Selecting elements satisfying a condition












2















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







share|improve this question



























    2















    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







    share|improve this question

























      2












      2








      2


      2






      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







      share|improve this question














      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




      list-manipulation






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      asked 10 hours ago









      amator2357amator2357

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          1 Answer
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          4














          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}}}







          share|improve this answer
























          • 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
            9 hours ago











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          4














          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}}}







          share|improve this answer
























          • 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
            9 hours ago
















          4














          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}}}







          share|improve this answer
























          • 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
            9 hours ago














          4












          4








          4







          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}}}







          share|improve this answer













          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}}}








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 9 hours ago









          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
            9 hours ago



















          • 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
            9 hours ago

















          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
          9 hours ago





          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
          9 hours ago


















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