How to avoid Replace substituting subscripts?Replacing powers of $n$-th partial sums in the output of a...
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How to avoid Replace substituting subscripts?
Replacing powers of $n$-th partial sums in the output of a function with $S[i]$ for $i = 1,ldots,n$Looking for an elegant way to solve a (system of) ODEs/functional equations with undetermined coefficientsHow do I write subscripts in the middle of a string?Subscripts in variable names: Bug?Replacing parts of a complicated expression with shorthand symbolsAn alternative to subscripts?Making subscripts with variableHow to sort subscripts of symbols in an expression?List all subscripts in an expressionSubstituting values
$begingroup$
I have a polynomial in $rho$ with many parameters (I will omit the polynomial code in the Mathematica expressions below because it is too long):
$$
rho ^3 J (mu +1) left[(A delta tau _0+B (1-delta ) tau _1right]+
rho ^2 left[left(A delta tau _0+B (1-delta ) tau _1right) left((mu +1) V_B-Jright)+A (1-delta ) J (mu +1) tau _0+B delta J (mu +1) tau _1+delta mu ^2 tau _0+(1-delta ) mu ^2 tau _1right]+
rho left[A (1-delta ) tau _0 left((mu +1) V_B-Jright)-V_B left(A delta tau _0+B (1-delta ) tau _1right)+B delta tau _1 left((mu +1) V_B-Jright)+(1-delta ) mu ^2 tau _0+delta mu ^2 tau _1right]
-A (1-delta ) tau _0 V_B-B delta tau _1 V_B
$$
Notice that $B$ is both a parameter and an index for $V_B$.
Say I saved this polynomial in p, then when I use Replace using levelspec, such as:
Replace[p, {δ -> 0, B -> 0}, 6]
Mathematica substitutes both $B$ in the coefficients and in the subscripts (why would anyone want this?):
$$
rho left(A tau _0 left((mu +1) V_B-Jright)+mu ^2 tau _0right)+rho ^2 left(A J (mu +1) tau _0+mu ^2 tau _1right)-A tau _0 V_0
$$
Notice now there's a $V_0$ and a $V_B$. If I don't use levelspec, and use /. instead, all $B$ are replaced, in all indices (which is obvisouly what I do not want).
How to avoid subscripts substitution?
polynomial code is here:
p=-A (1 - δ) Subscript[V, B] Subscript[τ, 0] - B δ Subscript[V, B] Subscript[τ, 1] + J (1 + μ) ρ^3 (A δ Subscript[τ, 0] + B (1 - δ) Subscript[τ, 1]) + ρ ((1 - δ) μ^2 Subscript[τ, 0] + A (1 - δ) (-J + (1 + μ) Subscript[V, B]) Subscript[τ, 0] + δ μ^2 Subscript[τ, 1] + B δ (-J + (1 + μ) Subscript[V, B]) Subscript[τ, 1] - Subscript[V, B] (A δ Subscript[τ, 0] + B (1 - δ) Subscript[τ,1])) + ρ^2 (δ μ^2 Subscript[τ, 0] + A J (1 - δ) (1 + μ) Subscript[τ, 0] + (1 - δ) μ^2 Subscript[τ, 1] + B J δ (1 + μ) Subscript[τ, 1] + (-J + (1 + μ) Subscript[V, B]) (A δ Subscript[τ, 0] + B (1 - δ) Subscript[τ, 1]))
replacement subscript
edited 14 hours ago
MarcoB
36.4k556112
asked 14 hours ago
GirardiGirardi
1134
New contributor
Girardi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I have a polynomial in $rho$ with many parameters (I will omit the polynomial code in the Mathematica expressions below because it is too long):
$$
rho ^3 J (mu +1) left[(A delta tau _0+B (1-delta ) tau _1right]+
rho ^2 left[left(A delta tau _0+B (1-delta ) tau _1right) left((mu +1) V_B-Jright)+A (1-delta ) J (mu +1) tau _0+B delta J (mu +1) tau _1+delta mu ^2 tau _0+(1-delta ) mu ^2 tau _1right]+
rho left[A (1-delta ) tau _0 left((mu +1) V_B-Jright)-V_B left(A delta tau _0+B (1-delta ) tau _1right)+B delta tau _1 left((mu +1) V_B-Jright)+(1-delta ) mu ^2 tau _0+delta mu ^2 tau _1right]
-A (1-delta ) tau _0 V_B-B delta tau _1 V_B
$$
Notice that $B$ is both a parameter and an index for $V_B$.
Say I saved this polynomial in p, then when I use Replace using levelspec, such as:
Replace[p, {δ -> 0, B -> 0}, 6]
Mathematica substitutes both $B$ in the coefficients and in the subscripts (why would anyone want this?):
$$
rho left(A tau _0 left((mu +1) V_B-Jright)+mu ^2 tau _0right)+rho ^2 left(A J (mu +1) tau _0+mu ^2 tau _1right)-A tau _0 V_0
$$
Notice now there's a $V_0$ and a $V_B$. If I don't use levelspec, and use /. instead, all $B$ are replaced, in all indices (which is obvisouly what I do not want).
How to avoid subscripts substitution?
polynomial code is here:
p=-A (1 - δ) Subscript[V, B] Subscript[τ, 0] - B δ Subscript[V, B] Subscript[τ, 1] + J (1 + μ) ρ^3 (A δ Subscript[τ, 0] + B (1 - δ) Subscript[τ, 1]) + ρ ((1 - δ) μ^2 Subscript[τ, 0] + A (1 - δ) (-J + (1 + μ) Subscript[V, B]) Subscript[τ, 0] + δ μ^2 Subscript[τ, 1] + B δ (-J + (1 + μ) Subscript[V, B]) Subscript[τ, 1] - Subscript[V, B] (A δ Subscript[τ, 0] + B (1 - δ) Subscript[τ,1])) + ρ^2 (δ μ^2 Subscript[τ, 0] + A J (1 - δ) (1 + μ) Subscript[τ, 0] + (1 - δ) μ^2 Subscript[τ, 1] + B J δ (1 + μ) Subscript[τ, 1] + (-J + (1 + μ) Subscript[V, B]) (A δ Subscript[τ, 0] + B (1 - δ) Subscript[τ, 1]))
replacement subscript
edited 14 hours ago
MarcoB
36.4k556112
asked 14 hours ago
GirardiGirardi
1134
New contributor
Girardi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I have a polynomial in $rho$ with many parameters (I will omit the polynomial code in the Mathematica expressions below because it is too long):
$$
rho ^3 J (mu +1) left[(A delta tau _0+B (1-delta ) tau _1right]+
rho ^2 left[left(A delta tau _0+B (1-delta ) tau _1right) left((mu +1) V_B-Jright)+A (1-delta ) J (mu +1) tau _0+B delta J (mu +1) tau _1+delta mu ^2 tau _0+(1-delta ) mu ^2 tau _1right]+
rho left[A (1-delta ) tau _0 left((mu +1) V_B-Jright)-V_B left(A delta tau _0+B (1-delta ) tau _1right)+B delta tau _1 left((mu +1) V_B-Jright)+(1-delta ) mu ^2 tau _0+delta mu ^2 tau _1right]
-A (1-delta ) tau _0 V_B-B delta tau _1 V_B
$$
Notice that $B$ is both a parameter and an index for $V_B$.
Say I saved this polynomial in p, then when I use Replace using levelspec, such as:
Replace[p, {δ -> 0, B -> 0}, 6]
Mathematica substitutes both $B$ in the coefficients and in the subscripts (why would anyone want this?):
$$
rho left(A tau _0 left((mu +1) V_B-Jright)+mu ^2 tau _0right)+rho ^2 left(A J (mu +1) tau _0+mu ^2 tau _1right)-A tau _0 V_0
$$
Notice now there's a $V_0$ and a $V_B$. If I don't use levelspec, and use /. instead, all $B$ are replaced, in all indices (which is obvisouly what I do not want).
How to avoid subscripts substitution?
polynomial code is here:
p=-A (1 - δ) Subscript[V, B] Subscript[τ, 0] - B δ Subscript[V, B] Subscript[τ, 1] + J (1 + μ) ρ^3 (A δ Subscript[τ, 0] + B (1 - δ) Subscript[τ, 1]) + ρ ((1 - δ) μ^2 Subscript[τ, 0] + A (1 - δ) (-J + (1 + μ) Subscript[V, B]) Subscript[τ, 0] + δ μ^2 Subscript[τ, 1] + B δ (-J + (1 + μ) Subscript[V, B]) Subscript[τ, 1] - Subscript[V, B] (A δ Subscript[τ, 0] + B (1 - δ) Subscript[τ,1])) + ρ^2 (δ μ^2 Subscript[τ, 0] + A J (1 - δ) (1 + μ) Subscript[τ, 0] + (1 - δ) μ^2 Subscript[τ, 1] + B J δ (1 + μ) Subscript[τ, 1] + (-J + (1 + μ) Subscript[V, B]) (A δ Subscript[τ, 0] + B (1 - δ) Subscript[τ, 1]))
replacement subscript
edited 14 hours ago
MarcoB
36.4k556112
asked 14 hours ago
GirardiGirardi
1134
New contributor
Girardi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have a polynomial in $rho$ with many parameters (I will omit the polynomial code in the Mathematica expressions below because it is too long):
$$
rho ^3 J (mu +1) left[(A delta tau _0+B (1-delta ) tau _1right]+
rho ^2 left[left(A delta tau _0+B (1-delta ) tau _1right) left((mu +1) V_B-Jright)+A (1-delta ) J (mu +1) tau _0+B delta J (mu +1) tau _1+delta mu ^2 tau _0+(1-delta ) mu ^2 tau _1right]+
rho left[A (1-delta ) tau _0 left((mu +1) V_B-Jright)-V_B left(A delta tau _0+B (1-delta ) tau _1right)+B delta tau _1 left((mu +1) V_B-Jright)+(1-delta ) mu ^2 tau _0+delta mu ^2 tau _1right]
-A (1-delta ) tau _0 V_B-B delta tau _1 V_B
$$
Notice that $B$ is both a parameter and an index for $V_B$.
Say I saved this polynomial in p, then when I use Replace using levelspec, such as:
Replace[p, {δ -> 0, B -> 0}, 6]
Mathematica substitutes both $B$ in the coefficients and in the subscripts (why would anyone want this?):
$$
rho left(A tau _0 left((mu +1) V_B-Jright)+mu ^2 tau _0right)+rho ^2 left(A J (mu +1) tau _0+mu ^2 tau _1right)-A tau _0 V_0
$$
Notice now there's a $V_0$ and a $V_B$. If I don't use levelspec, and use /. instead, all $B$ are replaced, in all indices (which is obvisouly what I do not want).
How to avoid subscripts substitution?
polynomial code is here:
p=-A (1 - δ) Subscript[V, B] Subscript[τ, 0] - B δ Subscript[V, B] Subscript[τ, 1] + J (1 + μ) ρ^3 (A δ Subscript[τ, 0] + B (1 - δ) Subscript[τ, 1]) + ρ ((1 - δ) μ^2 Subscript[τ, 0] + A (1 - δ) (-J + (1 + μ) Subscript[V, B]) Subscript[τ, 0] + δ μ^2 Subscript[τ, 1] + B δ (-J + (1 + μ) Subscript[V, B]) Subscript[τ, 1] - Subscript[V, B] (A δ Subscript[τ, 0] + B (1 - δ) Subscript[τ,1])) + ρ^2 (δ μ^2 Subscript[τ, 0] + A J (1 - δ) (1 + μ) Subscript[τ, 0] + (1 - δ) μ^2 Subscript[τ, 1] + B J δ (1 + μ) Subscript[τ, 1] + (-J + (1 + μ) Subscript[V, B]) (A δ Subscript[τ, 0] + B (1 - δ) Subscript[τ, 1]))
replacement subscript
replacement subscript
edited 14 hours ago
MarcoB
36.4k556112
asked 14 hours ago
GirardiGirardi
1134
New contributor
Girardi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 14 hours ago
MarcoB
36.4k556112
asked 14 hours ago
GirardiGirardi
1134
New contributor
Girardi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 14 hours ago
MarcoB
36.4k556112
edited 14 hours ago
MarcoB
36.4k556112
edited 14 hours ago
MarcoB
36.4k556112
36.4k556112
asked 14 hours ago
GirardiGirardi
1134
New contributor
Girardi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 14 hours ago
GirardiGirardi
1134
asked 14 hours ago
GirardiGirardi
1134
1134
New contributor
Girardi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Girardi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Girardi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The real solution here is not to use B in a subscript, or better: not to use Subscript at all.
If you can't or don't want to do that, you can add an extra transformation rule for all Subscript expressions. Once ReplaceAll processes a subexpression, it won't touch it anymore.
expr = Subscript[x, B] + B;
expr /. B -> 1
(* 1 + Subscript[x, 1] *)
expr /. {s_Subscript :> s, B -> 1}
(* 1 + Subscript[x, B] *)
answered 14 hours ago
SzabolcsSzabolcs
161k14438936
$endgroup$
add a comment |
$begingroup$
Mathematica does not concern itself with why you might want to do something, it does what you indicate that you want done.
Convert the subscripts to strings prior to replacing B
p /. {Subscript[var_, sub__] :> Subscript[var, ToString[sub]], δ -> 0,
B -> 0}
Or, if you don't want the subscripts left as strings
p /. {Subscript[var_, sub__] :> Subscript[var, ToString[sub]],
δ -> 0, B -> 0} /.
{Subscript[var_, sub__] :> Subscript[var, ToExpression[sub]]}
answered 14 hours ago
Bob HanlonBob Hanlon
60.7k33597
$endgroup$
add a comment |
$begingroup$
Try this:
p /. Subscript[a_, B] -> Subscript[a, x] /. {δ -> 0,
B -> 0} /. Subscript[a_, x] -> Subscript[a, B]
yielding the following
Have fun!
answered 13 hours ago
Alexei BoulbitchAlexei Boulbitch
21.8k2470
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The real solution here is not to use B in a subscript, or better: not to use Subscript at all.
If you can't or don't want to do that, you can add an extra transformation rule for all Subscript expressions. Once ReplaceAll processes a subexpression, it won't touch it anymore.
expr = Subscript[x, B] + B;
expr /. B -> 1
(* 1 + Subscript[x, 1] *)
expr /. {s_Subscript :> s, B -> 1}
(* 1 + Subscript[x, B] *)
answered 14 hours ago
SzabolcsSzabolcs
161k14438936
$endgroup$
add a comment |
$begingroup$
The real solution here is not to use B in a subscript, or better: not to use Subscript at all.
If you can't or don't want to do that, you can add an extra transformation rule for all Subscript expressions. Once ReplaceAll processes a subexpression, it won't touch it anymore.
expr = Subscript[x, B] + B;
expr /. B -> 1
(* 1 + Subscript[x, 1] *)
expr /. {s_Subscript :> s, B -> 1}
(* 1 + Subscript[x, B] *)
answered 14 hours ago
SzabolcsSzabolcs
161k14438936
$endgroup$
add a comment |
$begingroup$
The real solution here is not to use B in a subscript, or better: not to use Subscript at all.
If you can't or don't want to do that, you can add an extra transformation rule for all Subscript expressions. Once ReplaceAll processes a subexpression, it won't touch it anymore.
expr = Subscript[x, B] + B;
expr /. B -> 1
(* 1 + Subscript[x, 1] *)
expr /. {s_Subscript :> s, B -> 1}
(* 1 + Subscript[x, B] *)
answered 14 hours ago
SzabolcsSzabolcs
161k14438936
$endgroup$
The real solution here is not to use B in a subscript, or better: not to use Subscript at all.
If you can't or don't want to do that, you can add an extra transformation rule for all Subscript expressions. Once ReplaceAll processes a subexpression, it won't touch it anymore.
expr = Subscript[x, B] + B;
expr /. B -> 1
(* 1 + Subscript[x, 1] *)
expr /. {s_Subscript :> s, B -> 1}
(* 1 + Subscript[x, B] *)
answered 14 hours ago
SzabolcsSzabolcs
161k14438936
answered 14 hours ago
SzabolcsSzabolcs
161k14438936
answered 14 hours ago
SzabolcsSzabolcs
161k14438936
answered 14 hours ago
SzabolcsSzabolcs
161k14438936
161k14438936
add a comment |
add a comment |
$begingroup$
Mathematica does not concern itself with why you might want to do something, it does what you indicate that you want done.
Convert the subscripts to strings prior to replacing B
p /. {Subscript[var_, sub__] :> Subscript[var, ToString[sub]], δ -> 0,
B -> 0}
Or, if you don't want the subscripts left as strings
p /. {Subscript[var_, sub__] :> Subscript[var, ToString[sub]],
δ -> 0, B -> 0} /.
{Subscript[var_, sub__] :> Subscript[var, ToExpression[sub]]}
answered 14 hours ago
Bob HanlonBob Hanlon
60.7k33597
$endgroup$
add a comment |
$begingroup$
Mathematica does not concern itself with why you might want to do something, it does what you indicate that you want done.
Convert the subscripts to strings prior to replacing B
p /. {Subscript[var_, sub__] :> Subscript[var, ToString[sub]], δ -> 0,
B -> 0}
Or, if you don't want the subscripts left as strings
p /. {Subscript[var_, sub__] :> Subscript[var, ToString[sub]],
δ -> 0, B -> 0} /.
{Subscript[var_, sub__] :> Subscript[var, ToExpression[sub]]}
answered 14 hours ago
Bob HanlonBob Hanlon
60.7k33597
$endgroup$
add a comment |
$begingroup$
Mathematica does not concern itself with why you might want to do something, it does what you indicate that you want done.
Convert the subscripts to strings prior to replacing B
p /. {Subscript[var_, sub__] :> Subscript[var, ToString[sub]], δ -> 0,
B -> 0}
Or, if you don't want the subscripts left as strings
p /. {Subscript[var_, sub__] :> Subscript[var, ToString[sub]],
δ -> 0, B -> 0} /.
{Subscript[var_, sub__] :> Subscript[var, ToExpression[sub]]}
answered 14 hours ago
Bob HanlonBob Hanlon
60.7k33597
$endgroup$
Mathematica does not concern itself with why you might want to do something, it does what you indicate that you want done.
Convert the subscripts to strings prior to replacing B
p /. {Subscript[var_, sub__] :> Subscript[var, ToString[sub]], δ -> 0,
B -> 0}
Or, if you don't want the subscripts left as strings
p /. {Subscript[var_, sub__] :> Subscript[var, ToString[sub]],
δ -> 0, B -> 0} /.
{Subscript[var_, sub__] :> Subscript[var, ToExpression[sub]]}
answered 14 hours ago
Bob HanlonBob Hanlon
60.7k33597
answered 14 hours ago
Bob HanlonBob Hanlon
60.7k33597
answered 14 hours ago
Bob HanlonBob Hanlon
60.7k33597
answered 14 hours ago
Bob HanlonBob Hanlon
60.7k33597
60.7k33597
add a comment |
add a comment |
$begingroup$
Try this:
p /. Subscript[a_, B] -> Subscript[a, x] /. {δ -> 0,
B -> 0} /. Subscript[a_, x] -> Subscript[a, B]
yielding the following
Have fun!
answered 13 hours ago
Alexei BoulbitchAlexei Boulbitch
21.8k2470
$endgroup$
add a comment |
$begingroup$
Try this:
p /. Subscript[a_, B] -> Subscript[a, x] /. {δ -> 0,
B -> 0} /. Subscript[a_, x] -> Subscript[a, B]
yielding the following
Have fun!
answered 13 hours ago
Alexei BoulbitchAlexei Boulbitch
21.8k2470
$endgroup$
add a comment |
$begingroup$
Try this:
p /. Subscript[a_, B] -> Subscript[a, x] /. {δ -> 0,
B -> 0} /. Subscript[a_, x] -> Subscript[a, B]
yielding the following
Have fun!
answered 13 hours ago
Alexei BoulbitchAlexei Boulbitch
21.8k2470
$endgroup$
Try this:
p /. Subscript[a_, B] -> Subscript[a, x] /. {δ -> 0,
B -> 0} /. Subscript[a_, x] -> Subscript[a, B]
yielding the following
Have fun!
answered 13 hours ago
Alexei BoulbitchAlexei Boulbitch
21.8k2470
answered 13 hours ago
Alexei BoulbitchAlexei Boulbitch
21.8k2470
answered 13 hours ago
Alexei BoulbitchAlexei Boulbitch
21.8k2470
answered 13 hours ago
Alexei BoulbitchAlexei Boulbitch
21.8k2470
21.8k2470
add a comment |
add a comment |
Girardi is a new contributor. Be nice, and check out our Code of Conduct.
Girardi is a new contributor. Be nice, and check out our Code of Conduct.
Girardi is a new contributor. Be nice, and check out our Code of Conduct.
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