Is it possible to set values for a list of variables using a for loop? [duplicate]

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3












$begingroup$



This question already has an answer here:



  • How to create symbols from strings and set values for them?

    5 answers



Let me preface this question by noting that this is a simple example meant to clarify what I am asking. The actual context in which I want to implement this involves associations containing hundreds of variables, nested for loops, and writing to/reading from files. Because of this, the traditional method of setting a list of variables to a list of values (e.g. a,b,c=1,2,3) is not practical.



Here is my simplified example:



Suppose I have a list of variables which have already been defined: a,b,c,d,e,f. I want to set these variables to values of 1,2,3,4,5,6 using a for loop.



To do this, I define the list:



listOfVar = "a","b","c","d","e","f"


And naively use the following for loop



For[i=1,i<=listOfVar,i++,
listOfVar[[i]] = i
]


Clearly, this won't work, as it will just set



listOfVar = 1,2,3,4,5


My question is thus:



How do I replace listOfVar[[i]] with the variable name which its string represents? (e.g. how do I replace listOfVar[[3]] with the variable c so that the for loop correctly sets c to the value of 3?)



I have tried many combinations of Hold,ToExpression,ToString,etc., but I do not seem to have enough knowledge of the low-level operations of Mathematica to solve this problem (or perhaps I am overlooking an obvious solution for whatever reason). I would greatly appreciate the help, as this is a problem I have given up on multiple times.










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marked as duplicate by Kuba Feb 24 at 13:57


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






















    3












    $begingroup$



    This question already has an answer here:



    • How to create symbols from strings and set values for them?

      5 answers



    Let me preface this question by noting that this is a simple example meant to clarify what I am asking. The actual context in which I want to implement this involves associations containing hundreds of variables, nested for loops, and writing to/reading from files. Because of this, the traditional method of setting a list of variables to a list of values (e.g. a,b,c=1,2,3) is not practical.



    Here is my simplified example:



    Suppose I have a list of variables which have already been defined: a,b,c,d,e,f. I want to set these variables to values of 1,2,3,4,5,6 using a for loop.



    To do this, I define the list:



    listOfVar = "a","b","c","d","e","f"


    And naively use the following for loop



    For[i=1,i<=listOfVar,i++,
    listOfVar[[i]] = i
    ]


    Clearly, this won't work, as it will just set



    listOfVar = 1,2,3,4,5


    My question is thus:



    How do I replace listOfVar[[i]] with the variable name which its string represents? (e.g. how do I replace listOfVar[[3]] with the variable c so that the for loop correctly sets c to the value of 3?)



    I have tried many combinations of Hold,ToExpression,ToString,etc., but I do not seem to have enough knowledge of the low-level operations of Mathematica to solve this problem (or perhaps I am overlooking an obvious solution for whatever reason). I would greatly appreciate the help, as this is a problem I have given up on multiple times.










    share|improve this question











    $endgroup$



    marked as duplicate by Kuba Feb 24 at 13:57


    This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.




















      3












      3








      3





      $begingroup$



      This question already has an answer here:



      • How to create symbols from strings and set values for them?

        5 answers



      Let me preface this question by noting that this is a simple example meant to clarify what I am asking. The actual context in which I want to implement this involves associations containing hundreds of variables, nested for loops, and writing to/reading from files. Because of this, the traditional method of setting a list of variables to a list of values (e.g. a,b,c=1,2,3) is not practical.



      Here is my simplified example:



      Suppose I have a list of variables which have already been defined: a,b,c,d,e,f. I want to set these variables to values of 1,2,3,4,5,6 using a for loop.



      To do this, I define the list:



      listOfVar = "a","b","c","d","e","f"


      And naively use the following for loop



      For[i=1,i<=listOfVar,i++,
      listOfVar[[i]] = i
      ]


      Clearly, this won't work, as it will just set



      listOfVar = 1,2,3,4,5


      My question is thus:



      How do I replace listOfVar[[i]] with the variable name which its string represents? (e.g. how do I replace listOfVar[[3]] with the variable c so that the for loop correctly sets c to the value of 3?)



      I have tried many combinations of Hold,ToExpression,ToString,etc., but I do not seem to have enough knowledge of the low-level operations of Mathematica to solve this problem (or perhaps I am overlooking an obvious solution for whatever reason). I would greatly appreciate the help, as this is a problem I have given up on multiple times.










      share|improve this question











      $endgroup$





      This question already has an answer here:



      • How to create symbols from strings and set values for them?

        5 answers



      Let me preface this question by noting that this is a simple example meant to clarify what I am asking. The actual context in which I want to implement this involves associations containing hundreds of variables, nested for loops, and writing to/reading from files. Because of this, the traditional method of setting a list of variables to a list of values (e.g. a,b,c=1,2,3) is not practical.



      Here is my simplified example:



      Suppose I have a list of variables which have already been defined: a,b,c,d,e,f. I want to set these variables to values of 1,2,3,4,5,6 using a for loop.



      To do this, I define the list:



      listOfVar = "a","b","c","d","e","f"


      And naively use the following for loop



      For[i=1,i<=listOfVar,i++,
      listOfVar[[i]] = i
      ]


      Clearly, this won't work, as it will just set



      listOfVar = 1,2,3,4,5


      My question is thus:



      How do I replace listOfVar[[i]] with the variable name which its string represents? (e.g. how do I replace listOfVar[[3]] with the variable c so that the for loop correctly sets c to the value of 3?)



      I have tried many combinations of Hold,ToExpression,ToString,etc., but I do not seem to have enough knowledge of the low-level operations of Mathematica to solve this problem (or perhaps I am overlooking an obvious solution for whatever reason). I would greatly appreciate the help, as this is a problem I have given up on multiple times.





      This question already has an answer here:



      • How to create symbols from strings and set values for them?

        5 answers







      assignment






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Feb 24 at 7:27









      Henrik Schumacher

      57.4k578158




      57.4k578158










      asked Feb 24 at 6:46









      user63017user63017

      575




      575




      marked as duplicate by Kuba Feb 24 at 13:57


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









      marked as duplicate by Kuba Feb 24 at 13:57


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






















          2 Answers
          2






          active

          oldest

          votes


















          5












          $begingroup$

          listOfVar = "a", "b", "c", "d", "e", "f";
          ClearAll @@ listOfVar;
          Do[
          ToExpression[listOfVar[[i]], InputForm, Set[#, i] &],
          i, 1, Length[listOfVar]
          ];
          Symbol /@ listOfVar



          1, 2, 3, 4, 5, 6







          share|improve this answer









          $endgroup$








          • 1




            $begingroup$
            Notice that Set[#, i] & does not hold its arguments. So if a has value it will break.
            $endgroup$
            – Kuba
            Feb 24 at 13:58


















          1












          $begingroup$

          Here is a second answer, if anyone is curious. It is a modification of an answer to another question I found after searching for an hour or two.



          ClearAll @@ listOfVar;
          For[i = 1, i <= Length[listOfVar], i++,
          Evaluate[Symbol[listOfSavedVariables[[i]]]] = i;
          ]





          share|improve this answer











          $endgroup$








          • 2




            $begingroup$
            It will work only once. If a e.g. has any value it will fail.
            $endgroup$
            – Kuba
            Feb 24 at 13:58

















          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5












          $begingroup$

          listOfVar = "a", "b", "c", "d", "e", "f";
          ClearAll @@ listOfVar;
          Do[
          ToExpression[listOfVar[[i]], InputForm, Set[#, i] &],
          i, 1, Length[listOfVar]
          ];
          Symbol /@ listOfVar



          1, 2, 3, 4, 5, 6







          share|improve this answer









          $endgroup$








          • 1




            $begingroup$
            Notice that Set[#, i] & does not hold its arguments. So if a has value it will break.
            $endgroup$
            – Kuba
            Feb 24 at 13:58















          5












          $begingroup$

          listOfVar = "a", "b", "c", "d", "e", "f";
          ClearAll @@ listOfVar;
          Do[
          ToExpression[listOfVar[[i]], InputForm, Set[#, i] &],
          i, 1, Length[listOfVar]
          ];
          Symbol /@ listOfVar



          1, 2, 3, 4, 5, 6







          share|improve this answer









          $endgroup$








          • 1




            $begingroup$
            Notice that Set[#, i] & does not hold its arguments. So if a has value it will break.
            $endgroup$
            – Kuba
            Feb 24 at 13:58













          5












          5








          5





          $begingroup$

          listOfVar = "a", "b", "c", "d", "e", "f";
          ClearAll @@ listOfVar;
          Do[
          ToExpression[listOfVar[[i]], InputForm, Set[#, i] &],
          i, 1, Length[listOfVar]
          ];
          Symbol /@ listOfVar



          1, 2, 3, 4, 5, 6







          share|improve this answer









          $endgroup$



          listOfVar = "a", "b", "c", "d", "e", "f";
          ClearAll @@ listOfVar;
          Do[
          ToExpression[listOfVar[[i]], InputForm, Set[#, i] &],
          i, 1, Length[listOfVar]
          ];
          Symbol /@ listOfVar



          1, 2, 3, 4, 5, 6








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Feb 24 at 7:30









          Henrik SchumacherHenrik Schumacher

          57.4k578158




          57.4k578158







          • 1




            $begingroup$
            Notice that Set[#, i] & does not hold its arguments. So if a has value it will break.
            $endgroup$
            – Kuba
            Feb 24 at 13:58












          • 1




            $begingroup$
            Notice that Set[#, i] & does not hold its arguments. So if a has value it will break.
            $endgroup$
            – Kuba
            Feb 24 at 13:58







          1




          1




          $begingroup$
          Notice that Set[#, i] & does not hold its arguments. So if a has value it will break.
          $endgroup$
          – Kuba
          Feb 24 at 13:58




          $begingroup$
          Notice that Set[#, i] & does not hold its arguments. So if a has value it will break.
          $endgroup$
          – Kuba
          Feb 24 at 13:58











          1












          $begingroup$

          Here is a second answer, if anyone is curious. It is a modification of an answer to another question I found after searching for an hour or two.



          ClearAll @@ listOfVar;
          For[i = 1, i <= Length[listOfVar], i++,
          Evaluate[Symbol[listOfSavedVariables[[i]]]] = i;
          ]





          share|improve this answer











          $endgroup$








          • 2




            $begingroup$
            It will work only once. If a e.g. has any value it will fail.
            $endgroup$
            – Kuba
            Feb 24 at 13:58















          1












          $begingroup$

          Here is a second answer, if anyone is curious. It is a modification of an answer to another question I found after searching for an hour or two.



          ClearAll @@ listOfVar;
          For[i = 1, i <= Length[listOfVar], i++,
          Evaluate[Symbol[listOfSavedVariables[[i]]]] = i;
          ]





          share|improve this answer











          $endgroup$








          • 2




            $begingroup$
            It will work only once. If a e.g. has any value it will fail.
            $endgroup$
            – Kuba
            Feb 24 at 13:58













          1












          1








          1





          $begingroup$

          Here is a second answer, if anyone is curious. It is a modification of an answer to another question I found after searching for an hour or two.



          ClearAll @@ listOfVar;
          For[i = 1, i <= Length[listOfVar], i++,
          Evaluate[Symbol[listOfSavedVariables[[i]]]] = i;
          ]





          share|improve this answer











          $endgroup$



          Here is a second answer, if anyone is curious. It is a modification of an answer to another question I found after searching for an hour or two.



          ClearAll @@ listOfVar;
          For[i = 1, i <= Length[listOfVar], i++,
          Evaluate[Symbol[listOfSavedVariables[[i]]]] = i;
          ]






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Feb 24 at 8:34

























          answered Feb 24 at 8:29









          user63017user63017

          575




          575







          • 2




            $begingroup$
            It will work only once. If a e.g. has any value it will fail.
            $endgroup$
            – Kuba
            Feb 24 at 13:58












          • 2




            $begingroup$
            It will work only once. If a e.g. has any value it will fail.
            $endgroup$
            – Kuba
            Feb 24 at 13:58







          2




          2




          $begingroup$
          It will work only once. If a e.g. has any value it will fail.
          $endgroup$
          – Kuba
          Feb 24 at 13:58




          $begingroup$
          It will work only once. If a e.g. has any value it will fail.
          $endgroup$
          – Kuba
          Feb 24 at 13:58


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