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Specific list manipulation
Implementing a function which generalizes the merging step in merge sortWhat is the correct way to conditionally add elements to lists?How to create a Table of Tables with indexed variablesDealing with a huge datasetThreading elements over corresponding elements in the second listPick the maximum element of each nested listFinding the most common n-tuples of a list of lists where order does not matterGet sublists by pattern?Can a simple Part expression produce a list of elements from nested lists?Interpolating and plotting from a list
$begingroup$
Suppose I have two lists each of different size, e.g toy case ListX = {x1,x2,x3} and ListY = {y1,y2} and all possible combinations of elements give rise to an element in another list ListZ = {z1,z2,z3,z4,z5,z6} of dimension dim(ListX) x dim(ListY), i.e the tuple (x1,y1) is associated with z1, say.
How to merge all three lists such that I obtain the following combined list?
{{{x1,y1},z1}, {{x1,y2},z2}, {{x2,y1},z3}, {{x2,y2},z4}, {{x3,y1},z5},{{x3,y2},z6}}
I've tried nested tables but I always generate additional items that I don't want. The actual lists I'm dealing with are of dimension O(50) such that the equivalent of ListZ is of dimension O(2500).
list-manipulation table array
asked 3 hours ago
CAFCAF
255110
$endgroup$
add a comment |
$begingroup$
Suppose I have two lists each of different size, e.g toy case ListX = {x1,x2,x3} and ListY = {y1,y2} and all possible combinations of elements give rise to an element in another list ListZ = {z1,z2,z3,z4,z5,z6} of dimension dim(ListX) x dim(ListY), i.e the tuple (x1,y1) is associated with z1, say.
How to merge all three lists such that I obtain the following combined list?
{{{x1,y1},z1}, {{x1,y2},z2}, {{x2,y1},z3}, {{x2,y2},z4}, {{x3,y1},z5},{{x3,y2},z6}}
I've tried nested tables but I always generate additional items that I don't want. The actual lists I'm dealing with are of dimension O(50) such that the equivalent of ListZ is of dimension O(2500).
list-manipulation table array
asked 3 hours ago
CAFCAF
255110
$endgroup$
add a comment |
$begingroup$
Suppose I have two lists each of different size, e.g toy case ListX = {x1,x2,x3} and ListY = {y1,y2} and all possible combinations of elements give rise to an element in another list ListZ = {z1,z2,z3,z4,z5,z6} of dimension dim(ListX) x dim(ListY), i.e the tuple (x1,y1) is associated with z1, say.
How to merge all three lists such that I obtain the following combined list?
{{{x1,y1},z1}, {{x1,y2},z2}, {{x2,y1},z3}, {{x2,y2},z4}, {{x3,y1},z5},{{x3,y2},z6}}
I've tried nested tables but I always generate additional items that I don't want. The actual lists I'm dealing with are of dimension O(50) such that the equivalent of ListZ is of dimension O(2500).
list-manipulation table array
asked 3 hours ago
CAFCAF
255110
$endgroup$
Suppose I have two lists each of different size, e.g toy case ListX = {x1,x2,x3} and ListY = {y1,y2} and all possible combinations of elements give rise to an element in another list ListZ = {z1,z2,z3,z4,z5,z6} of dimension dim(ListX) x dim(ListY), i.e the tuple (x1,y1) is associated with z1, say.
How to merge all three lists such that I obtain the following combined list?
{{{x1,y1},z1}, {{x1,y2},z2}, {{x2,y1},z3}, {{x2,y2},z4}, {{x3,y1},z5},{{x3,y2},z6}}
I've tried nested tables but I always generate additional items that I don't want. The actual lists I'm dealing with are of dimension O(50) such that the equivalent of ListZ is of dimension O(2500).
list-manipulation table array
list-manipulation table array
asked 3 hours ago
CAFCAF
255110
asked 3 hours ago
CAFCAF
255110
asked 3 hours ago
CAFCAF
255110
asked 3 hours ago
CAFCAF
255110
asked 3 hours ago
CAFCAF
255110
255110
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This seems to do what you are looking for:
ListX = {x1, x2, x3};
ListY = {y1, y2};
ListZ = {z1, z2, z3, z4, z5, z6};
Partition[Riffle[Flatten[Outer[List, ListX, ListY], 1], ListZ], 2]
The Outer product gives all the pairs, but they are nested funny so you need to Flatten. Riffle does the interleaving of the Outer product with ListZ, and again it's not quite got your desired structure, but Partition fixes that.
answered 3 hours ago
bill sbill s
53.6k376152
$endgroup$
add a comment |
$begingroup$
Transpose@{Tuples@{ListX, ListY}, ListZ}
answered 3 hours ago
ukarukar
191
New contributor
ukar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Very nice solution! You can also useThreadinstead ofTranspose:Thread@{Tuples@{ListX, ListY}, ListZ}
$endgroup$
– rmw
2 hours ago
$begingroup$
Yes. Thanks for reminding me that.
$endgroup$
– ukar
10 mins ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This seems to do what you are looking for:
ListX = {x1, x2, x3};
ListY = {y1, y2};
ListZ = {z1, z2, z3, z4, z5, z6};
Partition[Riffle[Flatten[Outer[List, ListX, ListY], 1], ListZ], 2]
The Outer product gives all the pairs, but they are nested funny so you need to Flatten. Riffle does the interleaving of the Outer product with ListZ, and again it's not quite got your desired structure, but Partition fixes that.
answered 3 hours ago
bill sbill s
53.6k376152
$endgroup$
add a comment |
$begingroup$
This seems to do what you are looking for:
ListX = {x1, x2, x3};
ListY = {y1, y2};
ListZ = {z1, z2, z3, z4, z5, z6};
Partition[Riffle[Flatten[Outer[List, ListX, ListY], 1], ListZ], 2]
The Outer product gives all the pairs, but they are nested funny so you need to Flatten. Riffle does the interleaving of the Outer product with ListZ, and again it's not quite got your desired structure, but Partition fixes that.
answered 3 hours ago
bill sbill s
53.6k376152
$endgroup$
add a comment |
$begingroup$
This seems to do what you are looking for:
ListX = {x1, x2, x3};
ListY = {y1, y2};
ListZ = {z1, z2, z3, z4, z5, z6};
Partition[Riffle[Flatten[Outer[List, ListX, ListY], 1], ListZ], 2]
The Outer product gives all the pairs, but they are nested funny so you need to Flatten. Riffle does the interleaving of the Outer product with ListZ, and again it's not quite got your desired structure, but Partition fixes that.
answered 3 hours ago
bill sbill s
53.6k376152
$endgroup$
This seems to do what you are looking for:
ListX = {x1, x2, x3};
ListY = {y1, y2};
ListZ = {z1, z2, z3, z4, z5, z6};
Partition[Riffle[Flatten[Outer[List, ListX, ListY], 1], ListZ], 2]
The Outer product gives all the pairs, but they are nested funny so you need to Flatten. Riffle does the interleaving of the Outer product with ListZ, and again it's not quite got your desired structure, but Partition fixes that.
answered 3 hours ago
bill sbill s
53.6k376152
answered 3 hours ago
bill sbill s
53.6k376152
answered 3 hours ago
bill sbill s
53.6k376152
answered 3 hours ago
bill sbill s
53.6k376152
53.6k376152
add a comment |
add a comment |
$begingroup$
Transpose@{Tuples@{ListX, ListY}, ListZ}
answered 3 hours ago
ukarukar
191
New contributor
ukar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Very nice solution! You can also useThreadinstead ofTranspose:Thread@{Tuples@{ListX, ListY}, ListZ}
$endgroup$
– rmw
2 hours ago
$begingroup$
Yes. Thanks for reminding me that.
$endgroup$
– ukar
10 mins ago
add a comment |
$begingroup$
Transpose@{Tuples@{ListX, ListY}, ListZ}
answered 3 hours ago
ukarukar
191
New contributor
ukar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Very nice solution! You can also useThreadinstead ofTranspose:Thread@{Tuples@{ListX, ListY}, ListZ}
$endgroup$
– rmw
2 hours ago
$begingroup$
Yes. Thanks for reminding me that.
$endgroup$
– ukar
10 mins ago
add a comment |
$begingroup$
Transpose@{Tuples@{ListX, ListY}, ListZ}
answered 3 hours ago
ukarukar
191
New contributor
ukar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Transpose@{Tuples@{ListX, ListY}, ListZ}
answered 3 hours ago
ukarukar
191
New contributor
ukar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 3 hours ago
ukarukar
191
New contributor
ukar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 3 hours ago
ukarukar
191
answered 3 hours ago
ukarukar
191
191
New contributor
ukar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
ukar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
ukar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Very nice solution! You can also useThreadinstead ofTranspose:Thread@{Tuples@{ListX, ListY}, ListZ}
$endgroup$
– rmw
2 hours ago
$begingroup$
Yes. Thanks for reminding me that.
$endgroup$
– ukar
10 mins ago
add a comment |
1
$begingroup$
Very nice solution! You can also useThreadinstead ofTranspose:Thread@{Tuples@{ListX, ListY}, ListZ}
$endgroup$
– rmw
2 hours ago
$begingroup$
Yes. Thanks for reminding me that.
$endgroup$
– ukar
10 mins ago
1
1
$begingroup$
Very nice solution! You can also use
Thread instead of Transpose:Thread@{Tuples@{ListX, ListY}, ListZ}$endgroup$
– rmw
2 hours ago
$begingroup$
Very nice solution! You can also use
Thread instead of Transpose:Thread@{Tuples@{ListX, ListY}, ListZ}$endgroup$
– rmw
2 hours ago
$begingroup$
Yes. Thanks for reminding me that.
$endgroup$
– ukar
10 mins ago
$begingroup$
Yes. Thanks for reminding me that.
$endgroup$
– ukar
10 mins ago
add a comment |
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