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4

$begingroup$

I suspect this is a duplicate, but I can't seem to find what I'm looking for.

A routine problem I have is the following.

I have a set of data in three (or two, or more) lists:

l1={a1, a2, a3}

l2={b1, b2, b3, b4}

l3={{c1, c2, c3, c4}, {d1, d2, d3, d4}, {e1, e2, e3, e4}}

where c1 is a result under condition {a1, b1}, c2 is a result under condition {a1, b2}, etc.

I want to create the list:

{{a1, b1, c1}, {a1, b2, c2}, {a1, b3, c3},{a1, b4, c4}, {a2, b1, d1}, ...}

in preparation for creating a string to export to a text file.

My current solution:

Map[Transpose[{l2, #}] &, l3]

MapIndexed[Prepend[#1, l1[[#2[[1]]]]] &, %, {2}]

Flatten[%, 1]

This works, but the solution isn't intuitive to me, which makes me think there's a better way.

Is there a preferred approach for this task?

list-manipulation

share|improve this question

edited 1 hour ago

Coolwater

15k32553

asked 1 hour ago

BrianBrian

212

New contributor

Brian 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 | 

4

$begingroup$

I suspect this is a duplicate, but I can't seem to find what I'm looking for.

A routine problem I have is the following.

I have a set of data in three (or two, or more) lists:

l1={a1, a2, a3}

l2={b1, b2, b3, b4}

l3={{c1, c2, c3, c4}, {d1, d2, d3, d4}, {e1, e2, e3, e4}}

where c1 is a result under condition {a1, b1}, c2 is a result under condition {a1, b2}, etc.

I want to create the list:

{{a1, b1, c1}, {a1, b2, c2}, {a1, b3, c3},{a1, b4, c4}, {a2, b1, d1}, ...}

in preparation for creating a string to export to a text file.

My current solution:

Map[Transpose[{l2, #}] &, l3]

MapIndexed[Prepend[#1, l1[[#2[[1]]]]] &, %, {2}]

Flatten[%, 1]

This works, but the solution isn't intuitive to me, which makes me think there's a better way.

Is there a preferred approach for this task?

list-manipulation

share|improve this question

edited 1 hour ago

Coolwater

15k32553

asked 1 hour ago

BrianBrian

212

New contributor

Brian 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 suspect this is a duplicate, but I can't seem to find what I'm looking for.

A routine problem I have is the following.

I have a set of data in three (or two, or more) lists:

l1={a1, a2, a3}

l2={b1, b2, b3, b4}

l3={{c1, c2, c3, c4}, {d1, d2, d3, d4}, {e1, e2, e3, e4}}

where c1 is a result under condition {a1, b1}, c2 is a result under condition {a1, b2}, etc.

I want to create the list:

{{a1, b1, c1}, {a1, b2, c2}, {a1, b3, c3},{a1, b4, c4}, {a2, b1, d1}, ...}

in preparation for creating a string to export to a text file.

My current solution:

Map[Transpose[{l2, #}] &, l3]

MapIndexed[Prepend[#1, l1[[#2[[1]]]]] &, %, {2}]

Flatten[%, 1]

This works, but the solution isn't intuitive to me, which makes me think there's a better way.

Is there a preferred approach for this task?

list-manipulation

share|improve this question

edited 1 hour ago

Coolwater

15k32553

asked 1 hour ago

BrianBrian

212

New contributor

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

$endgroup$

l1={a1, a2, a3}

l2={b1, b2, b3, b4}

l3={{c1, c2, c3, c4}, {d1, d2, d3, d4}, {e1, e2, e3, e4}}

{{a1, b1, c1}, {a1, b2, c2}, {a1, b3, c3},{a1, b4, c4}, {a2, b1, d1}, ...}

Map[Transpose[{l2, #}] &, l3]

MapIndexed[Prepend[#1, l1[[#2[[1]]]]] &, %, {2}]

Flatten[%, 1]

share|improve this question

edited 1 hour ago

Coolwater

15k32553

asked 1 hour ago

BrianBrian

212

New contributor

Brian 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

edited 1 hour ago

Coolwater

15k32553

asked 1 hour ago

BrianBrian

212

New contributor

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

edited 1 hour ago

Coolwater

15k32553

edited 1 hour ago

Coolwater

15k32553

asked 1 hour ago

BrianBrian

212

New contributor

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

asked 1 hour ago

BrianBrian

212

2 Answers
2

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2

$begingroup$

Join[Tuples[{l1, l2}], ArrayReshape[l3, {Times @@ Dimensions[l3], 1}], 2]

{{a1, b1, c1}, {a1, b2, c2}, {a1, b3, c3}, ...}

Though if the elements of l3 are list of equal length, then ArrayReshape/Dimension won't work.

To avoid that problem you could write

ArrayReshape[Riffle[Tuples[{l1, l2}], #], {Length[#], 3}] &[Catenate[l3]]

share|improve this answer

edited 55 mins ago

answered 1 hour ago

CoolwaterCoolwater

15k32553

$endgroup$

add a comment | 

2

$begingroup$

This will generate a nested list, in accordance with l3:

MapThread[Append, {Outer[List, l1, l2], l3}, 2]

{{{a1, b1, c1}, {a1, b2, c2}, {a1, b3, c3}, {a1, b4, c4}}, {{a2, b1, d1}, {a2, b2, d2}, {a2, b3, d3}, {a2, b4, d4}}, {{a3, b1, e1}, {a3, b2, e2}, {a3, b3, e3}, {a3, b4, e4}}}

Flattening once will give you what you want:

Flatten[MapThread[Append, {Outer[List, l1, l2], l3}, 2], 1]

{{a1, b1, c1}, {a1, b2, c2}, {a1, b3, c3}, {a1, b4, c4}, {a2, b1, d1}, {a2, b2, d2}, {a2, b3, d3}, {a2, b4, d4}, {a3, b1, e1}, {a3, b2, e2}, {a3, b3, e3}, {a3, b4, e4}}

If you're not interested in the nested list above, then you can get straight to the result with

MapThread[Append, {Tuples[{l1, l2}], Flatten[l3]}]

{{a1, b1, c1}, {a1, b2, c2}, {a1, b3, c3}, {a1, b4, c4}, {a2, b1, d1}, {a2, b2, d2}, {a2, b3, d3}, {a2, b4, d4}, {a3, b1, e1}, {a3, b2, e2}, {a3, b3, e3}, {a3, b4, e4}}

(in effect, Flattening before MapThreading).

share|improve this answer

edited 24 mins ago

answered 42 mins ago

RomanRoman

1,626614

$endgroup$

add a comment | 

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2

$begingroup$

Join[Tuples[{l1, l2}], ArrayReshape[l3, {Times @@ Dimensions[l3], 1}], 2]

{{a1, b1, c1}, {a1, b2, c2}, {a1, b3, c3}, ...}

Though if the elements of l3 are list of equal length, then ArrayReshape/Dimension won't work.

To avoid that problem you could write

ArrayReshape[Riffle[Tuples[{l1, l2}], #], {Length[#], 3}] &[Catenate[l3]]

share|improve this answer

edited 55 mins ago

answered 1 hour ago

CoolwaterCoolwater

15k32553

$endgroup$

add a comment | 

2

$begingroup$

Join[Tuples[{l1, l2}], ArrayReshape[l3, {Times @@ Dimensions[l3], 1}], 2]

{{a1, b1, c1}, {a1, b2, c2}, {a1, b3, c3}, ...}

Though if the elements of l3 are list of equal length, then ArrayReshape/Dimension won't work.

To avoid that problem you could write

ArrayReshape[Riffle[Tuples[{l1, l2}], #], {Length[#], 3}] &[Catenate[l3]]

share|improve this answer

edited 55 mins ago

answered 1 hour ago

CoolwaterCoolwater

15k32553

$endgroup$

add a comment | 

$begingroup$

Join[Tuples[{l1, l2}], ArrayReshape[l3, {Times @@ Dimensions[l3], 1}], 2]

{{a1, b1, c1}, {a1, b2, c2}, {a1, b3, c3}, ...}

Though if the elements of l3 are list of equal length, then ArrayReshape/Dimension won't work.

To avoid that problem you could write

ArrayReshape[Riffle[Tuples[{l1, l2}], #], {Length[#], 3}] &[Catenate[l3]]

share|improve this answer

edited 55 mins ago

answered 1 hour ago

CoolwaterCoolwater

15k32553

$endgroup$

Join[Tuples[{l1, l2}], ArrayReshape[l3, {Times @@ Dimensions[l3], 1}], 2]

ArrayReshape[Riffle[Tuples[{l1, l2}], #], {Length[#], 3}] &[Catenate[l3]]

share|improve this answer

edited 55 mins ago

answered 1 hour ago

CoolwaterCoolwater

15k32553

answered 1 hour ago

CoolwaterCoolwater

15k32553

answered 1 hour ago

CoolwaterCoolwater

15k32553

2

$begingroup$

This will generate a nested list, in accordance with l3:

MapThread[Append, {Outer[List, l1, l2], l3}, 2]

{{{a1, b1, c1}, {a1, b2, c2}, {a1, b3, c3}, {a1, b4, c4}}, {{a2, b1, d1}, {a2, b2, d2}, {a2, b3, d3}, {a2, b4, d4}}, {{a3, b1, e1}, {a3, b2, e2}, {a3, b3, e3}, {a3, b4, e4}}}

Flattening once will give you what you want:

Flatten[MapThread[Append, {Outer[List, l1, l2], l3}, 2], 1]

{{a1, b1, c1}, {a1, b2, c2}, {a1, b3, c3}, {a1, b4, c4}, {a2, b1, d1}, {a2, b2, d2}, {a2, b3, d3}, {a2, b4, d4}, {a3, b1, e1}, {a3, b2, e2}, {a3, b3, e3}, {a3, b4, e4}}

If you're not interested in the nested list above, then you can get straight to the result with

MapThread[Append, {Tuples[{l1, l2}], Flatten[l3]}]

{{a1, b1, c1}, {a1, b2, c2}, {a1, b3, c3}, {a1, b4, c4}, {a2, b1, d1}, {a2, b2, d2}, {a2, b3, d3}, {a2, b4, d4}, {a3, b1, e1}, {a3, b2, e2}, {a3, b3, e3}, {a3, b4, e4}}

(in effect, Flattening before MapThreading).

share|improve this answer

edited 24 mins ago

answered 42 mins ago

RomanRoman

1,626614

$endgroup$

add a comment | 

2

$begingroup$

This will generate a nested list, in accordance with l3:

MapThread[Append, {Outer[List, l1, l2], l3}, 2]

{{{a1, b1, c1}, {a1, b2, c2}, {a1, b3, c3}, {a1, b4, c4}}, {{a2, b1, d1}, {a2, b2, d2}, {a2, b3, d3}, {a2, b4, d4}}, {{a3, b1, e1}, {a3, b2, e2}, {a3, b3, e3}, {a3, b4, e4}}}

Flattening once will give you what you want:

Flatten[MapThread[Append, {Outer[List, l1, l2], l3}, 2], 1]

{{a1, b1, c1}, {a1, b2, c2}, {a1, b3, c3}, {a1, b4, c4}, {a2, b1, d1}, {a2, b2, d2}, {a2, b3, d3}, {a2, b4, d4}, {a3, b1, e1}, {a3, b2, e2}, {a3, b3, e3}, {a3, b4, e4}}

If you're not interested in the nested list above, then you can get straight to the result with

MapThread[Append, {Tuples[{l1, l2}], Flatten[l3]}]

{{a1, b1, c1}, {a1, b2, c2}, {a1, b3, c3}, {a1, b4, c4}, {a2, b1, d1}, {a2, b2, d2}, {a2, b3, d3}, {a2, b4, d4}, {a3, b1, e1}, {a3, b2, e2}, {a3, b3, e3}, {a3, b4, e4}}

(in effect, Flattening before MapThreading).

share|improve this answer

edited 24 mins ago

answered 42 mins ago

RomanRoman

1,626614

$endgroup$

add a comment | 

$begingroup$

This will generate a nested list, in accordance with l3:

MapThread[Append, {Outer[List, l1, l2], l3}, 2]

{{{a1, b1, c1}, {a1, b2, c2}, {a1, b3, c3}, {a1, b4, c4}}, {{a2, b1, d1}, {a2, b2, d2}, {a2, b3, d3}, {a2, b4, d4}}, {{a3, b1, e1}, {a3, b2, e2}, {a3, b3, e3}, {a3, b4, e4}}}

Flattening once will give you what you want:

Flatten[MapThread[Append, {Outer[List, l1, l2], l3}, 2], 1]

{{a1, b1, c1}, {a1, b2, c2}, {a1, b3, c3}, {a1, b4, c4}, {a2, b1, d1}, {a2, b2, d2}, {a2, b3, d3}, {a2, b4, d4}, {a3, b1, e1}, {a3, b2, e2}, {a3, b3, e3}, {a3, b4, e4}}

If you're not interested in the nested list above, then you can get straight to the result with

MapThread[Append, {Tuples[{l1, l2}], Flatten[l3]}]

{{a1, b1, c1}, {a1, b2, c2}, {a1, b3, c3}, {a1, b4, c4}, {a2, b1, d1}, {a2, b2, d2}, {a2, b3, d3}, {a2, b4, d4}, {a3, b1, e1}, {a3, b2, e2}, {a3, b3, e3}, {a3, b4, e4}}

(in effect, Flattening before MapThreading).

share|improve this answer

edited 24 mins ago

answered 42 mins ago

RomanRoman

1,626614

$endgroup$

MapThread[Append, {Outer[List, l1, l2], l3}, 2]

Flatten[MapThread[Append, {Outer[List, l1, l2], l3}, 2], 1]

MapThread[Append, {Tuples[{l1, l2}], Flatten[l3]}]

share|improve this answer

edited 24 mins ago

answered 42 mins ago

RomanRoman

1,626614

answered 42 mins ago

RomanRoman

1,626614

answered 42 mins ago

RomanRoman

1,626614