Generate list of strings from a list of assigned variables
Here's a way:
var1 = 10;
var2 = 11;
var3 = 17;
var4 = 5;
compvar := {var1, var2, var3, var4}
compvar; (*all variables assigned*)
ClearAll[f];
SetAttributes[f, {HoldAll}];
f[x_, y__] := Flatten@{f[x], f[y]}
f[x_] := SymbolName@Unevaluated@x
OwnValues[compvar] /. {HoldPattern[y_] :> {x__}} :> f[x]
(*
{"var1", "var2", "var3", "var4"}
*)
You must introduce some form of holding in you definition of compvar
as otherwise, assuming it is defined after var1
, var2
, etc., there is no information to retrieve:
var1 = 10;
var2 = 11;
var3 = 17;
var4 = 5;
compvar = {var1, var2, var3, var4};
Definition[compvar]
compvar = {10, 11, 17, 5}
You could use Hold
but then you would need to ReleaseHold
(or similar) every time you used compvar
. Instead I suggest you use SetDelayed
and then recover the definition using my step
function from:
- How do I evaluate only one step of an expression?
It returns an expression wrapped in HoldForm
:
compvar := {var1, var2, var3, var4};
step[compvar] // InputForm
HoldForm[{var1, var2, var3, var4}]
To convert to a list of strings:
Cases[step[compvar], s_Symbol :> SymbolName @ Unevaluated @ s, {2}]
{"var1", "var2", "var3", "var4"}
Or:
StringSplit[ToString @ step[compvar], ("{" | "," | " " | "}") ..]
{"var1", "var2", "var3", "var4"}
The first method will return Symbols (as strings) only while the second will convert other expressions as well.
Incidentally if you do not need to store your Symbols in a List
you could use a more direct form:
compHeld = Hold[var1, var2, var3, var4];
List @@ SymbolName /@ Unevaluated /@ compHeld
{"var1", "var2", "var3", "var4"}
one way is to make a replacement rule seperately and use that.
Clear[var1, var2, var3, var4];
vars = {var1, var2, var3, var4};
values = {var1 -> 10, var2 -> 11, var3 -> 17, var4 -> 5};
compvar = vars /. values
compvarstr = ToString[#] & /@ vars
FullForm[compvarstr]
Otherwise, the way you had it:
var1 = 10; var2 = 11; var3 = 17; var4 = 5;
compvar = {var1, var2, var3, var4}; (*all variables assigned*)
Now the var1
name itself is replaced by 10 right away by the evaluator. Hence compvar
will always be {10, 11, 17, 5}
and the name of the variables is not known inside compvar
since their value is used.