Recursively turn expression into nested list

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4
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I've been trying to write a simple function that turns an expression into a nested list. So for example, a+2b+3c becomes a,2,b,3,c. This is what I wrote:

toList[x_] := Module[, 
 x = List@@x; 
 For[i=1, i <= Length[x]; i++, 
 If[x[[i]] != List@@x[[i]], x[[i]] = toList[x[[i]]]]
 ];
 x
 ]

However, I keep getting errors like "Tag Plus in a+2b+3c is protected" and "a+2b+3c in the part assignment is not symbol." How can I fix what's going wrong?

list-manipulation simplifying-expressions

share|improve this question

edited Sep 2 at 13:29

asked Sep 2 at 13:24

user2520385

1944

add a comment | 

up vote
4
down vote

favorite

I've been trying to write a simple function that turns an expression into a nested list. So for example, a+2b+3c becomes a,2,b,3,c. This is what I wrote:

toList[x_] := Module[, 
 x = List@@x; 
 For[i=1, i <= Length[x]; i++, 
 If[x[[i]] != List@@x[[i]], x[[i]] = toList[x[[i]]]]
 ];
 x
 ]

However, I keep getting errors like "Tag Plus in a+2b+3c is protected" and "a+2b+3c in the part assignment is not symbol." How can I fix what's going wrong?

list-manipulation simplifying-expressions

share|improve this question

edited Sep 2 at 13:29

asked Sep 2 at 13:24

user2520385

1944

add a comment | 

up vote
4
down vote

favorite

up vote
4
down vote

favorite

I've been trying to write a simple function that turns an expression into a nested list. So for example, a+2b+3c becomes a,2,b,3,c. This is what I wrote:

toList[x_] := Module[, 
 x = List@@x; 
 For[i=1, i <= Length[x]; i++, 
 If[x[[i]] != List@@x[[i]], x[[i]] = toList[x[[i]]]]
 ];
 x
 ]

However, I keep getting errors like "Tag Plus in a+2b+3c is protected" and "a+2b+3c in the part assignment is not symbol." How can I fix what's going wrong?

list-manipulation simplifying-expressions

share|improve this question

edited Sep 2 at 13:29

asked Sep 2 at 13:24

user2520385

1944

I've been trying to write a simple function that turns an expression into a nested list. So for example, a+2b+3c becomes a,2,b,3,c. This is what I wrote:

toList[x_] := Module[, 
 x = List@@x; 
 For[i=1, i <= Length[x]; i++, 
 If[x[[i]] != List@@x[[i]], x[[i]] = toList[x[[i]]]]
 ];
 x
 ]

However, I keep getting errors like "Tag Plus in a+2b+3c is protected" and "a+2b+3c in the part assignment is not symbol." How can I fix what's going wrong?

list-manipulation simplifying-expressions

share|improve this question

edited Sep 2 at 13:29

asked Sep 2 at 13:24

user2520385

1944

share|improve this question

share|improve this question

edited Sep 2 at 13:29

edited Sep 2 at 13:29

edited Sep 2 at 13:29

asked Sep 2 at 13:24

user2520385

1944

asked Sep 2 at 13:24

user2520385

1944

asked Sep 2 at 13:24

user2520385

1944

1944

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
11
down vote



accepted

The following does what you want:

Apply[List, a + 2 b + 3 c, All]

This simply applies List to all parts of the expression in one go.

Alternatively:

Replace[a + 2 b + 3 c, _[args___] :> args, All]
(* a, 2, b, 3, c *)

The idea here is to replace the head of any expression with List. Note that Replace[…,…,All] is not equivalent to ReplaceAll[…,…] (i.e. …/.…). The former replaces from the inside out, while the latter from the outside in (and ignores parts that have already been touched)

Your attempt

As to why your attempt doesn't work:

• Function arguments are not variables (as in their value can't be modified). This is the reason for the "Tag … is protected" errors. To fix this, copy the value to a local variable


• 
  For does not localize the iteration variable, so i is shared across recursive calls. To fix this, localize i explicitly.


• 
  Equal/Unequal (==/!=) is for value equality, and doesn't evaluate in many cases (e.g. a==1 is not False). To do structural comparisons, use SameQ/UnsameQ (===/=!=).

Changing only what's necessary in your code yields the following working version:

toList[y_] := Module[
 x = y, i,
 x = List @@ x;
 For[i = 1, i <= Length[x], i++,
 If[x[[i]] =!= List @@ x[[i]],
 x[[i]] = toList[x[[i]]]
 ]
 ];
 x
 ]

toList[a + 2 b + 3 c]
(* a, 2 b, 3 c *)

Further simplifications could be:

• 
  

  Replace the For loop with Do (which does localize the variable)

  
  
  

  toList[y_] := Module[
   x = y,
   x = List @@ x;
   Do[
   If[x[[i]] =!= List @@ x[[i]],
   x[[i]] = toList[x[[i]]]
   ],
   i, Length@x
   ];
   x
   ]
  

  
  


• 
  

  Replace the For/Do loop with Map (the If now needs to return something in both cases, as the output is used):

  
  
  

  toList[y_] := Module[
   x = y,
   x = List @@ x;
   x = Map[
   If[# =!= List @@ #,
   toList[#],
   #
   ] &,
   x
   ];
   x
   ]
  

  
  


• 
  

  Remove the unnecessary local variable x:

  
  
  

  toList[y_] := Map[
   If[# =!= List @@ #,
   toList[#],
   #
   ] &,
   List @@ y
   ]
  

  
  


• 
  

  Move the termination condition for the recursion into the function definition itself using Condition (/;):

  
  
  

  toList[y_] /; y =!= List @@ y := Map[
   toList[#] &,
   List @@ y
   ]
  toList[y_] := y
  

  
  


• 
  

  Use shorthands for Map (/@):

  
  
  

  toList[y_] /; y =!= List @@ y := toList /@ List @@ y
  toList[y_] := y
  

  
  


• 
  

  Remove the termination condition completely (it's not needed, as Map/Apply don't do anything on expressions that have "zero length" (i.e. atoms)

  
  
  

  toList[y_] := toList /@ List @@ y
  

  
  

share|improve this answer

answered Sep 2 at 14:06

Lukas Lang

5,1531525

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, it is a matter of replacement of heads.
  – Î‘λέξανδρος Ζεγγ
  Sep 2 at 14:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wow, thanks! That is a really detailed explanation
  – user2520385
  Sep 2 at 14:22
  

  
  
  
  

add a comment | 

up vote
3
down vote

expr = a + 2 b + 3 (4 E^x + 3 x);

ReplaceRepeated[expr, f_[args___] /; (TrueQ[f =!= List]) :> List[args]]

(* a, 2, b, 3, 4, E, x, 3, x *)

share|improve this answer

answered Sep 2 at 13:49

Anton Antonov

21.5k164107

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
11
down vote



accepted

The following does what you want:

Apply[List, a + 2 b + 3 c, All]

This simply applies List to all parts of the expression in one go.

Alternatively:

Replace[a + 2 b + 3 c, _[args___] :> args, All]
(* a, 2, b, 3, c *)

The idea here is to replace the head of any expression with List. Note that Replace[…,…,All] is not equivalent to ReplaceAll[…,…] (i.e. …/.…). The former replaces from the inside out, while the latter from the outside in (and ignores parts that have already been touched)

Your attempt

As to why your attempt doesn't work:

• Function arguments are not variables (as in their value can't be modified). This is the reason for the "Tag … is protected" errors. To fix this, copy the value to a local variable


• 
  For does not localize the iteration variable, so i is shared across recursive calls. To fix this, localize i explicitly.


• 
  Equal/Unequal (==/!=) is for value equality, and doesn't evaluate in many cases (e.g. a==1 is not False). To do structural comparisons, use SameQ/UnsameQ (===/=!=).

Changing only what's necessary in your code yields the following working version:

toList[y_] := Module[
 x = y, i,
 x = List @@ x;
 For[i = 1, i <= Length[x], i++,
 If[x[[i]] =!= List @@ x[[i]],
 x[[i]] = toList[x[[i]]]
 ]
 ];
 x
 ]

toList[a + 2 b + 3 c]
(* a, 2 b, 3 c *)

Further simplifications could be:

• 
  

  Replace the For loop with Do (which does localize the variable)

  
  
  

  toList[y_] := Module[
   x = y,
   x = List @@ x;
   Do[
   If[x[[i]] =!= List @@ x[[i]],
   x[[i]] = toList[x[[i]]]
   ],
   i, Length@x
   ];
   x
   ]
  

  
  


• 
  

  Replace the For/Do loop with Map (the If now needs to return something in both cases, as the output is used):

  
  
  

  toList[y_] := Module[
   x = y,
   x = List @@ x;
   x = Map[
   If[# =!= List @@ #,
   toList[#],
   #
   ] &,
   x
   ];
   x
   ]
  

  
  


• 
  

  Remove the unnecessary local variable x:

  
  
  

  toList[y_] := Map[
   If[# =!= List @@ #,
   toList[#],
   #
   ] &,
   List @@ y
   ]
  

  
  


• 
  

  Move the termination condition for the recursion into the function definition itself using Condition (/;):

  
  
  

  toList[y_] /; y =!= List @@ y := Map[
   toList[#] &,
   List @@ y
   ]
  toList[y_] := y
  

  
  


• 
  

  Use shorthands for Map (/@):

  
  
  

  toList[y_] /; y =!= List @@ y := toList /@ List @@ y
  toList[y_] := y
  

  
  


• 
  

  Remove the termination condition completely (it's not needed, as Map/Apply don't do anything on expressions that have "zero length" (i.e. atoms)

  
  
  

  toList[y_] := toList /@ List @@ y
  

  
  

share|improve this answer

answered Sep 2 at 14:06

Lukas Lang

5,1531525

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, it is a matter of replacement of heads.
  – Î‘λέξανδρος Ζεγγ
  Sep 2 at 14:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wow, thanks! That is a really detailed explanation
  – user2520385
  Sep 2 at 14:22
  

  
  
  
  

add a comment | 

up vote
11
down vote



accepted

The following does what you want:

Apply[List, a + 2 b + 3 c, All]

This simply applies List to all parts of the expression in one go.

Alternatively:

Replace[a + 2 b + 3 c, _[args___] :> args, All]
(* a, 2, b, 3, c *)

The idea here is to replace the head of any expression with List. Note that Replace[…,…,All] is not equivalent to ReplaceAll[…,…] (i.e. …/.…). The former replaces from the inside out, while the latter from the outside in (and ignores parts that have already been touched)

Your attempt

As to why your attempt doesn't work:

• Function arguments are not variables (as in their value can't be modified). This is the reason for the "Tag … is protected" errors. To fix this, copy the value to a local variable


• 
  For does not localize the iteration variable, so i is shared across recursive calls. To fix this, localize i explicitly.


• 
  Equal/Unequal (==/!=) is for value equality, and doesn't evaluate in many cases (e.g. a==1 is not False). To do structural comparisons, use SameQ/UnsameQ (===/=!=).

Changing only what's necessary in your code yields the following working version:

toList[y_] := Module[
 x = y, i,
 x = List @@ x;
 For[i = 1, i <= Length[x], i++,
 If[x[[i]] =!= List @@ x[[i]],
 x[[i]] = toList[x[[i]]]
 ]
 ];
 x
 ]

toList[a + 2 b + 3 c]
(* a, 2 b, 3 c *)

Further simplifications could be:

• 
  

  Replace the For loop with Do (which does localize the variable)

  
  
  

  toList[y_] := Module[
   x = y,
   x = List @@ x;
   Do[
   If[x[[i]] =!= List @@ x[[i]],
   x[[i]] = toList[x[[i]]]
   ],
   i, Length@x
   ];
   x
   ]
  

  
  


• 
  

  Replace the For/Do loop with Map (the If now needs to return something in both cases, as the output is used):

  
  
  

  toList[y_] := Module[
   x = y,
   x = List @@ x;
   x = Map[
   If[# =!= List @@ #,
   toList[#],
   #
   ] &,
   x
   ];
   x
   ]
  

  
  


• 
  

  Remove the unnecessary local variable x:

  
  
  

  toList[y_] := Map[
   If[# =!= List @@ #,
   toList[#],
   #
   ] &,
   List @@ y
   ]
  

  
  


• 
  

  Move the termination condition for the recursion into the function definition itself using Condition (/;):

  
  
  

  toList[y_] /; y =!= List @@ y := Map[
   toList[#] &,
   List @@ y
   ]
  toList[y_] := y
  

  
  


• 
  

  Use shorthands for Map (/@):

  
  
  

  toList[y_] /; y =!= List @@ y := toList /@ List @@ y
  toList[y_] := y
  

  
  


• 
  

  Remove the termination condition completely (it's not needed, as Map/Apply don't do anything on expressions that have "zero length" (i.e. atoms)

  
  
  

  toList[y_] := toList /@ List @@ y
  

  
  

share|improve this answer

answered Sep 2 at 14:06

Lukas Lang

5,1531525

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, it is a matter of replacement of heads.
  – Î‘λέξανδρος Ζεγγ
  Sep 2 at 14:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wow, thanks! That is a really detailed explanation
  – user2520385
  Sep 2 at 14:22
  

  
  
  
  

add a comment | 

up vote
11
down vote



accepted

up vote
11
down vote



accepted

The following does what you want:

Apply[List, a + 2 b + 3 c, All]

This simply applies List to all parts of the expression in one go.

Alternatively:

Replace[a + 2 b + 3 c, _[args___] :> args, All]
(* a, 2, b, 3, c *)

The idea here is to replace the head of any expression with List. Note that Replace[…,…,All] is not equivalent to ReplaceAll[…,…] (i.e. …/.…). The former replaces from the inside out, while the latter from the outside in (and ignores parts that have already been touched)

Your attempt

As to why your attempt doesn't work:

• Function arguments are not variables (as in their value can't be modified). This is the reason for the "Tag … is protected" errors. To fix this, copy the value to a local variable


• 
  For does not localize the iteration variable, so i is shared across recursive calls. To fix this, localize i explicitly.


• 
  Equal/Unequal (==/!=) is for value equality, and doesn't evaluate in many cases (e.g. a==1 is not False). To do structural comparisons, use SameQ/UnsameQ (===/=!=).

Changing only what's necessary in your code yields the following working version:

toList[y_] := Module[
 x = y, i,
 x = List @@ x;
 For[i = 1, i <= Length[x], i++,
 If[x[[i]] =!= List @@ x[[i]],
 x[[i]] = toList[x[[i]]]
 ]
 ];
 x
 ]

toList[a + 2 b + 3 c]
(* a, 2 b, 3 c *)

Further simplifications could be:

• 
  

  Replace the For loop with Do (which does localize the variable)

  
  
  

  toList[y_] := Module[
   x = y,
   x = List @@ x;
   Do[
   If[x[[i]] =!= List @@ x[[i]],
   x[[i]] = toList[x[[i]]]
   ],
   i, Length@x
   ];
   x
   ]
  

  
  


• 
  

  Replace the For/Do loop with Map (the If now needs to return something in both cases, as the output is used):

  
  
  

  toList[y_] := Module[
   x = y,
   x = List @@ x;
   x = Map[
   If[# =!= List @@ #,
   toList[#],
   #
   ] &,
   x
   ];
   x
   ]
  

  
  


• 
  

  Remove the unnecessary local variable x:

  
  
  

  toList[y_] := Map[
   If[# =!= List @@ #,
   toList[#],
   #
   ] &,
   List @@ y
   ]
  

  
  


• 
  

  Move the termination condition for the recursion into the function definition itself using Condition (/;):

  
  
  

  toList[y_] /; y =!= List @@ y := Map[
   toList[#] &,
   List @@ y
   ]
  toList[y_] := y
  

  
  


• 
  

  Use shorthands for Map (/@):

  
  
  

  toList[y_] /; y =!= List @@ y := toList /@ List @@ y
  toList[y_] := y
  

  
  


• 
  

  Remove the termination condition completely (it's not needed, as Map/Apply don't do anything on expressions that have "zero length" (i.e. atoms)

  
  
  

  toList[y_] := toList /@ List @@ y
  

  
  

share|improve this answer

answered Sep 2 at 14:06

Lukas Lang

5,1531525

The following does what you want:

Apply[List, a + 2 b + 3 c, All]

This simply applies List to all parts of the expression in one go.

Alternatively:

Replace[a + 2 b + 3 c, _[args___] :> args, All]
(* a, 2, b, 3, c *)

The idea here is to replace the head of any expression with List. Note that Replace[…,…,All] is not equivalent to ReplaceAll[…,…] (i.e. …/.…). The former replaces from the inside out, while the latter from the outside in (and ignores parts that have already been touched)

Your attempt

As to why your attempt doesn't work:

• Function arguments are not variables (as in their value can't be modified). This is the reason for the "Tag … is protected" errors. To fix this, copy the value to a local variable


• 
  For does not localize the iteration variable, so i is shared across recursive calls. To fix this, localize i explicitly.


• 
  Equal/Unequal (==/!=) is for value equality, and doesn't evaluate in many cases (e.g. a==1 is not False). To do structural comparisons, use SameQ/UnsameQ (===/=!=).

Changing only what's necessary in your code yields the following working version:

toList[y_] := Module[
 x = y, i,
 x = List @@ x;
 For[i = 1, i <= Length[x], i++,
 If[x[[i]] =!= List @@ x[[i]],
 x[[i]] = toList[x[[i]]]
 ]
 ];
 x
 ]

toList[a + 2 b + 3 c]
(* a, 2 b, 3 c *)

Further simplifications could be:

• 
  

  Replace the For loop with Do (which does localize the variable)

  
  
  

  toList[y_] := Module[
   x = y,
   x = List @@ x;
   Do[
   If[x[[i]] =!= List @@ x[[i]],
   x[[i]] = toList[x[[i]]]
   ],
   i, Length@x
   ];
   x
   ]
  

  
  


• 
  

  Replace the For/Do loop with Map (the If now needs to return something in both cases, as the output is used):

  
  
  

  toList[y_] := Module[
   x = y,
   x = List @@ x;
   x = Map[
   If[# =!= List @@ #,
   toList[#],
   #
   ] &,
   x
   ];
   x
   ]
  

  
  


• 
  

  Remove the unnecessary local variable x:

  
  
  

  toList[y_] := Map[
   If[# =!= List @@ #,
   toList[#],
   #
   ] &,
   List @@ y
   ]
  

  
  


• 
  

  Move the termination condition for the recursion into the function definition itself using Condition (/;):

  
  
  

  toList[y_] /; y =!= List @@ y := Map[
   toList[#] &,
   List @@ y
   ]
  toList[y_] := y
  

  
  


• 
  

  Use shorthands for Map (/@):

  
  
  

  toList[y_] /; y =!= List @@ y := toList /@ List @@ y
  toList[y_] := y
  

  
  


• 
  

  Remove the termination condition completely (it's not needed, as Map/Apply don't do anything on expressions that have "zero length" (i.e. atoms)

  
  
  

  toList[y_] := toList /@ List @@ y
  

  
  

share|improve this answer

answered Sep 2 at 14:06

Lukas Lang

5,1531525

share|improve this answer

share|improve this answer

answered Sep 2 at 14:06

Lukas Lang

5,1531525

answered Sep 2 at 14:06

Lukas Lang

5,1531525

answered Sep 2 at 14:06

Lukas Lang

5,1531525

5,1531525

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, it is a matter of replacement of heads.
  – Î‘λέξανδρος Ζεγγ
  Sep 2 at 14:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wow, thanks! That is a really detailed explanation
  – user2520385
  Sep 2 at 14:22
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, it is a matter of replacement of heads.
  – Î‘λέξανδρος Ζεγγ
  Sep 2 at 14:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wow, thanks! That is a really detailed explanation
  – user2520385
  Sep 2 at 14:22
  

  
  
  
  

Yes, it is a matter of replacement of heads.
– Î‘λέξανδρος Ζεγγ
Sep 2 at 14:10

Yes, it is a matter of replacement of heads.
– Î‘λέξανδρος Ζεγγ
Sep 2 at 14:10

Wow, thanks! That is a really detailed explanation
– user2520385
Sep 2 at 14:22

Wow, thanks! That is a really detailed explanation
– user2520385
Sep 2 at 14:22

add a comment | 

up vote
3
down vote

expr = a + 2 b + 3 (4 E^x + 3 x);

ReplaceRepeated[expr, f_[args___] /; (TrueQ[f =!= List]) :> List[args]]

(* a, 2, b, 3, 4, E, x, 3, x *)

share|improve this answer

answered Sep 2 at 13:49

Anton Antonov

21.5k164107

add a comment | 

up vote
3
down vote

expr = a + 2 b + 3 (4 E^x + 3 x);

ReplaceRepeated[expr, f_[args___] /; (TrueQ[f =!= List]) :> List[args]]

(* a, 2, b, 3, 4, E, x, 3, x *)

share|improve this answer

answered Sep 2 at 13:49

Anton Antonov

21.5k164107

add a comment | 

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expr = a + 2 b + 3 (4 E^x + 3 x);

ReplaceRepeated[expr, f_[args___] /; (TrueQ[f =!= List]) :> List[args]]

(* a, 2, b, 3, 4, E, x, 3, x *)

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answered Sep 2 at 13:49

Anton Antonov

21.5k164107

expr = a + 2 b + 3 (4 E^x + 3 x);

ReplaceRepeated[expr, f_[args___] /; (TrueQ[f =!= List]) :> List[args]]

(* a, 2, b, 3, 4, E, x, 3, x *)

share|improve this answer

answered Sep 2 at 13:49

Anton Antonov

21.5k164107

share|improve this answer

share|improve this answer

answered Sep 2 at 13:49

Anton Antonov

21.5k164107

answered Sep 2 at 13:49

Anton Antonov

21.5k164107

answered Sep 2 at 13:49

Anton Antonov

21.5k164107

21.5k164107

add a comment | 

add a comment | 

 

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