Truncate a set of lists until a subset of entries matches?
Consider lists of vectors of elements like
list[1]={{a1,b1,c11},{a2,b2,c12},{a3,b3,c13},{a4,b4,c14},{a5,b5,c15}};
list[2]={{a1,b1,c21} ,{a3,b3,c23},{a4,b4,c24},{a5,b5,c25}};
list[3]={{a1,b1,c31},{a2,b2,c32},{a3,b3,c33} ,{a5,b5,c35}};
where the length 3 of vectors in lists is just an example and could be longer, and there may be more than 3 such lists.
I would like to have a function truncate[listoflists_] that takes all these lists and truncates them to only have the vectors containing ai,bi entries that are present in all of them:
{list[1],list[2],list[3]}=truncate[{list[1],list[2],list[3]}];
list[1]
list[2]
list[3]
{{a1,b1,c11},{a3,b3,c13},{a5,b5,c15}}
{{a1,b1,c21},{a3,b3,c23},{a5,b5,c25}}
{{a1,b1,c31},{a3,b3,c33},{a5,b5,c35}}
Is there a quick way to do this in Mathematica? Thanks for any suggestion.
list-manipulation function-construction
asked Nov 23 '18 at 21:04
KagaratschKagaratsch
4,59231246
add a comment |
Consider lists of vectors of elements like
list[1]={{a1,b1,c11},{a2,b2,c12},{a3,b3,c13},{a4,b4,c14},{a5,b5,c15}};
list[2]={{a1,b1,c21} ,{a3,b3,c23},{a4,b4,c24},{a5,b5,c25}};
list[3]={{a1,b1,c31},{a2,b2,c32},{a3,b3,c33} ,{a5,b5,c35}};
where the length 3 of vectors in lists is just an example and could be longer, and there may be more than 3 such lists.
I would like to have a function truncate[listoflists_] that takes all these lists and truncates them to only have the vectors containing ai,bi entries that are present in all of them:
{list[1],list[2],list[3]}=truncate[{list[1],list[2],list[3]}];
list[1]
list[2]
list[3]
{{a1,b1,c11},{a3,b3,c13},{a5,b5,c15}}
{{a1,b1,c21},{a3,b3,c23},{a5,b5,c25}}
{{a1,b1,c31},{a3,b3,c33},{a5,b5,c35}}
Is there a quick way to do this in Mathematica? Thanks for any suggestion.
list-manipulation function-construction
asked Nov 23 '18 at 21:04
KagaratschKagaratsch
4,59231246
add a comment |
Consider lists of vectors of elements like
list[1]={{a1,b1,c11},{a2,b2,c12},{a3,b3,c13},{a4,b4,c14},{a5,b5,c15}};
list[2]={{a1,b1,c21} ,{a3,b3,c23},{a4,b4,c24},{a5,b5,c25}};
list[3]={{a1,b1,c31},{a2,b2,c32},{a3,b3,c33} ,{a5,b5,c35}};
where the length 3 of vectors in lists is just an example and could be longer, and there may be more than 3 such lists.
I would like to have a function truncate[listoflists_] that takes all these lists and truncates them to only have the vectors containing ai,bi entries that are present in all of them:
{list[1],list[2],list[3]}=truncate[{list[1],list[2],list[3]}];
list[1]
list[2]
list[3]
{{a1,b1,c11},{a3,b3,c13},{a5,b5,c15}}
{{a1,b1,c21},{a3,b3,c23},{a5,b5,c25}}
{{a1,b1,c31},{a3,b3,c33},{a5,b5,c35}}
Is there a quick way to do this in Mathematica? Thanks for any suggestion.
list-manipulation function-construction
asked Nov 23 '18 at 21:04
KagaratschKagaratsch
4,59231246
Consider lists of vectors of elements like
list[1]={{a1,b1,c11},{a2,b2,c12},{a3,b3,c13},{a4,b4,c14},{a5,b5,c15}};
list[2]={{a1,b1,c21} ,{a3,b3,c23},{a4,b4,c24},{a5,b5,c25}};
list[3]={{a1,b1,c31},{a2,b2,c32},{a3,b3,c33} ,{a5,b5,c35}};
where the length 3 of vectors in lists is just an example and could be longer, and there may be more than 3 such lists.
I would like to have a function truncate[listoflists_] that takes all these lists and truncates them to only have the vectors containing ai,bi entries that are present in all of them:
{list[1],list[2],list[3]}=truncate[{list[1],list[2],list[3]}];
list[1]
list[2]
list[3]
{{a1,b1,c11},{a3,b3,c13},{a5,b5,c15}}
{{a1,b1,c21},{a3,b3,c23},{a5,b5,c25}}
{{a1,b1,c31},{a3,b3,c33},{a5,b5,c35}}
Is there a quick way to do this in Mathematica? Thanks for any suggestion.
list-manipulation function-construction
list-manipulation function-construction
asked Nov 23 '18 at 21:04
KagaratschKagaratsch
4,59231246
asked Nov 23 '18 at 21:04
KagaratschKagaratsch
4,59231246
asked Nov 23 '18 at 21:04
KagaratschKagaratsch
4,59231246
asked Nov 23 '18 at 21:04
KagaratschKagaratsch
4,59231246
asked Nov 23 '18 at 21:04
KagaratschKagaratsch
4,59231246
4,59231246
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
It is short (does not mean quick) to get it via associations:
truncate = Values @ KeyIntersection @ Map[#[[;; 2]] -> # &, #, {2}] &
We convert {a1,b1,c11} to {a1,b1}->{a1,b1,c11} and everything else is straightforward.
answered Nov 23 '18 at 21:23
Kuba♦Kuba
104k12201518
1
Love it when a mod ends up in the LQ review. :)
– Michael E2
Nov 23 '18 at 21:45
@MichaelE2 ok, explanation added.
– Kuba♦
Nov 23 '18 at 21:53
I approved it anyway. It's not always clear that excessive exposition would be an improvement.
– Michael E2
Nov 23 '18 at 22:00
I have changed it totruncate = Values@KeyIntersection@Map[{#[[1]], #[[2]]} -> # &, #, {2}] &to make it compatible with longer vector cases.
– Kagaratsch
Nov 23 '18 at 23:35
1
@Kagaratsch right, you mentioned it. I edited the code.
– Kuba♦
Nov 24 '18 at 0:27
add a comment |
Also
tF = Module[{i = Intersection@@#[[;;, ;;, #2]], k = #2}, Select[MemberQ[i, #[[k]]]&]/@ #]&
lists = list /@ {1, 2, 3};
tF[lists, {1, 2}]
{{{a1, b1, c11}, {a3, b3, c13}, {a5, b5, c15}},
{{a1, b1, c21}, {a3, b3, c23}, {a5, b5, c25}},
{{a1, b1, c31}, {a3, b3, c33}, {a5, b5, c35}}}
answered Nov 24 '18 at 19:32
kglrkglr
178k9198409
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
It is short (does not mean quick) to get it via associations:
truncate = Values @ KeyIntersection @ Map[#[[;; 2]] -> # &, #, {2}] &
We convert {a1,b1,c11} to {a1,b1}->{a1,b1,c11} and everything else is straightforward.
answered Nov 23 '18 at 21:23
Kuba♦Kuba
104k12201518
1
Love it when a mod ends up in the LQ review. :)
– Michael E2
Nov 23 '18 at 21:45
@MichaelE2 ok, explanation added.
– Kuba♦
Nov 23 '18 at 21:53
I approved it anyway. It's not always clear that excessive exposition would be an improvement.
– Michael E2
Nov 23 '18 at 22:00
I have changed it totruncate = Values@KeyIntersection@Map[{#[[1]], #[[2]]} -> # &, #, {2}] &to make it compatible with longer vector cases.
– Kagaratsch
Nov 23 '18 at 23:35
1
@Kagaratsch right, you mentioned it. I edited the code.
– Kuba♦
Nov 24 '18 at 0:27
add a comment |
It is short (does not mean quick) to get it via associations:
truncate = Values @ KeyIntersection @ Map[#[[;; 2]] -> # &, #, {2}] &
We convert {a1,b1,c11} to {a1,b1}->{a1,b1,c11} and everything else is straightforward.
answered Nov 23 '18 at 21:23
Kuba♦Kuba
104k12201518
1
Love it when a mod ends up in the LQ review. :)
– Michael E2
Nov 23 '18 at 21:45
@MichaelE2 ok, explanation added.
– Kuba♦
Nov 23 '18 at 21:53
I approved it anyway. It's not always clear that excessive exposition would be an improvement.
– Michael E2
Nov 23 '18 at 22:00
I have changed it totruncate = Values@KeyIntersection@Map[{#[[1]], #[[2]]} -> # &, #, {2}] &to make it compatible with longer vector cases.
– Kagaratsch
Nov 23 '18 at 23:35
1
@Kagaratsch right, you mentioned it. I edited the code.
– Kuba♦
Nov 24 '18 at 0:27
add a comment |
It is short (does not mean quick) to get it via associations:
truncate = Values @ KeyIntersection @ Map[#[[;; 2]] -> # &, #, {2}] &
We convert {a1,b1,c11} to {a1,b1}->{a1,b1,c11} and everything else is straightforward.
answered Nov 23 '18 at 21:23
Kuba♦Kuba
104k12201518
It is short (does not mean quick) to get it via associations:
truncate = Values @ KeyIntersection @ Map[#[[;; 2]] -> # &, #, {2}] &
We convert {a1,b1,c11} to {a1,b1}->{a1,b1,c11} and everything else is straightforward.
answered Nov 23 '18 at 21:23
Kuba♦Kuba
104k12201518
edited Nov 24 '18 at 0:27
answered Nov 23 '18 at 21:23
Kuba♦Kuba
104k12201518
answered Nov 23 '18 at 21:23
Kuba♦Kuba
104k12201518
answered Nov 23 '18 at 21:23
Kuba♦Kuba
104k12201518
104k12201518
1
Love it when a mod ends up in the LQ review. :)
– Michael E2
Nov 23 '18 at 21:45
@MichaelE2 ok, explanation added.
– Kuba♦
Nov 23 '18 at 21:53
I approved it anyway. It's not always clear that excessive exposition would be an improvement.
– Michael E2
Nov 23 '18 at 22:00
I have changed it totruncate = Values@KeyIntersection@Map[{#[[1]], #[[2]]} -> # &, #, {2}] &to make it compatible with longer vector cases.
– Kagaratsch
Nov 23 '18 at 23:35
1
@Kagaratsch right, you mentioned it. I edited the code.
– Kuba♦
Nov 24 '18 at 0:27
add a comment |
1
Love it when a mod ends up in the LQ review. :)
– Michael E2
Nov 23 '18 at 21:45
@MichaelE2 ok, explanation added.
– Kuba♦
Nov 23 '18 at 21:53
I approved it anyway. It's not always clear that excessive exposition would be an improvement.
– Michael E2
Nov 23 '18 at 22:00
I have changed it totruncate = Values@KeyIntersection@Map[{#[[1]], #[[2]]} -> # &, #, {2}] &to make it compatible with longer vector cases.
– Kagaratsch
Nov 23 '18 at 23:35
1
@Kagaratsch right, you mentioned it. I edited the code.
– Kuba♦
Nov 24 '18 at 0:27
1
1
Love it when a mod ends up in the LQ review. :)
– Michael E2
Nov 23 '18 at 21:45
Love it when a mod ends up in the LQ review. :)
– Michael E2
Nov 23 '18 at 21:45
@MichaelE2 ok, explanation added.
– Kuba♦
Nov 23 '18 at 21:53
@MichaelE2 ok, explanation added.
– Kuba♦
Nov 23 '18 at 21:53
I approved it anyway. It's not always clear that excessive exposition would be an improvement.
– Michael E2
Nov 23 '18 at 22:00
I approved it anyway. It's not always clear that excessive exposition would be an improvement.
– Michael E2
Nov 23 '18 at 22:00
I have changed it to
truncate = Values@KeyIntersection@Map[{#[[1]], #[[2]]} -> # &, #, {2}] & to make it compatible with longer vector cases.– Kagaratsch
Nov 23 '18 at 23:35
I have changed it to
truncate = Values@KeyIntersection@Map[{#[[1]], #[[2]]} -> # &, #, {2}] & to make it compatible with longer vector cases.– Kagaratsch
Nov 23 '18 at 23:35
1
1
@Kagaratsch right, you mentioned it. I edited the code.
– Kuba♦
Nov 24 '18 at 0:27
@Kagaratsch right, you mentioned it. I edited the code.
– Kuba♦
Nov 24 '18 at 0:27
add a comment |
Also
tF = Module[{i = Intersection@@#[[;;, ;;, #2]], k = #2}, Select[MemberQ[i, #[[k]]]&]/@ #]&
lists = list /@ {1, 2, 3};
tF[lists, {1, 2}]
{{{a1, b1, c11}, {a3, b3, c13}, {a5, b5, c15}},
{{a1, b1, c21}, {a3, b3, c23}, {a5, b5, c25}},
{{a1, b1, c31}, {a3, b3, c33}, {a5, b5, c35}}}
answered Nov 24 '18 at 19:32
kglrkglr
178k9198409
add a comment |
Also
tF = Module[{i = Intersection@@#[[;;, ;;, #2]], k = #2}, Select[MemberQ[i, #[[k]]]&]/@ #]&
lists = list /@ {1, 2, 3};
tF[lists, {1, 2}]
{{{a1, b1, c11}, {a3, b3, c13}, {a5, b5, c15}},
{{a1, b1, c21}, {a3, b3, c23}, {a5, b5, c25}},
{{a1, b1, c31}, {a3, b3, c33}, {a5, b5, c35}}}
answered Nov 24 '18 at 19:32
kglrkglr
178k9198409
add a comment |
Also
tF = Module[{i = Intersection@@#[[;;, ;;, #2]], k = #2}, Select[MemberQ[i, #[[k]]]&]/@ #]&
lists = list /@ {1, 2, 3};
tF[lists, {1, 2}]
{{{a1, b1, c11}, {a3, b3, c13}, {a5, b5, c15}},
{{a1, b1, c21}, {a3, b3, c23}, {a5, b5, c25}},
{{a1, b1, c31}, {a3, b3, c33}, {a5, b5, c35}}}
answered Nov 24 '18 at 19:32
kglrkglr
178k9198409
Also
tF = Module[{i = Intersection@@#[[;;, ;;, #2]], k = #2}, Select[MemberQ[i, #[[k]]]&]/@ #]&
lists = list /@ {1, 2, 3};
tF[lists, {1, 2}]
{{{a1, b1, c11}, {a3, b3, c13}, {a5, b5, c15}},
{{a1, b1, c21}, {a3, b3, c23}, {a5, b5, c25}},
{{a1, b1, c31}, {a3, b3, c33}, {a5, b5, c35}}}
answered Nov 24 '18 at 19:32
kglrkglr
178k9198409
answered Nov 24 '18 at 19:32
kglrkglr
178k9198409
answered Nov 24 '18 at 19:32
kglrkglr
178k9198409
answered Nov 24 '18 at 19:32
kglrkglr
178k9198409
178k9198409
add a comment |
add a comment |
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