Selecting elements satisfying a condition
I have a list of lists(of lists):
{{{1}, {2}, {1, 2, 3}}, {{1}, {3}, {1, 2, 3}}, {{1}, {1, 2}, {1,2,
3}}, {{1}, {2, 3}, {1, 2, 3}}, {{2}, {3}, {1, 2, 3}}, {{2}, {1,
2}, {1, 2, 3}}, {{2}, {2, 3}, {1, 2, 3}}, {{3}, {1, 2}, {1, 2,
3}}, {{3}, {2, 3}, {1, 2, 3}}, {{1, 2}, {2, 3}, {1, 2, 3}}}
From the list above I want to select those lists in which two consecutive elements share either a first element or last element. It also counts if the first element of the list shares the first entry with the second element of that list, but the second element shares the last entry with the third element, etc. So from the list above I would get:
{{1},{1,2},{1,2,3}},{{2},{1,2},{1,2,3}},{{2},{2,3},{1,2,3}},{{3},{2,3},{1,2,3}}
So I did the following:test4 = Table[Select[test3,First[#[[i+1]]]==First[#[[i]]] [Or] Last[#[[i+1]]]==Last[#[[i]]] &],{i,n-1}]
where test3 is my list of lists from above and n is the maximum length that the element in one of the list of lists can have, so in our case it is 3 and that number is known in advance. Unforunately the above doesn't work. It spits out more elements then there are in the initial list. I think it picks the elements which satify that condition for at least one pair of consecutive elements, which is not what I'm after. I'd really appreciate some help with this. Thanks!
list-manipulation filtering
edited 21 hours ago
Mr.Wizard♦
230k294741040
asked yesterday
amator2357amator2357
353
add a comment |
I have a list of lists(of lists):
{{{1}, {2}, {1, 2, 3}}, {{1}, {3}, {1, 2, 3}}, {{1}, {1, 2}, {1,2,
3}}, {{1}, {2, 3}, {1, 2, 3}}, {{2}, {3}, {1, 2, 3}}, {{2}, {1,
2}, {1, 2, 3}}, {{2}, {2, 3}, {1, 2, 3}}, {{3}, {1, 2}, {1, 2,
3}}, {{3}, {2, 3}, {1, 2, 3}}, {{1, 2}, {2, 3}, {1, 2, 3}}}
From the list above I want to select those lists in which two consecutive elements share either a first element or last element. It also counts if the first element of the list shares the first entry with the second element of that list, but the second element shares the last entry with the third element, etc. So from the list above I would get:
{{1},{1,2},{1,2,3}},{{2},{1,2},{1,2,3}},{{2},{2,3},{1,2,3}},{{3},{2,3},{1,2,3}}
So I did the following:test4 = Table[Select[test3,First[#[[i+1]]]==First[#[[i]]] [Or] Last[#[[i+1]]]==Last[#[[i]]] &],{i,n-1}]
where test3 is my list of lists from above and n is the maximum length that the element in one of the list of lists can have, so in our case it is 3 and that number is known in advance. Unforunately the above doesn't work. It spits out more elements then there are in the initial list. I think it picks the elements which satify that condition for at least one pair of consecutive elements, which is not what I'm after. I'd really appreciate some help with this. Thanks!
list-manipulation filtering
edited 21 hours ago
Mr.Wizard♦
230k294741040
asked yesterday
amator2357amator2357
353
add a comment |
I have a list of lists(of lists):
{{{1}, {2}, {1, 2, 3}}, {{1}, {3}, {1, 2, 3}}, {{1}, {1, 2}, {1,2,
3}}, {{1}, {2, 3}, {1, 2, 3}}, {{2}, {3}, {1, 2, 3}}, {{2}, {1,
2}, {1, 2, 3}}, {{2}, {2, 3}, {1, 2, 3}}, {{3}, {1, 2}, {1, 2,
3}}, {{3}, {2, 3}, {1, 2, 3}}, {{1, 2}, {2, 3}, {1, 2, 3}}}
From the list above I want to select those lists in which two consecutive elements share either a first element or last element. It also counts if the first element of the list shares the first entry with the second element of that list, but the second element shares the last entry with the third element, etc. So from the list above I would get:
{{1},{1,2},{1,2,3}},{{2},{1,2},{1,2,3}},{{2},{2,3},{1,2,3}},{{3},{2,3},{1,2,3}}
So I did the following:test4 = Table[Select[test3,First[#[[i+1]]]==First[#[[i]]] [Or] Last[#[[i+1]]]==Last[#[[i]]] &],{i,n-1}]
where test3 is my list of lists from above and n is the maximum length that the element in one of the list of lists can have, so in our case it is 3 and that number is known in advance. Unforunately the above doesn't work. It spits out more elements then there are in the initial list. I think it picks the elements which satify that condition for at least one pair of consecutive elements, which is not what I'm after. I'd really appreciate some help with this. Thanks!
list-manipulation filtering
edited 21 hours ago
Mr.Wizard♦
230k294741040
asked yesterday
amator2357amator2357
353
I have a list of lists(of lists):
{{{1}, {2}, {1, 2, 3}}, {{1}, {3}, {1, 2, 3}}, {{1}, {1, 2}, {1,2,
3}}, {{1}, {2, 3}, {1, 2, 3}}, {{2}, {3}, {1, 2, 3}}, {{2}, {1,
2}, {1, 2, 3}}, {{2}, {2, 3}, {1, 2, 3}}, {{3}, {1, 2}, {1, 2,
3}}, {{3}, {2, 3}, {1, 2, 3}}, {{1, 2}, {2, 3}, {1, 2, 3}}}
From the list above I want to select those lists in which two consecutive elements share either a first element or last element. It also counts if the first element of the list shares the first entry with the second element of that list, but the second element shares the last entry with the third element, etc. So from the list above I would get:
{{1},{1,2},{1,2,3}},{{2},{1,2},{1,2,3}},{{2},{2,3},{1,2,3}},{{3},{2,3},{1,2,3}}
So I did the following:test4 = Table[Select[test3,First[#[[i+1]]]==First[#[[i]]] [Or] Last[#[[i+1]]]==Last[#[[i]]] &],{i,n-1}]
where test3 is my list of lists from above and n is the maximum length that the element in one of the list of lists can have, so in our case it is 3 and that number is known in advance. Unforunately the above doesn't work. It spits out more elements then there are in the initial list. I think it picks the elements which satify that condition for at least one pair of consecutive elements, which is not what I'm after. I'd really appreciate some help with this. Thanks!
list-manipulation filtering
list-manipulation filtering
edited 21 hours ago
Mr.Wizard♦
230k294741040
asked yesterday
amator2357amator2357
353
edited 21 hours ago
Mr.Wizard♦
230k294741040
asked yesterday
amator2357amator2357
353
edited 21 hours ago
Mr.Wizard♦
230k294741040
edited 21 hours ago
Mr.Wizard♦
230k294741040
edited 21 hours ago
Mr.Wizard♦
230k294741040
230k294741040
asked yesterday
amator2357amator2357
353
asked yesterday
amator2357amator2357
353
asked yesterday
amator2357amator2357
353
353
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
When I'm stuck with a problem like that, I always try to decompose it into simple, testable functions, that I can understand more or less at one glance.
The first thing you want is to check if two lists share the first or last element:
shareFirstOrLast[{a_, b_}] :=
First[a] == First[b] || Last[a] == Last[b]
Then you want to extend that to more than two elements:
shareFirstOrLast[{a_, b_, rest__}] :=
shareFirstOrLast[{a, b}] && shareFirstOrLast[{b, rest}]
(this recursion will essentially call shareFirstOrLast[{a_, b_}] on every consecutive pair of elements.)
Finally, test that function on different samples and edge cases. If it fits your needs, use it with select:
Select[test3, shareFirstOrLast]
{{{1}, {1, 2}, {1, 2, 3}}, {{2}, {1, 2}, {1, 2, 3}}, {{2}, {2, 3}, {1,
2, 3}}, {{3}, {2, 3}, {1, 2, 3}}}
answered yesterday
Niki EstnerNiki Estner
30.9k374132
Brilliant, thank you so much! I really like your approach, I was looking at this problem from too wide of a perspective. I really appreciate it!
– amator2357
yesterday
add a comment |
ClearAll[chainedQ]
chainedQ = And @@ BlockMap[Or @@ Equal @@@ Transpose@#[[All, {1, -1}]] &, #, 2, 1] &;
Select[list, chainedQ]
{{{1}, {1, 2}, {1, 2, 3}}, {{2}, {1, 2}, {1, 2, 3}}, {{2}, {2, 3}, {1,
2, 3}}, {{3}, {2, 3}, {1, 2, 3}}}
Pick[list, chainedQ /@ list]
{{{1}, {1, 2}, {1, 2, 3}}, {{2}, {1, 2}, {1, 2, 3}}, {{2}, {2, 3}, {1,
2, 3}}, {{3}, {2, 3}, {1, 2, 3}}}
answered 18 hours ago
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
When I'm stuck with a problem like that, I always try to decompose it into simple, testable functions, that I can understand more or less at one glance.
The first thing you want is to check if two lists share the first or last element:
shareFirstOrLast[{a_, b_}] :=
First[a] == First[b] || Last[a] == Last[b]
Then you want to extend that to more than two elements:
shareFirstOrLast[{a_, b_, rest__}] :=
shareFirstOrLast[{a, b}] && shareFirstOrLast[{b, rest}]
(this recursion will essentially call shareFirstOrLast[{a_, b_}] on every consecutive pair of elements.)
Finally, test that function on different samples and edge cases. If it fits your needs, use it with select:
Select[test3, shareFirstOrLast]
{{{1}, {1, 2}, {1, 2, 3}}, {{2}, {1, 2}, {1, 2, 3}}, {{2}, {2, 3}, {1,
2, 3}}, {{3}, {2, 3}, {1, 2, 3}}}
answered yesterday
Niki EstnerNiki Estner
30.9k374132
Brilliant, thank you so much! I really like your approach, I was looking at this problem from too wide of a perspective. I really appreciate it!
– amator2357
yesterday
add a comment |
When I'm stuck with a problem like that, I always try to decompose it into simple, testable functions, that I can understand more or less at one glance.
The first thing you want is to check if two lists share the first or last element:
shareFirstOrLast[{a_, b_}] :=
First[a] == First[b] || Last[a] == Last[b]
Then you want to extend that to more than two elements:
shareFirstOrLast[{a_, b_, rest__}] :=
shareFirstOrLast[{a, b}] && shareFirstOrLast[{b, rest}]
(this recursion will essentially call shareFirstOrLast[{a_, b_}] on every consecutive pair of elements.)
Finally, test that function on different samples and edge cases. If it fits your needs, use it with select:
Select[test3, shareFirstOrLast]
{{{1}, {1, 2}, {1, 2, 3}}, {{2}, {1, 2}, {1, 2, 3}}, {{2}, {2, 3}, {1,
2, 3}}, {{3}, {2, 3}, {1, 2, 3}}}
answered yesterday
Niki EstnerNiki Estner
30.9k374132
Brilliant, thank you so much! I really like your approach, I was looking at this problem from too wide of a perspective. I really appreciate it!
– amator2357
yesterday
add a comment |
When I'm stuck with a problem like that, I always try to decompose it into simple, testable functions, that I can understand more or less at one glance.
The first thing you want is to check if two lists share the first or last element:
shareFirstOrLast[{a_, b_}] :=
First[a] == First[b] || Last[a] == Last[b]
Then you want to extend that to more than two elements:
shareFirstOrLast[{a_, b_, rest__}] :=
shareFirstOrLast[{a, b}] && shareFirstOrLast[{b, rest}]
(this recursion will essentially call shareFirstOrLast[{a_, b_}] on every consecutive pair of elements.)
Finally, test that function on different samples and edge cases. If it fits your needs, use it with select:
Select[test3, shareFirstOrLast]
{{{1}, {1, 2}, {1, 2, 3}}, {{2}, {1, 2}, {1, 2, 3}}, {{2}, {2, 3}, {1,
2, 3}}, {{3}, {2, 3}, {1, 2, 3}}}
answered yesterday
Niki EstnerNiki Estner
30.9k374132
When I'm stuck with a problem like that, I always try to decompose it into simple, testable functions, that I can understand more or less at one glance.
The first thing you want is to check if two lists share the first or last element:
shareFirstOrLast[{a_, b_}] :=
First[a] == First[b] || Last[a] == Last[b]
Then you want to extend that to more than two elements:
shareFirstOrLast[{a_, b_, rest__}] :=
shareFirstOrLast[{a, b}] && shareFirstOrLast[{b, rest}]
(this recursion will essentially call shareFirstOrLast[{a_, b_}] on every consecutive pair of elements.)
Finally, test that function on different samples and edge cases. If it fits your needs, use it with select:
Select[test3, shareFirstOrLast]
{{{1}, {1, 2}, {1, 2, 3}}, {{2}, {1, 2}, {1, 2, 3}}, {{2}, {2, 3}, {1,
2, 3}}, {{3}, {2, 3}, {1, 2, 3}}}
answered yesterday
Niki EstnerNiki Estner
30.9k374132
answered yesterday
Niki EstnerNiki Estner
30.9k374132
answered yesterday
Niki EstnerNiki Estner
30.9k374132
answered yesterday
Niki EstnerNiki Estner
30.9k374132
30.9k374132
Brilliant, thank you so much! I really like your approach, I was looking at this problem from too wide of a perspective. I really appreciate it!
– amator2357
yesterday
add a comment |
Brilliant, thank you so much! I really like your approach, I was looking at this problem from too wide of a perspective. I really appreciate it!
– amator2357
yesterday
Brilliant, thank you so much! I really like your approach, I was looking at this problem from too wide of a perspective. I really appreciate it!
– amator2357
yesterday
Brilliant, thank you so much! I really like your approach, I was looking at this problem from too wide of a perspective. I really appreciate it!
– amator2357
yesterday
add a comment |
ClearAll[chainedQ]
chainedQ = And @@ BlockMap[Or @@ Equal @@@ Transpose@#[[All, {1, -1}]] &, #, 2, 1] &;
Select[list, chainedQ]
{{{1}, {1, 2}, {1, 2, 3}}, {{2}, {1, 2}, {1, 2, 3}}, {{2}, {2, 3}, {1,
2, 3}}, {{3}, {2, 3}, {1, 2, 3}}}
Pick[list, chainedQ /@ list]
{{{1}, {1, 2}, {1, 2, 3}}, {{2}, {1, 2}, {1, 2, 3}}, {{2}, {2, 3}, {1,
2, 3}}, {{3}, {2, 3}, {1, 2, 3}}}
answered 18 hours ago
kglrkglr
178k9198409
add a comment |
ClearAll[chainedQ]
chainedQ = And @@ BlockMap[Or @@ Equal @@@ Transpose@#[[All, {1, -1}]] &, #, 2, 1] &;
Select[list, chainedQ]
{{{1}, {1, 2}, {1, 2, 3}}, {{2}, {1, 2}, {1, 2, 3}}, {{2}, {2, 3}, {1,
2, 3}}, {{3}, {2, 3}, {1, 2, 3}}}
Pick[list, chainedQ /@ list]
{{{1}, {1, 2}, {1, 2, 3}}, {{2}, {1, 2}, {1, 2, 3}}, {{2}, {2, 3}, {1,
2, 3}}, {{3}, {2, 3}, {1, 2, 3}}}
answered 18 hours ago
kglrkglr
178k9198409
add a comment |
ClearAll[chainedQ]
chainedQ = And @@ BlockMap[Or @@ Equal @@@ Transpose@#[[All, {1, -1}]] &, #, 2, 1] &;
Select[list, chainedQ]
{{{1}, {1, 2}, {1, 2, 3}}, {{2}, {1, 2}, {1, 2, 3}}, {{2}, {2, 3}, {1,
2, 3}}, {{3}, {2, 3}, {1, 2, 3}}}
Pick[list, chainedQ /@ list]
{{{1}, {1, 2}, {1, 2, 3}}, {{2}, {1, 2}, {1, 2, 3}}, {{2}, {2, 3}, {1,
2, 3}}, {{3}, {2, 3}, {1, 2, 3}}}
answered 18 hours ago
kglrkglr
178k9198409
ClearAll[chainedQ]
chainedQ = And @@ BlockMap[Or @@ Equal @@@ Transpose@#[[All, {1, -1}]] &, #, 2, 1] &;
Select[list, chainedQ]
{{{1}, {1, 2}, {1, 2, 3}}, {{2}, {1, 2}, {1, 2, 3}}, {{2}, {2, 3}, {1,
2, 3}}, {{3}, {2, 3}, {1, 2, 3}}}
Pick[list, chainedQ /@ list]
{{{1}, {1, 2}, {1, 2, 3}}, {{2}, {1, 2}, {1, 2, 3}}, {{2}, {2, 3}, {1,
2, 3}}, {{3}, {2, 3}, {1, 2, 3}}}
answered 18 hours ago
kglrkglr
178k9198409
answered 18 hours ago
kglrkglr
178k9198409
answered 18 hours ago
kglrkglr
178k9198409
answered 18 hours ago
kglrkglr
178k9198409
178k9198409
add a comment |
add a comment |
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