How can I replace non-reals in a multidimensional list?

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Ok. I'm obviously not understanding something about pattern matching.



I have a list called t like the following:



t = 3.2,5.9,Indeterminate,4.5,-3 - 100 (1 - 2/(Underflow + 1)) + 200 (1 - Underflow),7.9,1.2,9.1,2.3


Where I would like to use ReplaceAll to convert all non-real values to 0.0, like so:



3.2,5.9,0.0,4.5,0.0,7.9,1.2,9.1,2.3


How could I do this with ReplaceAll and pattern matching?



I have tried:



t /. r_ /; Head[r] != List | Real -> 0.0


But that does nothing to the list. I am trying to wrap my head around pattern matching, but I'm not quite there yet, and any help would be greatly appreciated.



Thanks!



P.S.- I would really like to understand this in context of pattern matching, but other more efficient ways of doing this are also appreciated!






list-manipulation pattern-matching replacement







share|improve this question



























    up vote
    2
    down vote

    favorite












    Ok. I'm obviously not understanding something about pattern matching.



    I have a list called t like the following:



    t = 3.2,5.9,Indeterminate,4.5,-3 - 100 (1 - 2/(Underflow + 1)) + 200 (1 - Underflow),7.9,1.2,9.1,2.3


    Where I would like to use ReplaceAll to convert all non-real values to 0.0, like so:



    3.2,5.9,0.0,4.5,0.0,7.9,1.2,9.1,2.3


    How could I do this with ReplaceAll and pattern matching?



    I have tried:



    t /. r_ /; Head[r] != List | Real -> 0.0


    But that does nothing to the list. I am trying to wrap my head around pattern matching, but I'm not quite there yet, and any help would be greatly appreciated.



    Thanks!



    P.S.- I would really like to understand this in context of pattern matching, but other more efficient ways of doing this are also appreciated!






    list-manipulation pattern-matching replacement







    share|improve this question

























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Ok. I'm obviously not understanding something about pattern matching.



      I have a list called t like the following:



      t = 3.2,5.9,Indeterminate,4.5,-3 - 100 (1 - 2/(Underflow + 1)) + 200 (1 - Underflow),7.9,1.2,9.1,2.3


      Where I would like to use ReplaceAll to convert all non-real values to 0.0, like so:



      3.2,5.9,0.0,4.5,0.0,7.9,1.2,9.1,2.3


      How could I do this with ReplaceAll and pattern matching?



      I have tried:



      t /. r_ /; Head[r] != List | Real -> 0.0


      But that does nothing to the list. I am trying to wrap my head around pattern matching, but I'm not quite there yet, and any help would be greatly appreciated.



      Thanks!



      P.S.- I would really like to understand this in context of pattern matching, but other more efficient ways of doing this are also appreciated!






      list-manipulation pattern-matching replacement







      share|improve this question















      Ok. I'm obviously not understanding something about pattern matching.



      I have a list called t like the following:



      t = 3.2,5.9,Indeterminate,4.5,-3 - 100 (1 - 2/(Underflow + 1)) + 200 (1 - Underflow),7.9,1.2,9.1,2.3


      Where I would like to use ReplaceAll to convert all non-real values to 0.0, like so:



      3.2,5.9,0.0,4.5,0.0,7.9,1.2,9.1,2.3


      How could I do this with ReplaceAll and pattern matching?



      I have tried:



      t /. r_ /; Head[r] != List | Real -> 0.0


      But that does nothing to the list. I am trying to wrap my head around pattern matching, but I'm not quite there yet, and any help would be greatly appreciated.



      Thanks!



      P.S.- I would really like to understand this in context of pattern matching, but other more efficient ways of doing this are also appreciated!






      list-manipulation pattern-matching replacement




      list-manipulation pattern-matching replacement






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 3 hours ago









      Carl Woll

      57.5k273149




      57.5k273149










      asked 3 hours ago









      Jmeeks29ig

      46228




      46228




















          1 Answer
          1






          active

          oldest

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          up vote
          4
          down vote



          accepted










          You had two issues.



          1. You can't use Unequal ("!=") with Alternatives ("|").

          2. Even if you correct your pattern, the pattern would match the whole expression.

          Using Replace with a level spec and a corrected pattern:



          t = 3.2,5.9,Indeterminate,4.5,-3-100 (1-2/(Underflow+1))+200 (1-Underflow),7.9,1.2,9.1,2.3;

          Replace[
           t,
           r_ /; !MatchQ[Head[r], List | Real] -> 0.0,
           2
          ]



          3.2, 5.9, 0., 4.5, 0., 7.9, 1.2, 9.1, 2.3




          Another possibility:



          Replace[
           t,
           Except[_?NumberQ] -> 0.,
           2
          ]



          3.2, 5.9, 0., 4.5, 0., 7.9, 1.2, 9.1, 2.3







          share|improve this answer




















          • Great, thanks! That helps me to understand it a lot better
            – Jmeeks29ig
            1 hour 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
          4
          down vote



          accepted










          You had two issues.



          1. You can't use Unequal ("!=") with Alternatives ("|").

          2. Even if you correct your pattern, the pattern would match the whole expression.

          Using Replace with a level spec and a corrected pattern:



          t = 3.2,5.9,Indeterminate,4.5,-3-100 (1-2/(Underflow+1))+200 (1-Underflow),7.9,1.2,9.1,2.3;

          Replace[
           t,
           r_ /; !MatchQ[Head[r], List | Real] -> 0.0,
           2
          ]



          3.2, 5.9, 0., 4.5, 0., 7.9, 1.2, 9.1, 2.3




          Another possibility:



          Replace[
           t,
           Except[_?NumberQ] -> 0.,
           2
          ]



          3.2, 5.9, 0., 4.5, 0., 7.9, 1.2, 9.1, 2.3







          share|improve this answer




















          • Great, thanks! That helps me to understand it a lot better
            – Jmeeks29ig
            1 hour ago














          up vote
          4
          down vote



          accepted










          You had two issues.



          1. You can't use Unequal ("!=") with Alternatives ("|").

          2. Even if you correct your pattern, the pattern would match the whole expression.

          Using Replace with a level spec and a corrected pattern:



          t = 3.2,5.9,Indeterminate,4.5,-3-100 (1-2/(Underflow+1))+200 (1-Underflow),7.9,1.2,9.1,2.3;

          Replace[
           t,
           r_ /; !MatchQ[Head[r], List | Real] -> 0.0,
           2
          ]



          3.2, 5.9, 0., 4.5, 0., 7.9, 1.2, 9.1, 2.3




          Another possibility:



          Replace[
           t,
           Except[_?NumberQ] -> 0.,
           2
          ]



          3.2, 5.9, 0., 4.5, 0., 7.9, 1.2, 9.1, 2.3







          share|improve this answer




















          • Great, thanks! That helps me to understand it a lot better
            – Jmeeks29ig
            1 hour ago












          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          You had two issues.



          1. You can't use Unequal ("!=") with Alternatives ("|").

          2. Even if you correct your pattern, the pattern would match the whole expression.

          Using Replace with a level spec and a corrected pattern:



          t = 3.2,5.9,Indeterminate,4.5,-3-100 (1-2/(Underflow+1))+200 (1-Underflow),7.9,1.2,9.1,2.3;

          Replace[
           t,
           r_ /; !MatchQ[Head[r], List | Real] -> 0.0,
           2
          ]



          3.2, 5.9, 0., 4.5, 0., 7.9, 1.2, 9.1, 2.3




          Another possibility:



          Replace[
           t,
           Except[_?NumberQ] -> 0.,
           2
          ]



          3.2, 5.9, 0., 4.5, 0., 7.9, 1.2, 9.1, 2.3







          share|improve this answer












          You had two issues.



          1. You can't use Unequal ("!=") with Alternatives ("|").

          2. Even if you correct your pattern, the pattern would match the whole expression.

          Using Replace with a level spec and a corrected pattern:



          t = 3.2,5.9,Indeterminate,4.5,-3-100 (1-2/(Underflow+1))+200 (1-Underflow),7.9,1.2,9.1,2.3;

          Replace[
           t,
           r_ /; !MatchQ[Head[r], List | Real] -> 0.0,
           2
          ]



          3.2, 5.9, 0., 4.5, 0., 7.9, 1.2, 9.1, 2.3




          Another possibility:



          Replace[
           t,
           Except[_?NumberQ] -> 0.,
           2
          ]



          3.2, 5.9, 0., 4.5, 0., 7.9, 1.2, 9.1, 2.3








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 3 hours ago









          Carl Woll

          57.5k273149




          57.5k273149











          • Great, thanks! That helps me to understand it a lot better
            – Jmeeks29ig
            1 hour ago
















          • Great, thanks! That helps me to understand it a lot better
            – Jmeeks29ig
            1 hour ago















          Great, thanks! That helps me to understand it a lot better
          – Jmeeks29ig
          1 hour ago




          Great, thanks! That helps me to understand it a lot better
          – Jmeeks29ig
          1 hour ago

















           

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