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Adding elements to some sublists of unequal length

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4

1

$begingroup$

Complicated title for a simple problem:
I have list

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

In which the first element of each sublist is basically an index. The following values are the actual data.
Then I want to add data to this list, e.g.,:

b = {{2, 1}, {4, 3}}

Which means that to the list with the index '2' the '1' should be added and the list with the index '4' the '3' should be added, so the result reads:

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

I found a couple of rather complicated, i.e., time consuming solutions involving loops. However, the actual datset is huge and is part of a numerical simulation, i.e., this procedure needs to be very fast.

list-manipulation symbolic

share|improve this question

edited 8 hours ago

Vitaliy Kaurov

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asked 9 hours ago

Mockup DungeonMockup Dungeon

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$endgroup$

add a comment
 | 

4

1

$begingroup$

Complicated title for a simple problem:
I have list

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

In which the first element of each sublist is basically an index. The following values are the actual data.
Then I want to add data to this list, e.g.,:

b = {{2, 1}, {4, 3}}

Which means that to the list with the index '2' the '1' should be added and the list with the index '4' the '3' should be added, so the result reads:

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

I found a couple of rather complicated, i.e., time consuming solutions involving loops. However, the actual datset is huge and is part of a numerical simulation, i.e., this procedure needs to be very fast.

list-manipulation symbolic

share|improve this question

edited 8 hours ago

Vitaliy Kaurov

59k66 gold badges166166 silver badges286286 bronze badges

asked 9 hours ago

Mockup DungeonMockup Dungeon

91766 silver badges1313 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Complicated title for a simple problem:
I have list

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

In which the first element of each sublist is basically an index. The following values are the actual data.
Then I want to add data to this list, e.g.,:

b = {{2, 1}, {4, 3}}

Which means that to the list with the index '2' the '1' should be added and the list with the index '4' the '3' should be added, so the result reads:

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

I found a couple of rather complicated, i.e., time consuming solutions involving loops. However, the actual datset is huge and is part of a numerical simulation, i.e., this procedure needs to be very fast.

list-manipulation symbolic

share|improve this question

edited 8 hours ago

Vitaliy Kaurov

59k66 gold badges166166 silver badges286286 bronze badges

asked 9 hours ago

Mockup DungeonMockup Dungeon

91766 silver badges1313 bronze badges

$endgroup$

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

b = {{2, 1}, {4, 3}}

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

share|improve this question

edited 8 hours ago

Vitaliy Kaurov

59k66 gold badges166166 silver badges286286 bronze badges

asked 9 hours ago

Mockup DungeonMockup Dungeon

91766 silver badges1313 bronze badges

share|improve this question

edited 8 hours ago

Vitaliy Kaurov

59k66 gold badges166166 silver badges286286 bronze badges

asked 9 hours ago

Mockup DungeonMockup Dungeon

91766 silver badges1313 bronze badges

edited 8 hours ago

Vitaliy Kaurov

59k66 gold badges166166 silver badges286286 bronze badges

edited 8 hours ago

Vitaliy Kaurov

59k66 gold badges166166 silver badges286286 bronze badges

asked 9 hours ago

Mockup DungeonMockup Dungeon

91766 silver badges1313 bronze badges

asked 9 hours ago

Mockup DungeonMockup Dungeon

91766 silver badges1313 bronze badges

3 Answers
3

active

oldest

votes

5

$begingroup$

Fold[Insert[#1, Last[#2], {First[#2], -1}] &, a, b]

share|improve this answer

answered 8 hours ago

Suba ThomasSuba Thomas

4,0561111 silver badges2020 bronze badges

$endgroup$

add a comment
 | 

2

$begingroup$

Perhaps something like this:

ReplacePart[a, #1 -> Append[a[[#1]], #2]& @@@ b]

If data are large and speedup is needed you might want to try Join

ReplacePart[a, #1 -> Join[a[[#1]], {#2}] & @@@ b]

In case you want value of a to be the new list with added elements, you whether can use AppendTo:

ReplacePart[a, #1 -> AppendTo[a[[#1]], #2] & @@@ b]

or simply reassign:

a = ReplacePart[a, #1 -> Append[a[[#1]], #2] & @@@ b]

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Vitaliy KaurovVitaliy Kaurov

59k66 gold badges166166 silver badges286286 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

You can make b into a list of rules

rules = {#, a__} :> {#, a, #2} & @@@ b;

and use it with ReplaceAll:

a /. rules

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

or with Replace:

Replace[a, rules, All]

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

This approach works for any ordering of the elements in the input list:

c = RandomSample[a]

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

c /. rules

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

Replace[c, rules, All]

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

share|improve this answer

edited 7 hours ago

answered 8 hours ago

kglrkglr

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$endgroup$

add a comment
 | 

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5

$begingroup$

Fold[Insert[#1, Last[#2], {First[#2], -1}] &, a, b]

share|improve this answer

answered 8 hours ago

Suba ThomasSuba Thomas

4,0561111 silver badges2020 bronze badges

$endgroup$

add a comment
 | 

5

$begingroup$

Fold[Insert[#1, Last[#2], {First[#2], -1}] &, a, b]

share|improve this answer

answered 8 hours ago

Suba ThomasSuba Thomas

4,0561111 silver badges2020 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Fold[Insert[#1, Last[#2], {First[#2], -1}] &, a, b]

share|improve this answer

answered 8 hours ago

Suba ThomasSuba Thomas

4,0561111 silver badges2020 bronze badges

$endgroup$

Fold[Insert[#1, Last[#2], {First[#2], -1}] &, a, b]

share|improve this answer

answered 8 hours ago

Suba ThomasSuba Thomas

4,0561111 silver badges2020 bronze badges

answered 8 hours ago

Suba ThomasSuba Thomas

4,0561111 silver badges2020 bronze badges

answered 8 hours ago

Suba ThomasSuba Thomas

4,0561111 silver badges2020 bronze badges

2

$begingroup$

Perhaps something like this:

ReplacePart[a, #1 -> Append[a[[#1]], #2]& @@@ b]

If data are large and speedup is needed you might want to try Join

ReplacePart[a, #1 -> Join[a[[#1]], {#2}] & @@@ b]

In case you want value of a to be the new list with added elements, you whether can use AppendTo:

ReplacePart[a, #1 -> AppendTo[a[[#1]], #2] & @@@ b]

or simply reassign:

a = ReplacePart[a, #1 -> Append[a[[#1]], #2] & @@@ b]

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Vitaliy KaurovVitaliy Kaurov

59k66 gold badges166166 silver badges286286 bronze badges

$endgroup$

add a comment
 | 

2

$begingroup$

Perhaps something like this:

ReplacePart[a, #1 -> Append[a[[#1]], #2]& @@@ b]

If data are large and speedup is needed you might want to try Join

ReplacePart[a, #1 -> Join[a[[#1]], {#2}] & @@@ b]

In case you want value of a to be the new list with added elements, you whether can use AppendTo:

ReplacePart[a, #1 -> AppendTo[a[[#1]], #2] & @@@ b]

or simply reassign:

a = ReplacePart[a, #1 -> Append[a[[#1]], #2] & @@@ b]

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Vitaliy KaurovVitaliy Kaurov

59k66 gold badges166166 silver badges286286 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Perhaps something like this:

ReplacePart[a, #1 -> Append[a[[#1]], #2]& @@@ b]

If data are large and speedup is needed you might want to try Join

ReplacePart[a, #1 -> Join[a[[#1]], {#2}] & @@@ b]

In case you want value of a to be the new list with added elements, you whether can use AppendTo:

ReplacePart[a, #1 -> AppendTo[a[[#1]], #2] & @@@ b]

or simply reassign:

a = ReplacePart[a, #1 -> Append[a[[#1]], #2] & @@@ b]

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Vitaliy KaurovVitaliy Kaurov

59k66 gold badges166166 silver badges286286 bronze badges

$endgroup$

ReplacePart[a, #1 -> Append[a[[#1]], #2]& @@@ b]

ReplacePart[a, #1 -> Join[a[[#1]], {#2}] & @@@ b]

ReplacePart[a, #1 -> AppendTo[a[[#1]], #2] & @@@ b]

a = ReplacePart[a, #1 -> Append[a[[#1]], #2] & @@@ b]

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Vitaliy KaurovVitaliy Kaurov

59k66 gold badges166166 silver badges286286 bronze badges

answered 8 hours ago

Vitaliy KaurovVitaliy Kaurov

59k66 gold badges166166 silver badges286286 bronze badges

answered 8 hours ago

Vitaliy KaurovVitaliy Kaurov

59k66 gold badges166166 silver badges286286 bronze badges

1

$begingroup$

You can make b into a list of rules

rules = {#, a__} :> {#, a, #2} & @@@ b;

and use it with ReplaceAll:

a /. rules

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

or with Replace:

Replace[a, rules, All]

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

This approach works for any ordering of the elements in the input list:

c = RandomSample[a]

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

c /. rules

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

Replace[c, rules, All]

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

share|improve this answer

edited 7 hours ago

answered 8 hours ago

kglrkglr

220k1010 gold badges250250 silver badges504504 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

You can make b into a list of rules

rules = {#, a__} :> {#, a, #2} & @@@ b;

and use it with ReplaceAll:

a /. rules

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

or with Replace:

Replace[a, rules, All]

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

This approach works for any ordering of the elements in the input list:

c = RandomSample[a]

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

c /. rules

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

Replace[c, rules, All]

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

share|improve this answer

edited 7 hours ago

answered 8 hours ago

kglrkglr

220k1010 gold badges250250 silver badges504504 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

You can make b into a list of rules

rules = {#, a__} :> {#, a, #2} & @@@ b;

and use it with ReplaceAll:

a /. rules

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

or with Replace:

Replace[a, rules, All]

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

This approach works for any ordering of the elements in the input list:

c = RandomSample[a]

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

c /. rules

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

Replace[c, rules, All]

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

share|improve this answer

edited 7 hours ago

answered 8 hours ago

kglrkglr

220k1010 gold badges250250 silver badges504504 bronze badges

$endgroup$

rules = {#, a__} :> {#, a, #2} & @@@ b;

a /. rules

Replace[a, rules, All]

c = RandomSample[a]

c /. rules

Replace[c, rules, All]

share|improve this answer

edited 7 hours ago

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