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






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 10 hours ago









      amator2357amator2357

      253




      253






















          1 Answer
          1






          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











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "387"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: false,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: null,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f189367%2fselecting-elements-satisfying-a-condition%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          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


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematica Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f189367%2fselecting-elements-satisfying-a-condition%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          How to make a Squid Proxy server?

          Image
          1 I want to set up squid proxy server for monitoring user activity in a network and Internet Web URL filtering for user. I also want to configure the user login, user log, user access group, and block unwanted web sites in my network. Please help me and guide me the full process to do this. server proxy squid share | improve this question edited Jul 31 '12 at 14:45 Jorge Castro 37.1k 107 422 617 asked Jul 31 '12 at 5:16 Goivind Tiwari Goivind Tiwari ...
          Read more

          Is this a new Fibonacci Identity?

          Image
          6 1 $begingroup$ I have found the following Fibonacci Identity (and proved it). If $F_n$ denotes the nth Fibonacci Number, we have the following identity begin{equation} F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1} end{equation} where $F_1 = F_2 = 1$ , $r leq n$ , $h leq g$ , and $n, g, k in mathbb{N}$ . It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few. I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see w...
          Read more

          19世紀

          Image
          この記事は検証可能な参考文献や出典が全く示されていないか、不十分です。 出典を追加して記事の信頼性向上にご協力ください。 （ 2012年3月 ） 千年紀: 2千年紀 世紀: 18世紀 - 19世紀 - 20世紀 十年紀: 1800年代 1810年代 1820年代 1830年代 1840年代 1850年代 1860年代 1870年代 1880年代 1890年代 19世紀に君臨した大英帝国。 19世紀 （じゅうきゅうせいき）は、西暦1801年から西暦1900年までの100年間を指す世紀。 目次 1 19世紀の歴史 1.1 国民国家の成立 1.2 帝国主義の興隆 1.3 列強の植民地争奪戦 2 できごと 2.1 1800年代 2.2 1810年代 2.3 1820年代 2.4 1830年代 2.5 1840年代 2.6 1850年代 2.7 1860年代 2.8 1870年代 2.9 1880年代 2.10 1890年代 2.11 1900年代 3 文化 3.1 文学 3.2 音楽 3.3 思想 3.4 科学 3.5 技術 4 人物 4.1 ヨーロッパ 4.1.1 政治家・王族 4.1.1.1 フランス 4.1.1.2 オーストリア＝ハンガリー 4.1.1.3 ロシア 4.1.1.4 イギリス 4.1.1.5 ドイツ（プロイセンほかドイツ領邦を含む） 4.1.1.6 北欧 4.1.1.7 イタリア 4.1.1.8 スペイン 4.1.1.9 ベルギー 4.1.1.10 ギリシア 4.1.2 軍人 4.1.3 実業家 4.1.4 科学と技術 4.1.5 思想と哲学・人文諸学 4.1.6 宗教 4.1.7 文学 4.1.8 美術 4.1.9 音楽 4.1.10 社会事業家 4.1.11 探検家・旅行家 4.1.12 料...
          Read more