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







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$endgroup$








  • 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.
    4 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.
    4 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




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share|improve this question




share|improve this question






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asked 5 hours ago









Mikel GarcíaMikel García

161




161




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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.






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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.
    4 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.
    4 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.
4 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.
4 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








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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









        $endgroup$













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          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.
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              share|improve this answer



              share|improve this answer






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              answered 4 hours ago









              OppenedeOppenede

              1211




              1211




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                  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






















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