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What's the easiest way to combine specific columns of two matrices?



Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?How to combine elements of two matrices?Best way to create symmetric matrices with block structure and additional symmetries of the submatricesChange entire column of matrix using its own elements for calculationsRiffle two matrices COLUMNWISE; where boh matices have the same number of rowsInterleaving the columns of two matrices horizontally in a particular wayFinding a column from Max comparing each element of two columns; subsequent column used for backward calculationCombine two matrices of unequal dimensionsObtaining all possible ways to concatenate matricesExtracting Elements of a Matrix of Matrices to Create two separate MatricesChanging the position of rows and columns in a matrix










3












$begingroup$


given matrices



A = 3, 10, 10, 5.`, 4, 10, 10, 6.`, 5, 10, 10, 7.`
B = 6, 10, 10, 5.`, 7, 10, 10, 6.`, 8, 10, 10, 7.`


how can i get a matrix which has the combines the first column of A with the first coulumn of B and the last column of A?
Resulting in



C = 3,6, 5.`, 4,7, 6.`, 5,8, 7.`


Thanky you in advance for any assistance










share|improve this question









$endgroup$
















    3












    $begingroup$


    given matrices



    A = 3, 10, 10, 5.`, 4, 10, 10, 6.`, 5, 10, 10, 7.`
    B = 6, 10, 10, 5.`, 7, 10, 10, 6.`, 8, 10, 10, 7.`


    how can i get a matrix which has the combines the first column of A with the first coulumn of B and the last column of A?
    Resulting in



    C = 3,6, 5.`, 4,7, 6.`, 5,8, 7.`


    Thanky you in advance for any assistance










    share|improve this question









    $endgroup$














      3












      3








      3





      $begingroup$


      given matrices



      A = 3, 10, 10, 5.`, 4, 10, 10, 6.`, 5, 10, 10, 7.`
      B = 6, 10, 10, 5.`, 7, 10, 10, 6.`, 8, 10, 10, 7.`


      how can i get a matrix which has the combines the first column of A with the first coulumn of B and the last column of A?
      Resulting in



      C = 3,6, 5.`, 4,7, 6.`, 5,8, 7.`


      Thanky you in advance for any assistance










      share|improve this question









      $endgroup$




      given matrices



      A = 3, 10, 10, 5.`, 4, 10, 10, 6.`, 5, 10, 10, 7.`
      B = 6, 10, 10, 5.`, 7, 10, 10, 6.`, 8, 10, 10, 7.`


      how can i get a matrix which has the combines the first column of A with the first coulumn of B and the last column of A?
      Resulting in



      C = 3,6, 5.`, 4,7, 6.`, 5,8, 7.`


      Thanky you in advance for any assistance







      matrix concatenation






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Apr 20 at 8:49









      AndreasAndreas

      1198




      1198




















          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$

          just try



           Transpose@First/@A,First/@B,Last/@A 


          or



           Transpose@A[[All,1]],B[[All,1]],A[[All,4]]



          3, 6, 5., 4, 7, 6., 5, 8, 7.







          share|improve this answer











          $endgroup$












          • $begingroup$
            Than you very much
            $endgroup$
            – Andreas
            Apr 20 at 9:57











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          1 Answer
          1






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          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          just try



           Transpose@First/@A,First/@B,Last/@A 


          or



           Transpose@A[[All,1]],B[[All,1]],A[[All,4]]



          3, 6, 5., 4, 7, 6., 5, 8, 7.







          share|improve this answer











          $endgroup$












          • $begingroup$
            Than you very much
            $endgroup$
            – Andreas
            Apr 20 at 9:57















          4












          $begingroup$

          just try



           Transpose@First/@A,First/@B,Last/@A 


          or



           Transpose@A[[All,1]],B[[All,1]],A[[All,4]]



          3, 6, 5., 4, 7, 6., 5, 8, 7.







          share|improve this answer











          $endgroup$












          • $begingroup$
            Than you very much
            $endgroup$
            – Andreas
            Apr 20 at 9:57













          4












          4








          4





          $begingroup$

          just try



           Transpose@First/@A,First/@B,Last/@A 


          or



           Transpose@A[[All,1]],B[[All,1]],A[[All,4]]



          3, 6, 5., 4, 7, 6., 5, 8, 7.







          share|improve this answer











          $endgroup$



          just try



           Transpose@First/@A,First/@B,Last/@A 


          or



           Transpose@A[[All,1]],B[[All,1]],A[[All,4]]



          3, 6, 5., 4, 7, 6., 5, 8, 7.








          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Apr 20 at 9:04

























          answered Apr 20 at 8:57









          J42161217J42161217

          4,723324




          4,723324











          • $begingroup$
            Than you very much
            $endgroup$
            – Andreas
            Apr 20 at 9:57
















          • $begingroup$
            Than you very much
            $endgroup$
            – Andreas
            Apr 20 at 9:57















          $begingroup$
          Than you very much
          $endgroup$
          – Andreas
          Apr 20 at 9:57




          $begingroup$
          Than you very much
          $endgroup$
          – Andreas
          Apr 20 at 9:57

















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