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Map a function that takes arguments in different levels of a list

Map a function across a list conditionallyMap a function over the columns of an M x N arraySpan a function across several consecutive elements in a listUsing MapIndexed only at certain elements of a listMapping a function over n levelsMap an Argument to a list of functionsA list of functions taking different argumentsApply a function to a list that is a subset of sublistsMap function over two lists

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4

$begingroup$

I would like to map a function that takes as an argument a list in a list of lists and an element of the list of lists.
So for example, this is my list of lists:

{{4,1}, {3,1,1}, {2,2,1}}

I would like to map the function to the list of lists to obtain the following outcome:

{ {f[{4,1},4], f[{4,1},1]}, {f[{3,1,1},3}], f[{3,1,1},1], f[{3,1,1},1] }, {f[{2,2,1},2], f[{2,2,1},2], f[{2,2,1},1]}

I would appreciate it if someone could tell me how to do it. Thank you in advance!

map

share|improve this question

asked 8 hours ago

MavatanetMavatanet

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add a comment | 

4

$begingroup$

I would like to map a function that takes as an argument a list in a list of lists and an element of the list of lists.
So for example, this is my list of lists:

{{4,1}, {3,1,1}, {2,2,1}}

I would like to map the function to the list of lists to obtain the following outcome:

{ {f[{4,1},4], f[{4,1},1]}, {f[{3,1,1},3}], f[{3,1,1},1], f[{3,1,1},1] }, {f[{2,2,1},2], f[{2,2,1},2], f[{2,2,1},1]}

I would appreciate it if someone could tell me how to do it. Thank you in advance!

map

share|improve this question

asked 8 hours ago

MavatanetMavatanet

2611 bronze badge

New contributor

Mavatanet 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 | 

$begingroup$

I would like to map a function that takes as an argument a list in a list of lists and an element of the list of lists.
So for example, this is my list of lists:

{{4,1}, {3,1,1}, {2,2,1}}

I would like to map the function to the list of lists to obtain the following outcome:

{ {f[{4,1},4], f[{4,1},1]}, {f[{3,1,1},3}], f[{3,1,1},1], f[{3,1,1},1] }, {f[{2,2,1},2], f[{2,2,1},2], f[{2,2,1},1]}

I would appreciate it if someone could tell me how to do it. Thank you in advance!

map

share|improve this question

asked 8 hours ago

MavatanetMavatanet

2611 bronze badge

New contributor

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

$endgroup$

share|improve this question

asked 8 hours ago

MavatanetMavatanet

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

share|improve this question

asked 8 hours ago

MavatanetMavatanet

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New contributor

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

asked 8 hours ago

MavatanetMavatanet

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New contributor

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

asked 8 hours ago

MavatanetMavatanet

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3 Answers
3

active

oldest

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3

$begingroup$

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

Table[Table[f[i, j], {j, i}], {i, lst}]

{{f[{4, 1}, 4], f[{4, 1}, 1]}, {f[{3, 1, 1}, 3], f[{3, 1, 1}, 1],
f[{3, 1, 1}, 1]}, {f[{2, 2, 1}, 2], f[{2, 2, 1}, 2],
f[{2, 2, 1}, 1]}}

share|improve this answer

answered 8 hours ago

MelaGoMelaGo

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$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +1. also: Table[f[i, j], {i, lst}, {j, i}]
  $endgroup$
  – WReach
  44 mins ago
  

  
  
  
  

add a comment | 

3

$begingroup$

ClearAll[g1,  f]

g1 = Function[x, f[x, #] & /@ x]



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



g1 /@ lst

{{f[{4, 1}, 4], f[{4, 1}, 1]},

{f[{3, 1, 1}, 3], f[{3, 1, 1}, 1], f[{3, 1, 1}, 1]},

{f[{2, 2, 1}, 2], f[{2, 2, 1}, 2], f[{2, 2, 1}, 1]}}

Also

ClearAll[g2]

g2[x_]:= Map[f[x, #] &] @ x



g2 /@ lst

{{f[{4, 1}, 4], f[{4, 1}, 1]},

{f[{3, 1, 1}, 3], f[{3, 1, 1}, 1], f[{3, 1, 1}, 1]},

{f[{2, 2, 1}, 2], f[{2, 2, 1}, 2], f[{2, 2, 1}, 1]}}

and

ClearAll[g3]

g3 = Thread[f[#, #], List, {2}] & 



g3 /@ lst 

{{f[{4, 1}, 4], f[{4, 1}, 1]},

{f[{3, 1, 1}, 3], f[{3, 1, 1}, 1], f[{3, 1, 1}, 1]},

{f[{2, 2, 1}, 2], f[{2, 2, 1}, 2], f[{2, 2, 1}, 1]}}

share|improve this answer

edited 7 hours ago

answered 8 hours ago

kglrkglr

215k1010 gold badges245245 silver badges490490 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

I like the techniques in the other answers (particularly the Table approach in the answer by MelaGo), but here are a few other options to add to the mix...

Given:

$list = {{4,1}, {3,1,1}, {2,2,1}};

Then any of the following expressions yield the desired result:

Curry[f, 2][#] /@ # & /@ $list

or

Cases[$list, l_ :> (f[l, #] & /@ l)]

or

Replace[$list, l_ :> (f[l, #] & /@ l), {1}]

or

MapIndexed[f[$list[[#2[[1]]]], #] &, $list, {2}]

or

ReplacePart[$list, {i_, j_} :> f[$list[[i]], $list[[i, j]]]]

share|improve this answer

answered 23 mins ago

WReachWReach

55.1k22 gold badges119119 silver badges219219 bronze badges

$endgroup$

add a comment | 

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3

$begingroup$

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

Table[Table[f[i, j], {j, i}], {i, lst}]

{{f[{4, 1}, 4], f[{4, 1}, 1]}, {f[{3, 1, 1}, 3], f[{3, 1, 1}, 1],
f[{3, 1, 1}, 1]}, {f[{2, 2, 1}, 2], f[{2, 2, 1}, 2],
f[{2, 2, 1}, 1]}}

share|improve this answer

answered 8 hours ago

MelaGoMelaGo

2,63611 gold badge11 silver badge77 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +1. also: Table[f[i, j], {i, lst}, {j, i}]
  $endgroup$
  – WReach
  44 mins ago
  

  
  
  
  

add a comment | 

3

$begingroup$

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

Table[Table[f[i, j], {j, i}], {i, lst}]

{{f[{4, 1}, 4], f[{4, 1}, 1]}, {f[{3, 1, 1}, 3], f[{3, 1, 1}, 1],
f[{3, 1, 1}, 1]}, {f[{2, 2, 1}, 2], f[{2, 2, 1}, 2],
f[{2, 2, 1}, 1]}}

share|improve this answer

answered 8 hours ago

MelaGoMelaGo

2,63611 gold badge11 silver badge77 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +1. also: Table[f[i, j], {i, lst}, {j, i}]
  $endgroup$
  – WReach
  44 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

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

Table[Table[f[i, j], {j, i}], {i, lst}]

{{f[{4, 1}, 4], f[{4, 1}, 1]}, {f[{3, 1, 1}, 3], f[{3, 1, 1}, 1],
f[{3, 1, 1}, 1]}, {f[{2, 2, 1}, 2], f[{2, 2, 1}, 2],
f[{2, 2, 1}, 1]}}

share|improve this answer

answered 8 hours ago

MelaGoMelaGo

2,63611 gold badge11 silver badge77 bronze badges

$endgroup$

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

Table[Table[f[i, j], {j, i}], {i, lst}]

share|improve this answer

answered 8 hours ago

MelaGoMelaGo

2,63611 gold badge11 silver badge77 bronze badges

answered 8 hours ago

MelaGoMelaGo

2,63611 gold badge11 silver badge77 bronze badges

answered 8 hours ago

MelaGoMelaGo

2,63611 gold badge11 silver badge77 bronze badges

3

$begingroup$

ClearAll[g1,  f]

g1 = Function[x, f[x, #] & /@ x]



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



g1 /@ lst

{{f[{4, 1}, 4], f[{4, 1}, 1]},

{f[{3, 1, 1}, 3], f[{3, 1, 1}, 1], f[{3, 1, 1}, 1]},

{f[{2, 2, 1}, 2], f[{2, 2, 1}, 2], f[{2, 2, 1}, 1]}}

Also

ClearAll[g2]

g2[x_]:= Map[f[x, #] &] @ x



g2 /@ lst

{{f[{4, 1}, 4], f[{4, 1}, 1]},

{f[{3, 1, 1}, 3], f[{3, 1, 1}, 1], f[{3, 1, 1}, 1]},

{f[{2, 2, 1}, 2], f[{2, 2, 1}, 2], f[{2, 2, 1}, 1]}}

and

ClearAll[g3]

g3 = Thread[f[#, #], List, {2}] & 



g3 /@ lst 

{{f[{4, 1}, 4], f[{4, 1}, 1]},

{f[{3, 1, 1}, 3], f[{3, 1, 1}, 1], f[{3, 1, 1}, 1]},

{f[{2, 2, 1}, 2], f[{2, 2, 1}, 2], f[{2, 2, 1}, 1]}}

share|improve this answer

edited 7 hours ago

answered 8 hours ago

kglrkglr

215k1010 gold badges245245 silver badges490490 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

ClearAll[g1,  f]

g1 = Function[x, f[x, #] & /@ x]



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



g1 /@ lst

{{f[{4, 1}, 4], f[{4, 1}, 1]},

{f[{3, 1, 1}, 3], f[{3, 1, 1}, 1], f[{3, 1, 1}, 1]},

{f[{2, 2, 1}, 2], f[{2, 2, 1}, 2], f[{2, 2, 1}, 1]}}

Also

ClearAll[g2]

g2[x_]:= Map[f[x, #] &] @ x



g2 /@ lst

{{f[{4, 1}, 4], f[{4, 1}, 1]},

{f[{3, 1, 1}, 3], f[{3, 1, 1}, 1], f[{3, 1, 1}, 1]},

{f[{2, 2, 1}, 2], f[{2, 2, 1}, 2], f[{2, 2, 1}, 1]}}

and

ClearAll[g3]

g3 = Thread[f[#, #], List, {2}] & 



g3 /@ lst 

{{f[{4, 1}, 4], f[{4, 1}, 1]},

{f[{3, 1, 1}, 3], f[{3, 1, 1}, 1], f[{3, 1, 1}, 1]},

{f[{2, 2, 1}, 2], f[{2, 2, 1}, 2], f[{2, 2, 1}, 1]}}

share|improve this answer

edited 7 hours ago

answered 8 hours ago

kglrkglr

215k1010 gold badges245245 silver badges490490 bronze badges

$endgroup$

add a comment | 

$begingroup$

ClearAll[g1,  f]

g1 = Function[x, f[x, #] & /@ x]



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



g1 /@ lst

{{f[{4, 1}, 4], f[{4, 1}, 1]},

{f[{3, 1, 1}, 3], f[{3, 1, 1}, 1], f[{3, 1, 1}, 1]},

{f[{2, 2, 1}, 2], f[{2, 2, 1}, 2], f[{2, 2, 1}, 1]}}

Also

ClearAll[g2]

g2[x_]:= Map[f[x, #] &] @ x



g2 /@ lst

{{f[{4, 1}, 4], f[{4, 1}, 1]},

{f[{3, 1, 1}, 3], f[{3, 1, 1}, 1], f[{3, 1, 1}, 1]},

{f[{2, 2, 1}, 2], f[{2, 2, 1}, 2], f[{2, 2, 1}, 1]}}

and

ClearAll[g3]

g3 = Thread[f[#, #], List, {2}] & 



g3 /@ lst 

{{f[{4, 1}, 4], f[{4, 1}, 1]},

{f[{3, 1, 1}, 3], f[{3, 1, 1}, 1], f[{3, 1, 1}, 1]},

{f[{2, 2, 1}, 2], f[{2, 2, 1}, 2], f[{2, 2, 1}, 1]}}

share|improve this answer

edited 7 hours ago

answered 8 hours ago

kglrkglr

215k1010 gold badges245245 silver badges490490 bronze badges

$endgroup$

ClearAll[g1,  f]

g1 = Function[x, f[x, #] & /@ x]



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



g1 /@ lst

ClearAll[g2]

g2[x_]:= Map[f[x, #] &] @ x



g2 /@ lst

ClearAll[g3]

g3 = Thread[f[#, #], List, {2}] & 



g3 /@ lst 

share|improve this answer

edited 7 hours ago

answered 8 hours ago

kglrkglr

215k1010 gold badges245245 silver badges490490 bronze badges

answered 8 hours ago

kglrkglr

215k1010 gold badges245245 silver badges490490 bronze badges

answered 8 hours ago

kglrkglr

215k1010 gold badges245245 silver badges490490 bronze badges

0

$begingroup$

I like the techniques in the other answers (particularly the Table approach in the answer by MelaGo), but here are a few other options to add to the mix...

Given:

$list = {{4,1}, {3,1,1}, {2,2,1}};

Then any of the following expressions yield the desired result:

Curry[f, 2][#] /@ # & /@ $list

or

Cases[$list, l_ :> (f[l, #] & /@ l)]

or

Replace[$list, l_ :> (f[l, #] & /@ l), {1}]

or

MapIndexed[f[$list[[#2[[1]]]], #] &, $list, {2}]

or

ReplacePart[$list, {i_, j_} :> f[$list[[i]], $list[[i, j]]]]

share|improve this answer

answered 23 mins ago

WReachWReach

55.1k22 gold badges119119 silver badges219219 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

I like the techniques in the other answers (particularly the Table approach in the answer by MelaGo), but here are a few other options to add to the mix...

Given:

$list = {{4,1}, {3,1,1}, {2,2,1}};

Then any of the following expressions yield the desired result:

Curry[f, 2][#] /@ # & /@ $list

or

Cases[$list, l_ :> (f[l, #] & /@ l)]

or

Replace[$list, l_ :> (f[l, #] & /@ l), {1}]

or

MapIndexed[f[$list[[#2[[1]]]], #] &, $list, {2}]

or

ReplacePart[$list, {i_, j_} :> f[$list[[i]], $list[[i, j]]]]

share|improve this answer

answered 23 mins ago

WReachWReach

55.1k22 gold badges119119 silver badges219219 bronze badges

$endgroup$

add a comment | 

$begingroup$

I like the techniques in the other answers (particularly the Table approach in the answer by MelaGo), but here are a few other options to add to the mix...

Given:

$list = {{4,1}, {3,1,1}, {2,2,1}};

Then any of the following expressions yield the desired result:

Curry[f, 2][#] /@ # & /@ $list

or

Cases[$list, l_ :> (f[l, #] & /@ l)]

or

Replace[$list, l_ :> (f[l, #] & /@ l), {1}]

or

MapIndexed[f[$list[[#2[[1]]]], #] &, $list, {2}]

or

ReplacePart[$list, {i_, j_} :> f[$list[[i]], $list[[i, j]]]]

share|improve this answer

answered 23 mins ago

WReachWReach

55.1k22 gold badges119119 silver badges219219 bronze badges

$endgroup$

$list = {{4,1}, {3,1,1}, {2,2,1}};

Curry[f, 2][#] /@ # & /@ $list

Cases[$list, l_ :> (f[l, #] & /@ l)]

Replace[$list, l_ :> (f[l, #] & /@ l), {1}]

MapIndexed[f[$list[[#2[[1]]]], #] &, $list, {2}]

ReplacePart[$list, {i_, j_} :> f[$list[[i]], $list[[i, j]]]]

share|improve this answer

answered 23 mins ago

WReachWReach

55.1k22 gold badges119119 silver badges219219 bronze badges

answered 23 mins ago

WReachWReach

55.1k22 gold badges119119 silver badges219219 bronze badges

answered 23 mins ago

WReachWReach

55.1k22 gold badges119119 silver badges219219 bronze badges