Replace targeted elements in first level of list only

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I have a list where some elements are strings or sublists:



list = "DTLCIGYHANNSTDT", "LCLGHHAVPNGTLVKTITNDQIEVTNATELVQSSSTGKI", 
"CLGHHAVPNGTLVKTITNDQIEVTNATELVQSSSTGKIC", "H25", " H45", " I361", 
 " D362", "N197", "H25", " H45", " S46", " V47", " T332", " V361", " D362",
 " G363", " W364", " Q381", " K382", " T384", " Q385", " I388", 
 " N389", " V395", " N396", " I399", "H25", " H45", " V47", " N48",
 " L49", " T332", " D362", " G363", " W364", " Q381", " T384", 
 " Q385", " I388", " N389", " T392", " V395", " N396", " I399", 
 " E400", "NSTDTVDTVLEKNVT", "D31", " S46", "S145", ...


I want to target the 1st level strings that look like "N197" or "S145" and replace them with N197 and S145. Transform from String to List, essentially.



I've tried with 



list /. 
 x_String /; 
 StringMatchQ[x, RegularExpression["^[A-Z]\d1,3$"]] :> List[x]


This command affects the desired elements, PLUS the first string element of every sublist. I know I'm missing something obvious, but how can I keep the changes to the 1st-level of the list?






list-manipulation replacement







share|improve this question



























    up vote
    2
    down vote

    favorite












    I have a list where some elements are strings or sublists:



    list = "DTLCIGYHANNSTDT", "LCLGHHAVPNGTLVKTITNDQIEVTNATELVQSSSTGKI", 
    "CLGHHAVPNGTLVKTITNDQIEVTNATELVQSSSTGKIC", "H25", " H45", " I361", 
     " D362", "N197", "H25", " H45", " S46", " V47", " T332", " V361", " D362",
     " G363", " W364", " Q381", " K382", " T384", " Q385", " I388", 
     " N389", " V395", " N396", " I399", "H25", " H45", " V47", " N48",
     " L49", " T332", " D362", " G363", " W364", " Q381", " T384", 
     " Q385", " I388", " N389", " T392", " V395", " N396", " I399", 
     " E400", "NSTDTVDTVLEKNVT", "D31", " S46", "S145", ...


    I want to target the 1st level strings that look like "N197" or "S145" and replace them with N197 and S145. Transform from String to List, essentially.



    I've tried with 



    list /. 
     x_String /; 
     StringMatchQ[x, RegularExpression["^[A-Z]\d1,3$"]] :> List[x]


    This command affects the desired elements, PLUS the first string element of every sublist. I know I'm missing something obvious, but how can I keep the changes to the 1st-level of the list?






    list-manipulation replacement







    share|improve this question

























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      I have a list where some elements are strings or sublists:



      list = "DTLCIGYHANNSTDT", "LCLGHHAVPNGTLVKTITNDQIEVTNATELVQSSSTGKI", 
      "CLGHHAVPNGTLVKTITNDQIEVTNATELVQSSSTGKIC", "H25", " H45", " I361", 
       " D362", "N197", "H25", " H45", " S46", " V47", " T332", " V361", " D362",
       " G363", " W364", " Q381", " K382", " T384", " Q385", " I388", 
       " N389", " V395", " N396", " I399", "H25", " H45", " V47", " N48",
       " L49", " T332", " D362", " G363", " W364", " Q381", " T384", 
       " Q385", " I388", " N389", " T392", " V395", " N396", " I399", 
       " E400", "NSTDTVDTVLEKNVT", "D31", " S46", "S145", ...


      I want to target the 1st level strings that look like "N197" or "S145" and replace them with N197 and S145. Transform from String to List, essentially.



      I've tried with 



      list /. 
       x_String /; 
       StringMatchQ[x, RegularExpression["^[A-Z]\d1,3$"]] :> List[x]


      This command affects the desired elements, PLUS the first string element of every sublist. I know I'm missing something obvious, but how can I keep the changes to the 1st-level of the list?






      list-manipulation replacement







      share|improve this question















      I have a list where some elements are strings or sublists:



      list = "DTLCIGYHANNSTDT", "LCLGHHAVPNGTLVKTITNDQIEVTNATELVQSSSTGKI", 
      "CLGHHAVPNGTLVKTITNDQIEVTNATELVQSSSTGKIC", "H25", " H45", " I361", 
       " D362", "N197", "H25", " H45", " S46", " V47", " T332", " V361", " D362",
       " G363", " W364", " Q381", " K382", " T384", " Q385", " I388", 
       " N389", " V395", " N396", " I399", "H25", " H45", " V47", " N48",
       " L49", " T332", " D362", " G363", " W364", " Q381", " T384", 
       " Q385", " I388", " N389", " T392", " V395", " N396", " I399", 
       " E400", "NSTDTVDTVLEKNVT", "D31", " S46", "S145", ...


      I want to target the 1st level strings that look like "N197" or "S145" and replace them with N197 and S145. Transform from String to List, essentially.



      I've tried with 



      list /. 
       x_String /; 
       StringMatchQ[x, RegularExpression["^[A-Z]\d1,3$"]] :> List[x]


      This command affects the desired elements, PLUS the first string element of every sublist. I know I'm missing something obvious, but how can I keep the changes to the 1st-level of the list?






      list-manipulation replacement




      list-manipulation replacement






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 12 mins ago

























      asked 23 mins ago









      briennakh

      2647




      2647




















          1 Answer
          1






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          down vote



          accepted










          Do not use ReplaceAll when you need a level spec. Use Replace. Avoid in general using ReplaceAll in favor of Replace in complex data structures unless you know for sure what is happening and what side-effects ReplaceAll can bring as you keep building your application and growing your code.



          pattern = x_String /; StringMatchQ[x,RegularExpression["^[A-Z]\d1,3$"]];

          Replace[list, pattern :> List[x], 1]





          share|improve this answer




















          • Thanks for the tip about ReplaceAll!
            – briennakh
            13 mins ago










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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








          up vote
          3
          down vote



          accepted










          Do not use ReplaceAll when you need a level spec. Use Replace. Avoid in general using ReplaceAll in favor of Replace in complex data structures unless you know for sure what is happening and what side-effects ReplaceAll can bring as you keep building your application and growing your code.



          pattern = x_String /; StringMatchQ[x,RegularExpression["^[A-Z]\d1,3$"]];

          Replace[list, pattern :> List[x], 1]





          share|improve this answer




















          • Thanks for the tip about ReplaceAll!
            – briennakh
            13 mins ago














          up vote
          3
          down vote



          accepted










          Do not use ReplaceAll when you need a level spec. Use Replace. Avoid in general using ReplaceAll in favor of Replace in complex data structures unless you know for sure what is happening and what side-effects ReplaceAll can bring as you keep building your application and growing your code.



          pattern = x_String /; StringMatchQ[x,RegularExpression["^[A-Z]\d1,3$"]];

          Replace[list, pattern :> List[x], 1]





          share|improve this answer




















          • Thanks for the tip about ReplaceAll!
            – briennakh
            13 mins ago












          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          Do not use ReplaceAll when you need a level spec. Use Replace. Avoid in general using ReplaceAll in favor of Replace in complex data structures unless you know for sure what is happening and what side-effects ReplaceAll can bring as you keep building your application and growing your code.



          pattern = x_String /; StringMatchQ[x,RegularExpression["^[A-Z]\d1,3$"]];

          Replace[list, pattern :> List[x], 1]





          share|improve this answer












          Do not use ReplaceAll when you need a level spec. Use Replace. Avoid in general using ReplaceAll in favor of Replace in complex data structures unless you know for sure what is happening and what side-effects ReplaceAll can bring as you keep building your application and growing your code.



          pattern = x_String /; StringMatchQ[x,RegularExpression["^[A-Z]\d1,3$"]];

          Replace[list, pattern :> List[x], 1]






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 16 mins ago









          Vitaliy Kaurov

          56.2k6157275




          56.2k6157275











          • Thanks for the tip about ReplaceAll!
            – briennakh
            13 mins ago
















          • Thanks for the tip about ReplaceAll!
            – briennakh
            13 mins ago















          Thanks for the tip about ReplaceAll!
          – briennakh
          13 mins ago




          Thanks for the tip about ReplaceAll!
          – briennakh
          13 mins ago

















           

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