How do I use drop to drop the second element from all these sublists?












2















{{1, 2}, {2, 3}, {5, 4}}




I tried to do this:



Drop[#, {2}] & @@ {{1, 2}, {2, 3}, {5, 4}}


but it gave




{1}




but what I want is




{{1},{2},{5}}







list-manipulation







share|improve this question









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  • 2




    Use Map(/@) instead of Apply (@@). Drop[#, {2}] & @@ {{1, 2}, {2, 3}, {5, 4}} is essentially equivalent to Drop[{1, 2}, {2, 3}, {5, 4}, {2}], which is clearly not what you want
    – Lukas Lang
    3 hours ago












  • I'd like to add to the answers here that the correct function for dropping a single element is Delete, not Drop. Drop is the opposite of Take.
    – Sjoerd Smit
    1 hour ago










  • Welcome to Mathematica.SE user62264! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
    – Chris K
    1 hour ago
















2















{{1, 2}, {2, 3}, {5, 4}}




I tried to do this:



Drop[#, {2}] & @@ {{1, 2}, {2, 3}, {5, 4}}


but it gave




{1}




but what I want is




{{1},{2},{5}}







list-manipulation







share|improve this question









New contributor




user62264 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 2




    Use Map(/@) instead of Apply (@@). Drop[#, {2}] & @@ {{1, 2}, {2, 3}, {5, 4}} is essentially equivalent to Drop[{1, 2}, {2, 3}, {5, 4}, {2}], which is clearly not what you want
    – Lukas Lang
    3 hours ago












  • I'd like to add to the answers here that the correct function for dropping a single element is Delete, not Drop. Drop is the opposite of Take.
    – Sjoerd Smit
    1 hour ago










  • Welcome to Mathematica.SE user62264! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
    – Chris K
    1 hour ago














2












2








2








{{1, 2}, {2, 3}, {5, 4}}




I tried to do this:



Drop[#, {2}] & @@ {{1, 2}, {2, 3}, {5, 4}}


but it gave




{1}




but what I want is




{{1},{2},{5}}







list-manipulation







share|improve this question









New contributor




user62264 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












{{1, 2}, {2, 3}, {5, 4}}




I tried to do this:



Drop[#, {2}] & @@ {{1, 2}, {2, 3}, {5, 4}}


but it gave




{1}




but what I want is




{{1},{2},{5}}







list-manipulation




list-manipulation






share|improve this question









New contributor




user62264 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




user62264 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








edited 1 hour ago









Glorfindel

1971211




1971211






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asked 3 hours ago









user62264user62264

111




111




New contributor




user62264 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





user62264 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






user62264 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 2




    Use Map(/@) instead of Apply (@@). Drop[#, {2}] & @@ {{1, 2}, {2, 3}, {5, 4}} is essentially equivalent to Drop[{1, 2}, {2, 3}, {5, 4}, {2}], which is clearly not what you want
    – Lukas Lang
    3 hours ago












  • I'd like to add to the answers here that the correct function for dropping a single element is Delete, not Drop. Drop is the opposite of Take.
    – Sjoerd Smit
    1 hour ago










  • Welcome to Mathematica.SE user62264! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
    – Chris K
    1 hour ago














  • 2




    Use Map(/@) instead of Apply (@@). Drop[#, {2}] & @@ {{1, 2}, {2, 3}, {5, 4}} is essentially equivalent to Drop[{1, 2}, {2, 3}, {5, 4}, {2}], which is clearly not what you want
    – Lukas Lang
    3 hours ago












  • I'd like to add to the answers here that the correct function for dropping a single element is Delete, not Drop. Drop is the opposite of Take.
    – Sjoerd Smit
    1 hour ago










  • Welcome to Mathematica.SE user62264! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
    – Chris K
    1 hour ago








2




2




Use Map(/@) instead of Apply (@@). Drop[#, {2}] & @@ {{1, 2}, {2, 3}, {5, 4}} is essentially equivalent to Drop[{1, 2}, {2, 3}, {5, 4}, {2}], which is clearly not what you want
– Lukas Lang
3 hours ago






Use Map(/@) instead of Apply (@@). Drop[#, {2}] & @@ {{1, 2}, {2, 3}, {5, 4}} is essentially equivalent to Drop[{1, 2}, {2, 3}, {5, 4}, {2}], which is clearly not what you want
– Lukas Lang
3 hours ago














I'd like to add to the answers here that the correct function for dropping a single element is Delete, not Drop. Drop is the opposite of Take.
– Sjoerd Smit
1 hour ago




I'd like to add to the answers here that the correct function for dropping a single element is Delete, not Drop. Drop is the opposite of Take.
– Sjoerd Smit
1 hour ago












Welcome to Mathematica.SE user62264! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
– Chris K
1 hour ago




Welcome to Mathematica.SE user62264! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
– Chris K
1 hour ago










2 Answers
2






active

oldest

votes


















3














lst = {{1, 2}, {2, 3}, {5, 4}};



Drop[lst, None, {2}]



{{1}, {2}, {5}}







share|improve this answer





























    1














    Your result is the first column of a matrix. If list={{1, 2}, {2, 3}, {5, 4}} ,



    res=list[[All, 1]]


    If you need the brackets around each element, Partition[res, 1]



    If your lists are of unequal length, the solution from Lukas Lang is fine. I include for completeness.



    Drop[#, {2}] & /@ {{1, 2}, {2, 3}, {5, 4}}





    share|improve this answer





















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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






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      3














      lst = {{1, 2}, {2, 3}, {5, 4}};
      

      
Drop[lst, None, {2}]



      {{1}, {2}, {5}}







      share|improve this answer


























        3














        lst = {{1, 2}, {2, 3}, {5, 4}};
        

        
Drop[lst, None, {2}]



        {{1}, {2}, {5}}







        share|improve this answer
























          3












          3








          3






          lst = {{1, 2}, {2, 3}, {5, 4}};
          

          
Drop[lst, None, {2}]



          {{1}, {2}, {5}}







          share|improve this answer












          lst = {{1, 2}, {2, 3}, {5, 4}};
          

          
Drop[lst, None, {2}]



          {{1}, {2}, {5}}








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 3 hours ago









          kglrkglr

          177k9198408




          177k9198408























              1














              Your result is the first column of a matrix. If list={{1, 2}, {2, 3}, {5, 4}} ,



              res=list[[All, 1]]


              If you need the brackets around each element, Partition[res, 1]



              If your lists are of unequal length, the solution from Lukas Lang is fine. I include for completeness.



              Drop[#, {2}] & /@ {{1, 2}, {2, 3}, {5, 4}}





              share|improve this answer


























                1














                Your result is the first column of a matrix. If list={{1, 2}, {2, 3}, {5, 4}} ,



                res=list[[All, 1]]


                If you need the brackets around each element, Partition[res, 1]



                If your lists are of unequal length, the solution from Lukas Lang is fine. I include for completeness.



                Drop[#, {2}] & /@ {{1, 2}, {2, 3}, {5, 4}}





                share|improve this answer
























                  1












                  1








                  1






                  Your result is the first column of a matrix. If list={{1, 2}, {2, 3}, {5, 4}} ,



                  res=list[[All, 1]]


                  If you need the brackets around each element, Partition[res, 1]



                  If your lists are of unequal length, the solution from Lukas Lang is fine. I include for completeness.



                  Drop[#, {2}] & /@ {{1, 2}, {2, 3}, {5, 4}}





                  share|improve this answer












                  Your result is the first column of a matrix. If list={{1, 2}, {2, 3}, {5, 4}} ,



                  res=list[[All, 1]]


                  If you need the brackets around each element, Partition[res, 1]



                  If your lists are of unequal length, the solution from Lukas Lang is fine. I include for completeness.



                  Drop[#, {2}] & /@ {{1, 2}, {2, 3}, {5, 4}}






                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 3 hours ago









                  TitusTitus

                  605417




                  605417






















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