Selecting elements satisfying a condition












4















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







share|improve this question





























    4















    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







    share|improve this question



























      4












      4








      4


      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 filtering







      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 filtering




      list-manipulation filtering






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 21 hours ago









      Mr.Wizard♦

      230k294741040




      230k294741040










      asked yesterday









      amator2357amator2357

      353




      353






















          2 Answers
          2






          active

          oldest

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          6














          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
            yesterday



















          1














          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}}}







          share|improve this answer























            Your Answer





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            2 Answers
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            active

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            2 Answers
            2






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            6














            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
              yesterday
















            6














            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
              yesterday














            6












            6








            6







            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 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



















            • 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











            1














            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}}}







            share|improve this answer




























              1














              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}}}







              share|improve this answer


























                1












                1








                1







                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}}}







                share|improve this answer













                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}}}








                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 18 hours ago









                kglrkglr

                178k9198409




                178k9198409






























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