Reshaping nested list












3












$begingroup$


I have the feeling this is a very basic question, but I can't seem to find the way to solve this easily.



I have imported a set of data from a txt file into a table so the resulting list has the following structure:



{{x1,y1,f1},{x2,y2,f2},...}


I want it to reshape it into this form:



{{{x1,y1},f1},{{x2,y2},f2},...}


I apologize because I know it must be trivial but I haven't found the way yet. I've always had problems manipulating lists in Mathematica so, if anyone has any resource I can use to learn about this it'd be very much appreciated.



Regards






list-manipulation







share|improve this question







New contributor




Mikel García is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    Look at the output for Apply[{{#1, #2}, #3} &, {a, b, c}] and it should help you. Check Apply for more details.
    $endgroup$
    – Jason B.
    5 hours ago










  • $begingroup$
    Also possible is the following: reshape = Curry[Replace, {3, 1, 2}][{s_, t_, u_} :> {{s, t}, u}, {1}]. Then {{x1, y1, f1}, {x2, y2, f2}} // reshape gives the desired result.
    $endgroup$
    – Shredderroy
    3 hours ago










  • $begingroup$
    @JasonB. this should be Apply[{{#1, #2}, #3} &, {{x1, y1, f1}, {x2, y2, f2}}, {1}]
    $endgroup$
    – GenericAccountName
    2 hours ago
















3












$begingroup$


I have the feeling this is a very basic question, but I can't seem to find the way to solve this easily.



I have imported a set of data from a txt file into a table so the resulting list has the following structure:



{{x1,y1,f1},{x2,y2,f2},...}


I want it to reshape it into this form:



{{{x1,y1},f1},{{x2,y2},f2},...}


I apologize because I know it must be trivial but I haven't found the way yet. I've always had problems manipulating lists in Mathematica so, if anyone has any resource I can use to learn about this it'd be very much appreciated.



Regards






list-manipulation







share|improve this question







New contributor




Mikel García is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    Look at the output for Apply[{{#1, #2}, #3} &, {a, b, c}] and it should help you. Check Apply for more details.
    $endgroup$
    – Jason B.
    5 hours ago










  • $begingroup$
    Also possible is the following: reshape = Curry[Replace, {3, 1, 2}][{s_, t_, u_} :> {{s, t}, u}, {1}]. Then {{x1, y1, f1}, {x2, y2, f2}} // reshape gives the desired result.
    $endgroup$
    – Shredderroy
    3 hours ago










  • $begingroup$
    @JasonB. this should be Apply[{{#1, #2}, #3} &, {{x1, y1, f1}, {x2, y2, f2}}, {1}]
    $endgroup$
    – GenericAccountName
    2 hours ago














3












3








3





$begingroup$


I have the feeling this is a very basic question, but I can't seem to find the way to solve this easily.



I have imported a set of data from a txt file into a table so the resulting list has the following structure:



{{x1,y1,f1},{x2,y2,f2},...}


I want it to reshape it into this form:



{{{x1,y1},f1},{{x2,y2},f2},...}


I apologize because I know it must be trivial but I haven't found the way yet. I've always had problems manipulating lists in Mathematica so, if anyone has any resource I can use to learn about this it'd be very much appreciated.



Regards






list-manipulation







share|improve this question







New contributor




Mikel García is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I have the feeling this is a very basic question, but I can't seem to find the way to solve this easily.



I have imported a set of data from a txt file into a table so the resulting list has the following structure:



{{x1,y1,f1},{x2,y2,f2},...}


I want it to reshape it into this form:



{{{x1,y1},f1},{{x2,y2},f2},...}


I apologize because I know it must be trivial but I haven't found the way yet. I've always had problems manipulating lists in Mathematica so, if anyone has any resource I can use to learn about this it'd be very much appreciated.



Regards






list-manipulation




list-manipulation






share|improve this question







New contributor




Mikel García is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question







New contributor




Mikel García is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question






New contributor




Mikel García is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 5 hours ago









Mikel GarcíaMikel García

161




161




New contributor




Mikel García is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Mikel García is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Mikel García is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    Look at the output for Apply[{{#1, #2}, #3} &, {a, b, c}] and it should help you. Check Apply for more details.
    $endgroup$
    – Jason B.
    5 hours ago










  • $begingroup$
    Also possible is the following: reshape = Curry[Replace, {3, 1, 2}][{s_, t_, u_} :> {{s, t}, u}, {1}]. Then {{x1, y1, f1}, {x2, y2, f2}} // reshape gives the desired result.
    $endgroup$
    – Shredderroy
    3 hours ago










  • $begingroup$
    @JasonB. this should be Apply[{{#1, #2}, #3} &, {{x1, y1, f1}, {x2, y2, f2}}, {1}]
    $endgroup$
    – GenericAccountName
    2 hours ago














  • 1




    $begingroup$
    Look at the output for Apply[{{#1, #2}, #3} &, {a, b, c}] and it should help you. Check Apply for more details.
    $endgroup$
    – Jason B.
    5 hours ago










  • $begingroup$
    Also possible is the following: reshape = Curry[Replace, {3, 1, 2}][{s_, t_, u_} :> {{s, t}, u}, {1}]. Then {{x1, y1, f1}, {x2, y2, f2}} // reshape gives the desired result.
    $endgroup$
    – Shredderroy
    3 hours ago










  • $begingroup$
    @JasonB. this should be Apply[{{#1, #2}, #3} &, {{x1, y1, f1}, {x2, y2, f2}}, {1}]
    $endgroup$
    – GenericAccountName
    2 hours ago








1




1




$begingroup$
Look at the output for Apply[{{#1, #2}, #3} &, {a, b, c}] and it should help you. Check Apply for more details.
$endgroup$
– Jason B.
5 hours ago




$begingroup$
Look at the output for Apply[{{#1, #2}, #3} &, {a, b, c}] and it should help you. Check Apply for more details.
$endgroup$
– Jason B.
5 hours ago












$begingroup$
Also possible is the following: reshape = Curry[Replace, {3, 1, 2}][{s_, t_, u_} :> {{s, t}, u}, {1}]. Then {{x1, y1, f1}, {x2, y2, f2}} // reshape gives the desired result.
$endgroup$
– Shredderroy
3 hours ago




$begingroup$
Also possible is the following: reshape = Curry[Replace, {3, 1, 2}][{s_, t_, u_} :> {{s, t}, u}, {1}]. Then {{x1, y1, f1}, {x2, y2, f2}} // reshape gives the desired result.
$endgroup$
– Shredderroy
3 hours ago












$begingroup$
@JasonB. this should be Apply[{{#1, #2}, #3} &, {{x1, y1, f1}, {x2, y2, f2}}, {1}]
$endgroup$
– GenericAccountName
2 hours ago




$begingroup$
@JasonB. this should be Apply[{{#1, #2}, #3} &, {{x1, y1, f1}, {x2, y2, f2}}, {1}]
$endgroup$
– GenericAccountName
2 hours ago










4 Answers
4






active

oldest

votes


















2












$begingroup$

list = {{x1, y1, f1}, {x2, y2, f2}};



Transpose[{Drop[list, 0, -1], list[[All, -1]]}]

(* {{{x1, y1}, f1}, {{x2, y2}, f2}} *)





share|improve this answer








New contributor




Oppenede is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$





















    2












    $begingroup$

    list = {{x1, y1, f1}, {x2, y2, f2}};
    

    
{Most[#], Last[#]} & /@ list
    

    
(* {{{x1, y1}, f1}, {{x2, y2}, f2}} *)





    share|improve this answer









    $endgroup$





















      1












      $begingroup$

      I think my vote is this so far, but I bet there's even cleaner/more clever ways to do this:



      data = {{x1, y1, f1}, {x2, y2, f2}}
      
Replace[data, {x_, y_, f_} :> {{x, y}, f}, {1}]


      I like the Apply solution just fine too: 



      Apply[{{#1, #2}, #3} &, {{x1, y1, f1}, {x2, y2, f2}}, {1}]


      For larger data they perform similarly, with Replace winning slightly:



      In[53]:= data=(ToExpression/@{"x"<>#,"y"<>#,"f"<>#})&/@(ToString[#]&/@Range[1000000]);
      
Apply[{{#1,#2},#3}&,data,{1}]//RepeatedTiming//First
      
Replace[data,{x_,y_,f_}:>{{x,y},f},{1}]//RepeatedTiming//First
      
Out[54]= 0.825
      
Out[55]= 0.740


      This is much better than the most naive approach:



      In[57]:= ({{#[[1]], #[[2]]}, #[[3]]} & /@ data) // 
      
  RepeatedTiming // First
      

      
Out[57]= 1.859


      The solution from @Oppenede is similar in timing to Apply



      In[77]:= Transpose[{Drop[data, 0, -1], data[[All, -1]]}] // RepeatedTiming // First
      

      
Out[77]= 0.836





      share|improve this answer









      $endgroup$





















        1












        $begingroup$

        Use a rule and replacement:



        {{x1, y1, f1}, {x2, y2, f2}} /. {x_, y_, f_} -> {{x, y}, f}
        

        
(* {{{x1,y1},f1},{{x2,y2},f2}} *)





        share|improve this answer









        $endgroup$













          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
          });


          }
          });






          Mikel García is a new contributor. Be nice, and check out our Code of Conduct.










          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f190877%2freshaping-nested-list%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          list = {{x1, y1, f1}, {x2, y2, f2}};
          

          
Transpose[{Drop[list, 0, -1], list[[All, -1]]}]
          
(* {{{x1, y1}, f1}, {{x2, y2}, f2}} *)





          share|improve this answer








          New contributor




          Oppenede is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$


















            2












            $begingroup$

            list = {{x1, y1, f1}, {x2, y2, f2}};
            

            
Transpose[{Drop[list, 0, -1], list[[All, -1]]}]
            
(* {{{x1, y1}, f1}, {{x2, y2}, f2}} *)





            share|improve this answer








            New contributor




            Oppenede is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$
















              2












              2








              2





              $begingroup$

              list = {{x1, y1, f1}, {x2, y2, f2}};
              

              
Transpose[{Drop[list, 0, -1], list[[All, -1]]}]
              
(* {{{x1, y1}, f1}, {{x2, y2}, f2}} *)





              share|improve this answer








              New contributor




              Oppenede is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






              $endgroup$



              list = {{x1, y1, f1}, {x2, y2, f2}};
              

              
Transpose[{Drop[list, 0, -1], list[[All, -1]]}]
              
(* {{{x1, y1}, f1}, {{x2, y2}, f2}} *)






              share|improve this answer








              New contributor




              Oppenede is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.









              share|improve this answer



              share|improve this answer






              New contributor




              Oppenede is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.









              answered 4 hours ago









              OppenedeOppenede

              1211




              1211




              New contributor




              Oppenede is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.





              New contributor





              Oppenede is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






              Oppenede is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.























                  2












                  $begingroup$

                  list = {{x1, y1, f1}, {x2, y2, f2}};
                  

                  
{Most[#], Last[#]} & /@ list
                  

                  
(* {{{x1, y1}, f1}, {{x2, y2}, f2}} *)





                  share|improve this answer









                  $endgroup$


















                    2












                    $begingroup$

                    list = {{x1, y1, f1}, {x2, y2, f2}};
                    

                    
{Most[#], Last[#]} & /@ list
                    

                    
(* {{{x1, y1}, f1}, {{x2, y2}, f2}} *)





                    share|improve this answer









                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      list = {{x1, y1, f1}, {x2, y2, f2}};
                      

                      
{Most[#], Last[#]} & /@ list
                      

                      
(* {{{x1, y1}, f1}, {{x2, y2}, f2}} *)





                      share|improve this answer









                      $endgroup$



                      list = {{x1, y1, f1}, {x2, y2, f2}};
                      

                      
{Most[#], Last[#]} & /@ list
                      

                      
(* {{{x1, y1}, f1}, {{x2, y2}, f2}} *)






                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered 2 hours ago









                      Rohit NamjoshiRohit Namjoshi

                      8351212




                      8351212























                          1












                          $begingroup$

                          I think my vote is this so far, but I bet there's even cleaner/more clever ways to do this:



                          data = {{x1, y1, f1}, {x2, y2, f2}}
                          
Replace[data, {x_, y_, f_} :> {{x, y}, f}, {1}]


                          I like the Apply solution just fine too: 



                          Apply[{{#1, #2}, #3} &, {{x1, y1, f1}, {x2, y2, f2}}, {1}]


                          For larger data they perform similarly, with Replace winning slightly:



                          In[53]:= data=(ToExpression/@{"x"<>#,"y"<>#,"f"<>#})&/@(ToString[#]&/@Range[1000000]);
                          
Apply[{{#1,#2},#3}&,data,{1}]//RepeatedTiming//First
                          
Replace[data,{x_,y_,f_}:>{{x,y},f},{1}]//RepeatedTiming//First
                          
Out[54]= 0.825
                          
Out[55]= 0.740


                          This is much better than the most naive approach:



                          In[57]:= ({{#[[1]], #[[2]]}, #[[3]]} & /@ data) // 
                          
  RepeatedTiming // First
                          

                          
Out[57]= 1.859


                          The solution from @Oppenede is similar in timing to Apply



                          In[77]:= Transpose[{Drop[data, 0, -1], data[[All, -1]]}] // RepeatedTiming // First
                          

                          
Out[77]= 0.836





                          share|improve this answer









                          $endgroup$


















                            1












                            $begingroup$

                            I think my vote is this so far, but I bet there's even cleaner/more clever ways to do this:



                            data = {{x1, y1, f1}, {x2, y2, f2}}
                            
Replace[data, {x_, y_, f_} :> {{x, y}, f}, {1}]


                            I like the Apply solution just fine too: 



                            Apply[{{#1, #2}, #3} &, {{x1, y1, f1}, {x2, y2, f2}}, {1}]


                            For larger data they perform similarly, with Replace winning slightly:



                            In[53]:= data=(ToExpression/@{"x"<>#,"y"<>#,"f"<>#})&/@(ToString[#]&/@Range[1000000]);
                            
Apply[{{#1,#2},#3}&,data,{1}]//RepeatedTiming//First
                            
Replace[data,{x_,y_,f_}:>{{x,y},f},{1}]//RepeatedTiming//First
                            
Out[54]= 0.825
                            
Out[55]= 0.740


                            This is much better than the most naive approach:



                            In[57]:= ({{#[[1]], #[[2]]}, #[[3]]} & /@ data) // 
                            
  RepeatedTiming // First
                            

                            
Out[57]= 1.859


                            The solution from @Oppenede is similar in timing to Apply



                            In[77]:= Transpose[{Drop[data, 0, -1], data[[All, -1]]}] // RepeatedTiming // First
                            

                            
Out[77]= 0.836





                            share|improve this answer









                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              I think my vote is this so far, but I bet there's even cleaner/more clever ways to do this:



                              data = {{x1, y1, f1}, {x2, y2, f2}}
                              
Replace[data, {x_, y_, f_} :> {{x, y}, f}, {1}]


                              I like the Apply solution just fine too: 



                              Apply[{{#1, #2}, #3} &, {{x1, y1, f1}, {x2, y2, f2}}, {1}]


                              For larger data they perform similarly, with Replace winning slightly:



                              In[53]:= data=(ToExpression/@{"x"<>#,"y"<>#,"f"<>#})&/@(ToString[#]&/@Range[1000000]);
                              
Apply[{{#1,#2},#3}&,data,{1}]//RepeatedTiming//First
                              
Replace[data,{x_,y_,f_}:>{{x,y},f},{1}]//RepeatedTiming//First
                              
Out[54]= 0.825
                              
Out[55]= 0.740


                              This is much better than the most naive approach:



                              In[57]:= ({{#[[1]], #[[2]]}, #[[3]]} & /@ data) // 
                              
  RepeatedTiming // First
                              

                              
Out[57]= 1.859


                              The solution from @Oppenede is similar in timing to Apply



                              In[77]:= Transpose[{Drop[data, 0, -1], data[[All, -1]]}] // RepeatedTiming // First
                              

                              
Out[77]= 0.836





                              share|improve this answer









                              $endgroup$



                              I think my vote is this so far, but I bet there's even cleaner/more clever ways to do this:



                              data = {{x1, y1, f1}, {x2, y2, f2}}
                              
Replace[data, {x_, y_, f_} :> {{x, y}, f}, {1}]


                              I like the Apply solution just fine too: 



                              Apply[{{#1, #2}, #3} &, {{x1, y1, f1}, {x2, y2, f2}}, {1}]


                              For larger data they perform similarly, with Replace winning slightly:



                              In[53]:= data=(ToExpression/@{"x"<>#,"y"<>#,"f"<>#})&/@(ToString[#]&/@Range[1000000]);
                              
Apply[{{#1,#2},#3}&,data,{1}]//RepeatedTiming//First
                              
Replace[data,{x_,y_,f_}:>{{x,y},f},{1}]//RepeatedTiming//First
                              
Out[54]= 0.825
                              
Out[55]= 0.740


                              This is much better than the most naive approach:



                              In[57]:= ({{#[[1]], #[[2]]}, #[[3]]} & /@ data) // 
                              
  RepeatedTiming // First
                              

                              
Out[57]= 1.859


                              The solution from @Oppenede is similar in timing to Apply



                              In[77]:= Transpose[{Drop[data, 0, -1], data[[All, -1]]}] // RepeatedTiming // First
                              

                              
Out[77]= 0.836






                              share|improve this answer












                              share|improve this answer



                              share|improve this answer










                              answered 2 hours ago









                              GenericAccountNameGenericAccountName

                              68227




                              68227























                                  1












                                  $begingroup$

                                  Use a rule and replacement:



                                  {{x1, y1, f1}, {x2, y2, f2}} /. {x_, y_, f_} -> {{x, y}, f}
                                  

                                  
(* {{{x1,y1},f1},{{x2,y2},f2}} *)





                                  share|improve this answer









                                  $endgroup$


















                                    1












                                    $begingroup$

                                    Use a rule and replacement:



                                    {{x1, y1, f1}, {x2, y2, f2}} /. {x_, y_, f_} -> {{x, y}, f}
                                    

                                    
(* {{{x1,y1},f1},{{x2,y2},f2}} *)





                                    share|improve this answer









                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      Use a rule and replacement:



                                      {{x1, y1, f1}, {x2, y2, f2}} /. {x_, y_, f_} -> {{x, y}, f}
                                      

                                      
(* {{{x1,y1},f1},{{x2,y2},f2}} *)





                                      share|improve this answer









                                      $endgroup$



                                      Use a rule and replacement:



                                      {{x1, y1, f1}, {x2, y2, f2}} /. {x_, y_, f_} -> {{x, y}, f}
                                      

                                      
(* {{{x1,y1},f1},{{x2,y2},f2}} *)






                                      share|improve this answer












                                      share|improve this answer



                                      share|improve this answer










                                      answered 2 hours ago









                                      David KeithDavid Keith

                                      1,151313




                                      1,151313






















                                          Mikel García is a new contributor. Be nice, and check out our Code of Conduct.










                                          draft saved

                                          draft discarded


















                                          Mikel García is a new contributor. Be nice, and check out our Code of Conduct.













                                          Mikel García is a new contributor. Be nice, and check out our Code of Conduct.












                                          Mikel García is a new contributor. Be nice, and check out our Code of Conduct.
















                                          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%2f190877%2freshaping-nested-list%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