a_1,a_2, a_3, …, a_n}

    b = {b_1,b_2, b_3, …, b_n}

    c = {c_1,c_2, c_3, …, c_n} 

and a function $f(x_1,x_2,x_3)$, obtain

     f(a_1,b_1,c_1)   f(a_1,b_1,c_2)   .....   f(a_1,b_1,c_n)  

     f(a_2,b_2,c_1)   f(a_2,b_2,c_2)   .....   f(a_2,b_2,c_n) 

     .....            .....            .....   .....

     f(a_n,b_n,c_1)   f(a_n,b_n,c_2)   .....   f(a_n,b_n,c_n) 

I cannot find a solution not using For.

list-manipulation

share|improve this question

edited 1 hour ago

corey979

20.9k64382

asked 2 hours ago

SmerdjakovSmerdjakov

1255

$endgroup$

add a comment | 

$begingroup$

I would like to know what is the most efficient to implement the following computation. Given three lists

    a = {a_1,a_2, a_3, …, a_n}

    b = {b_1,b_2, b_3, …, b_n}

    c = {c_1,c_2, c_3, …, c_n} 

and a function $f(x_1,x_2,x_3)$, obtain

     f(a_1,b_1,c_1)   f(a_1,b_1,c_2)   .....   f(a_1,b_1,c_n)  

     f(a_2,b_2,c_1)   f(a_2,b_2,c_2)   .....   f(a_2,b_2,c_n) 

     .....            .....            .....   .....

     f(a_n,b_n,c_1)   f(a_n,b_n,c_2)   .....   f(a_n,b_n,c_n) 

I cannot find a solution not using For.

list-manipulation

share|improve this question

edited 1 hour ago

corey979

20.9k64382

asked 2 hours ago

SmerdjakovSmerdjakov

1255

$endgroup$

    a = {a_1,a_2, a_3, …, a_n}

    b = {b_1,b_2, b_3, …, b_n}

    c = {c_1,c_2, c_3, …, c_n} 

     f(a_1,b_1,c_1)   f(a_1,b_1,c_2)   .....   f(a_1,b_1,c_n)  

     f(a_2,b_2,c_1)   f(a_2,b_2,c_2)   .....   f(a_2,b_2,c_n) 

     .....            .....            .....   .....

     f(a_n,b_n,c_1)   f(a_n,b_n,c_2)   .....   f(a_n,b_n,c_n) 

share|improve this question

edited 1 hour ago

corey979

20.9k64382

asked 2 hours ago

SmerdjakovSmerdjakov

1255

share|improve this question

edited 1 hour ago

corey979

20.9k64382

asked 2 hours ago

SmerdjakovSmerdjakov

1255

edited 1 hour ago

corey979

20.9k64382

edited 1 hour ago

corey979

20.9k64382

asked 2 hours ago

SmerdjakovSmerdjakov

1255

asked 2 hours ago

SmerdjakovSmerdjakov

1255

3 Answers
3

active

oldest

votes

3

$begingroup$

Here's one way to do it with Outer:

n = 3;

l1 = Array[a, n];

l2 = Array[b, n];

l3 = Array[c, n];



Outer[

  f[#1[[1]], #1[[2]], #2] &,

  Transpose @ {l1, l2},

  l3,

  1

]

Out[25]= {{f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]]}, {f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]]}, {f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]}}

share|improve this answer

answered 1 hour ago

Sjoerd SmitSjoerd Smit

4,600817

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Or Outer[f[Sequence @@ #1, #2] &, Transpose@{l1, l2}, l3, 1] so you don't need to unravel #1 manually.
  $endgroup$
  – Roman
  1 hour ago
  

  
  
  
  

add a comment | 

2

$begingroup$

a = {a1, a2, a3, a4, a5};

b = {b1, b2, b3, b4, b5};

c = {c1, c2, c3, c4, c5};



Table[f[a[[j]], b[[j]], c[[k]]], {j, 1, 5}, {k, 1, 5}]

share|improve this answer

answered 1 hour ago

corey979corey979

20.9k64382

$endgroup$

add a comment | 

1

$begingroup$

Another possibility is to use the 3-arg version of Thread. With Sjoerd's example:

n = 3;

l1 = Array[a,n];

l2 = Array[b,n];

l3 = Array[c,n];

Using Thread:

Thread /@ Thread[f[l1, l2, l3], List, 2]

{{f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]]}, {f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]]}, {f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]}}

share|improve this answer

answered 49 mins ago

Carl WollCarl Woll

75.9k3100198

$endgroup$

add a comment | 

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3

$begingroup$

Here's one way to do it with Outer:

n = 3;

l1 = Array[a, n];

l2 = Array[b, n];

l3 = Array[c, n];



Outer[

  f[#1[[1]], #1[[2]], #2] &,

  Transpose @ {l1, l2},

  l3,

  1

]

Out[25]= {{f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]]}, {f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]]}, {f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]}}

share|improve this answer

answered 1 hour ago

Sjoerd SmitSjoerd Smit

4,600817

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Or Outer[f[Sequence @@ #1, #2] &, Transpose@{l1, l2}, l3, 1] so you don't need to unravel #1 manually.
  $endgroup$
  – Roman
  1 hour ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Here's one way to do it with Outer:

n = 3;

l1 = Array[a, n];

l2 = Array[b, n];

l3 = Array[c, n];



Outer[

  f[#1[[1]], #1[[2]], #2] &,

  Transpose @ {l1, l2},

  l3,

  1

]

Out[25]= {{f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]]}, {f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]]}, {f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]}}

share|improve this answer

answered 1 hour ago

Sjoerd SmitSjoerd Smit

4,600817

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Or Outer[f[Sequence @@ #1, #2] &, Transpose@{l1, l2}, l3, 1] so you don't need to unravel #1 manually.
  $endgroup$
  – Roman
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

Here's one way to do it with Outer:

n = 3;

l1 = Array[a, n];

l2 = Array[b, n];

l3 = Array[c, n];



Outer[

  f[#1[[1]], #1[[2]], #2] &,

  Transpose @ {l1, l2},

  l3,

  1

]

Out[25]= {{f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]]}, {f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]]}, {f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]}}

share|improve this answer

answered 1 hour ago

Sjoerd SmitSjoerd Smit

4,600817

$endgroup$

n = 3;

l1 = Array[a, n];

l2 = Array[b, n];

l3 = Array[c, n];



Outer[

  f[#1[[1]], #1[[2]], #2] &,

  Transpose @ {l1, l2},

  l3,

  1

]

share|improve this answer

answered 1 hour ago

Sjoerd SmitSjoerd Smit

4,600817

answered 1 hour ago

Sjoerd SmitSjoerd Smit

4,600817

answered 1 hour ago

Sjoerd SmitSjoerd Smit

4,600817

2

$begingroup$

a = {a1, a2, a3, a4, a5};

b = {b1, b2, b3, b4, b5};

c = {c1, c2, c3, c4, c5};



Table[f[a[[j]], b[[j]], c[[k]]], {j, 1, 5}, {k, 1, 5}]

share|improve this answer

answered 1 hour ago

corey979corey979

20.9k64382

$endgroup$

add a comment | 

2

$begingroup$

a = {a1, a2, a3, a4, a5};

b = {b1, b2, b3, b4, b5};

c = {c1, c2, c3, c4, c5};



Table[f[a[[j]], b[[j]], c[[k]]], {j, 1, 5}, {k, 1, 5}]

share|improve this answer

answered 1 hour ago

corey979corey979

20.9k64382

$endgroup$

add a comment | 

$begingroup$

a = {a1, a2, a3, a4, a5};

b = {b1, b2, b3, b4, b5};

c = {c1, c2, c3, c4, c5};



Table[f[a[[j]], b[[j]], c[[k]]], {j, 1, 5}, {k, 1, 5}]

share|improve this answer

answered 1 hour ago

corey979corey979

20.9k64382

$endgroup$

a = {a1, a2, a3, a4, a5};

b = {b1, b2, b3, b4, b5};

c = {c1, c2, c3, c4, c5};



Table[f[a[[j]], b[[j]], c[[k]]], {j, 1, 5}, {k, 1, 5}]

share|improve this answer

answered 1 hour ago

corey979corey979

20.9k64382

answered 1 hour ago

corey979corey979

20.9k64382

answered 1 hour ago

corey979corey979

20.9k64382

1

$begingroup$

Another possibility is to use the 3-arg version of Thread. With Sjoerd's example:

n = 3;

l1 = Array[a,n];

l2 = Array[b,n];

l3 = Array[c,n];

Using Thread:

Thread /@ Thread[f[l1, l2, l3], List, 2]

{{f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]]}, {f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]]}, {f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]}}

share|improve this answer

answered 49 mins ago

Carl WollCarl Woll

75.9k3100198

$endgroup$

add a comment | 

1

$begingroup$

Another possibility is to use the 3-arg version of Thread. With Sjoerd's example:

n = 3;

l1 = Array[a,n];

l2 = Array[b,n];

l3 = Array[c,n];

Using Thread:

Thread /@ Thread[f[l1, l2, l3], List, 2]

{{f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]]}, {f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]]}, {f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]}}

share|improve this answer

answered 49 mins ago

Carl WollCarl Woll

75.9k3100198

$endgroup$

add a comment | 

$begingroup$

Another possibility is to use the 3-arg version of Thread. With Sjoerd's example:

n = 3;

l1 = Array[a,n];

l2 = Array[b,n];

l3 = Array[c,n];

Using Thread:

Thread /@ Thread[f[l1, l2, l3], List, 2]

{{f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]]}, {f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]]}, {f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]}}

share|improve this answer

answered 49 mins ago

Carl WollCarl Woll

75.9k3100198

$endgroup$

n = 3;

l1 = Array[a,n];

l2 = Array[b,n];

l3 = Array[c,n];

Thread /@ Thread[f[l1, l2, l3], List, 2]

share|improve this answer

answered 49 mins ago

Carl WollCarl Woll

75.9k3100198

answered 49 mins ago

Carl WollCarl Woll

75.9k3100198

answered 49 mins ago

Carl WollCarl Woll

75.9k3100198