Apply Max to each element of a list












1














I have a list



a = {1, 8, 0, 6, 5, 3, 5, 2, 2, 5}


I want to generate a new list whose elements are the same of a if it's bigger than 5, or 5 elsewhere.
I managed to achieve this using Map and a pure function doing



Map[(Max[#, 5]) &, a]


but this looks a bit clumsy to me. Is there a better way?



EDIT: I found this solution



a /. x_ /; x < 5 -> 5


but I cannot really understand why is working. Could someone give an insight into it?



Thanks






map







share|improve this question









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  • 1




    Why not Max[#, 5] & /@ a ?
    – Mauro Lacy
    1 hour ago
















1














I have a list



a = {1, 8, 0, 6, 5, 3, 5, 2, 2, 5}


I want to generate a new list whose elements are the same of a if it's bigger than 5, or 5 elsewhere.
I managed to achieve this using Map and a pure function doing



Map[(Max[#, 5]) &, a]


but this looks a bit clumsy to me. Is there a better way?



EDIT: I found this solution



a /. x_ /; x < 5 -> 5


but I cannot really understand why is working. Could someone give an insight into it?



Thanks






map







share|improve this question









New contributor




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
















  • 1




    Why not Max[#, 5] & /@ a ?
    – Mauro Lacy
    1 hour ago














1












1








1







I have a list



a = {1, 8, 0, 6, 5, 3, 5, 2, 2, 5}


I want to generate a new list whose elements are the same of a if it's bigger than 5, or 5 elsewhere.
I managed to achieve this using Map and a pure function doing



Map[(Max[#, 5]) &, a]


but this looks a bit clumsy to me. Is there a better way?



EDIT: I found this solution



a /. x_ /; x < 5 -> 5


but I cannot really understand why is working. Could someone give an insight into it?



Thanks






map







share|improve this question









New contributor




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











I have a list



a = {1, 8, 0, 6, 5, 3, 5, 2, 2, 5}


I want to generate a new list whose elements are the same of a if it's bigger than 5, or 5 elsewhere.
I managed to achieve this using Map and a pure function doing



Map[(Max[#, 5]) &, a]


but this looks a bit clumsy to me. Is there a better way?



EDIT: I found this solution



a /. x_ /; x < 5 -> 5


but I cannot really understand why is working. Could someone give an insight into it?



Thanks






map




map






share|improve this question









New contributor




Luca Amerio 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




Luca Amerio 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 2 hours ago





















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asked 2 hours ago









Luca Amerio

83




83




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Luca Amerio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





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






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








  • 1




    Why not Max[#, 5] & /@ a ?
    – Mauro Lacy
    1 hour ago














  • 1




    Why not Max[#, 5] & /@ a ?
    – Mauro Lacy
    1 hour ago








1




1




Why not Max[#, 5] & /@ a ?
– Mauro Lacy
1 hour ago




Why not Max[#, 5] & /@ a ?
– Mauro Lacy
1 hour ago










1 Answer
1






active

oldest

votes


















3














You can use Clip or Ramp:



Clip[a, {5, ∞}]



{5, 8, 5, 6, 5, 5, 5, 5, 5, 5}




5 + Ramp[a - 5]



{5, 8, 5, 6, 5, 5, 5, 5, 5, 5}







share|improve this answer





















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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    You can use Clip or Ramp:



    Clip[a, {5, ∞}]



    {5, 8, 5, 6, 5, 5, 5, 5, 5, 5}




    5 + Ramp[a - 5]



    {5, 8, 5, 6, 5, 5, 5, 5, 5, 5}







    share|improve this answer


























      3














      You can use Clip or Ramp:



      Clip[a, {5, ∞}]



      {5, 8, 5, 6, 5, 5, 5, 5, 5, 5}




      5 + Ramp[a - 5]



      {5, 8, 5, 6, 5, 5, 5, 5, 5, 5}







      share|improve this answer
























        3












        3








        3






        You can use Clip or Ramp:



        Clip[a, {5, ∞}]



        {5, 8, 5, 6, 5, 5, 5, 5, 5, 5}




        5 + Ramp[a - 5]



        {5, 8, 5, 6, 5, 5, 5, 5, 5, 5}







        share|improve this answer












        You can use Clip or Ramp:



        Clip[a, {5, ∞}]



        {5, 8, 5, 6, 5, 5, 5, 5, 5, 5}




        5 + Ramp[a - 5]



        {5, 8, 5, 6, 5, 5, 5, 5, 5, 5}








        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 2 hours ago









        kglr

        177k9198405




        177k9198405






















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