Change arguments of a product of functions
2
$begingroup$
I am trying to do the following with rule-based pattern matching
(f1[t] f2[t] f3[t]) /. {x__ -> n[
Product[x[[i, 0]][om[i]], {i, 3}]]} // Quiet
which gives me the output n[f1[om[1]] f2[om[2]] f3[om[3]]]. However, it seems as if one cannot specify parts of a pattern MMA gives the error messages
"Part specification x[[1,0]] is longer than depth of object"
I would like to do the operatoration for a product of n arbitrary functions f1[t]...fn[t] where the number of functions is automatically detected (I cannot use Length as well). How would one do it correctly?
pattern-matching
asked 3 hours ago
Display NameDisplay Name
56938
$endgroup$
add a comment |
2
$begingroup$
I am trying to do the following with rule-based pattern matching
(f1[t] f2[t] f3[t]) /. {x__ -> n[
Product[x[[i, 0]][om[i]], {i, 3}]]} // Quiet
which gives me the output n[f1[om[1]] f2[om[2]] f3[om[3]]]. However, it seems as if one cannot specify parts of a pattern MMA gives the error messages
"Part specification x[[1,0]] is longer than depth of object"
I would like to do the operatoration for a product of n arbitrary functions f1[t]...fn[t] where the number of functions is automatically detected (I cannot use Length as well). How would one do it correctly?
pattern-matching
asked 3 hours ago
Display NameDisplay Name
56938
$endgroup$
add a comment |
2
2
2
$begingroup$
I am trying to do the following with rule-based pattern matching
(f1[t] f2[t] f3[t]) /. {x__ -> n[
Product[x[[i, 0]][om[i]], {i, 3}]]} // Quiet
which gives me the output n[f1[om[1]] f2[om[2]] f3[om[3]]]. However, it seems as if one cannot specify parts of a pattern MMA gives the error messages
"Part specification x[[1,0]] is longer than depth of object"
I would like to do the operatoration for a product of n arbitrary functions f1[t]...fn[t] where the number of functions is automatically detected (I cannot use Length as well). How would one do it correctly?
pattern-matching
asked 3 hours ago
Display NameDisplay Name
56938
$endgroup$
I am trying to do the following with rule-based pattern matching
(f1[t] f2[t] f3[t]) /. {x__ -> n[
Product[x[[i, 0]][om[i]], {i, 3}]]} // Quiet
which gives me the output n[f1[om[1]] f2[om[2]] f3[om[3]]]. However, it seems as if one cannot specify parts of a pattern MMA gives the error messages
"Part specification x[[1,0]] is longer than depth of object"
I would like to do the operatoration for a product of n arbitrary functions f1[t]...fn[t] where the number of functions is automatically detected (I cannot use Length as well). How would one do it correctly?
pattern-matching
pattern-matching
asked 3 hours ago
Display NameDisplay Name
56938
asked 3 hours ago
Display NameDisplay Name
56938
asked 3 hours ago
Display NameDisplay Name
56938
asked 3 hours ago
Display NameDisplay Name
56938
asked 3 hours ago
Display NameDisplay Name
56938
56938
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
2
$begingroup$
f1[t] f2[t] f3[t] /. Times[x_, y__] :> n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, {x, y}]]
n[f1[om[1]] f2[om[2]] f3[om[3]]]
Also
Module[{i = 1}, f1[t] f2[t] f3[t] /. {t :> om[i++]} // n]
n[f1[om[1]] f2[om[2]] f3[om[3]]]
Update: "If all the functions are equal":
Inactivate[f1[t] f1[t] f1[t]] /.
Inactive[Times][x_, y__] :> Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, {x, y}]]
n[f1[om[1]] f1[om[2]] f1[om[3]]]
Activate@Module[{i = 1}, Inactivate[f1[t] f1[t] f1[t]] /. {t :> om[i++]} // n]
n[f1[om[1]] f1[om[2]] f1[om[3]]]
answered 1 hour ago
kglrkglr
179k9198410
$endgroup$
$begingroup$
Thank you for your answer. It works forf1 != f2 != f3but stops working if all the functions are equal.
$endgroup$
– Display Name
58 mins ago
$begingroup$
@DisplayName, please see the update.
$endgroup$
– kglr
48 mins ago
$begingroup$
Thanks again. I think I lack a little bit of knowledge to come up with sth like this. I just read into pattern matching but it needs some time to settle. But I also like the approach with theRuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem.nis then simply a functional involving integrations over theoms.
$endgroup$
– Display Name
36 mins ago
add a comment |
0
$begingroup$
A simple function can do it.
changeArg[p : Times[__], var_Symbol, head_Symbol] :=
head[MapIndexed[#1[var[#2[[1]]]] &, Head /@ p]]
changeArg[f1[t] f2[t] f3[t], om, n]
n[f1[om[1]] f2[om[2]] f3[om[3]]]
answered 36 mins ago
m_goldbergm_goldberg
84.7k872196
$endgroup$
$begingroup$
Thank you for another possible answer. I think in the current form it doesn't cover the case of equal functions.
$endgroup$
– Display Name
22 mins ago
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
2
$begingroup$
f1[t] f2[t] f3[t] /. Times[x_, y__] :> n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, {x, y}]]
n[f1[om[1]] f2[om[2]] f3[om[3]]]
Also
Module[{i = 1}, f1[t] f2[t] f3[t] /. {t :> om[i++]} // n]
n[f1[om[1]] f2[om[2]] f3[om[3]]]
Update: "If all the functions are equal":
Inactivate[f1[t] f1[t] f1[t]] /.
Inactive[Times][x_, y__] :> Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, {x, y}]]
n[f1[om[1]] f1[om[2]] f1[om[3]]]
Activate@Module[{i = 1}, Inactivate[f1[t] f1[t] f1[t]] /. {t :> om[i++]} // n]
n[f1[om[1]] f1[om[2]] f1[om[3]]]
answered 1 hour ago
kglrkglr
179k9198410
$endgroup$
$begingroup$
Thank you for your answer. It works forf1 != f2 != f3but stops working if all the functions are equal.
$endgroup$
– Display Name
58 mins ago
$begingroup$
@DisplayName, please see the update.
$endgroup$
– kglr
48 mins ago
$begingroup$
Thanks again. I think I lack a little bit of knowledge to come up with sth like this. I just read into pattern matching but it needs some time to settle. But I also like the approach with theRuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem.nis then simply a functional involving integrations over theoms.
$endgroup$
– Display Name
36 mins ago
add a comment |
2
$begingroup$
f1[t] f2[t] f3[t] /. Times[x_, y__] :> n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, {x, y}]]
n[f1[om[1]] f2[om[2]] f3[om[3]]]
Also
Module[{i = 1}, f1[t] f2[t] f3[t] /. {t :> om[i++]} // n]
n[f1[om[1]] f2[om[2]] f3[om[3]]]
Update: "If all the functions are equal":
Inactivate[f1[t] f1[t] f1[t]] /.
Inactive[Times][x_, y__] :> Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, {x, y}]]
n[f1[om[1]] f1[om[2]] f1[om[3]]]
Activate@Module[{i = 1}, Inactivate[f1[t] f1[t] f1[t]] /. {t :> om[i++]} // n]
n[f1[om[1]] f1[om[2]] f1[om[3]]]
answered 1 hour ago
kglrkglr
179k9198410
$endgroup$
$begingroup$
Thank you for your answer. It works forf1 != f2 != f3but stops working if all the functions are equal.
$endgroup$
– Display Name
58 mins ago
$begingroup$
@DisplayName, please see the update.
$endgroup$
– kglr
48 mins ago
$begingroup$
Thanks again. I think I lack a little bit of knowledge to come up with sth like this. I just read into pattern matching but it needs some time to settle. But I also like the approach with theRuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem.nis then simply a functional involving integrations over theoms.
$endgroup$
– Display Name
36 mins ago
add a comment |
2
2
2
$begingroup$
f1[t] f2[t] f3[t] /. Times[x_, y__] :> n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, {x, y}]]
n[f1[om[1]] f2[om[2]] f3[om[3]]]
Also
Module[{i = 1}, f1[t] f2[t] f3[t] /. {t :> om[i++]} // n]
n[f1[om[1]] f2[om[2]] f3[om[3]]]
Update: "If all the functions are equal":
Inactivate[f1[t] f1[t] f1[t]] /.
Inactive[Times][x_, y__] :> Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, {x, y}]]
n[f1[om[1]] f1[om[2]] f1[om[3]]]
Activate@Module[{i = 1}, Inactivate[f1[t] f1[t] f1[t]] /. {t :> om[i++]} // n]
n[f1[om[1]] f1[om[2]] f1[om[3]]]
answered 1 hour ago
kglrkglr
179k9198410
$endgroup$
f1[t] f2[t] f3[t] /. Times[x_, y__] :> n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, {x, y}]]
n[f1[om[1]] f2[om[2]] f3[om[3]]]
Also
Module[{i = 1}, f1[t] f2[t] f3[t] /. {t :> om[i++]} // n]
n[f1[om[1]] f2[om[2]] f3[om[3]]]
Update: "If all the functions are equal":
Inactivate[f1[t] f1[t] f1[t]] /.
Inactive[Times][x_, y__] :> Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, {x, y}]]
n[f1[om[1]] f1[om[2]] f1[om[3]]]
Activate@Module[{i = 1}, Inactivate[f1[t] f1[t] f1[t]] /. {t :> om[i++]} // n]
n[f1[om[1]] f1[om[2]] f1[om[3]]]
answered 1 hour ago
kglrkglr
179k9198410
edited 49 mins ago
answered 1 hour ago
kglrkglr
179k9198410
answered 1 hour ago
kglrkglr
179k9198410
answered 1 hour ago
kglrkglr
179k9198410
179k9198410
$begingroup$
Thank you for your answer. It works forf1 != f2 != f3but stops working if all the functions are equal.
$endgroup$
– Display Name
58 mins ago
$begingroup$
@DisplayName, please see the update.
$endgroup$
– kglr
48 mins ago
$begingroup$
Thanks again. I think I lack a little bit of knowledge to come up with sth like this. I just read into pattern matching but it needs some time to settle. But I also like the approach with theRuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem.nis then simply a functional involving integrations over theoms.
$endgroup$
– Display Name
36 mins ago
add a comment |
$begingroup$
Thank you for your answer. It works forf1 != f2 != f3but stops working if all the functions are equal.
$endgroup$
– Display Name
58 mins ago
$begingroup$
@DisplayName, please see the update.
$endgroup$
– kglr
48 mins ago
$begingroup$
Thanks again. I think I lack a little bit of knowledge to come up with sth like this. I just read into pattern matching but it needs some time to settle. But I also like the approach with theRuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem.nis then simply a functional involving integrations over theoms.
$endgroup$
– Display Name
36 mins ago
$begingroup$
Thank you for your answer. It works for
f1 != f2 != f3 but stops working if all the functions are equal.$endgroup$
– Display Name
58 mins ago
$begingroup$
Thank you for your answer. It works for
f1 != f2 != f3 but stops working if all the functions are equal.$endgroup$
– Display Name
58 mins ago
$begingroup$
@DisplayName, please see the update.
$endgroup$
– kglr
48 mins ago
$begingroup$
@DisplayName, please see the update.
$endgroup$
– kglr
48 mins ago
$begingroup$
Thanks again. I think I lack a little bit of knowledge to come up with sth like this. I just read into pattern matching but it needs some time to settle. But I also like the approach with the
RuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem. n is then simply a functional involving integrations over the oms.$endgroup$
– Display Name
36 mins ago
$begingroup$
Thanks again. I think I lack a little bit of knowledge to come up with sth like this. I just read into pattern matching but it needs some time to settle. But I also like the approach with the
RuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem. n is then simply a functional involving integrations over the oms.$endgroup$
– Display Name
36 mins ago
add a comment |
0
$begingroup$
A simple function can do it.
changeArg[p : Times[__], var_Symbol, head_Symbol] :=
head[MapIndexed[#1[var[#2[[1]]]] &, Head /@ p]]
changeArg[f1[t] f2[t] f3[t], om, n]
n[f1[om[1]] f2[om[2]] f3[om[3]]]
answered 36 mins ago
m_goldbergm_goldberg
84.7k872196
$endgroup$
$begingroup$
Thank you for another possible answer. I think in the current form it doesn't cover the case of equal functions.
$endgroup$
– Display Name
22 mins ago
add a comment |
0
$begingroup$
A simple function can do it.
changeArg[p : Times[__], var_Symbol, head_Symbol] :=
head[MapIndexed[#1[var[#2[[1]]]] &, Head /@ p]]
changeArg[f1[t] f2[t] f3[t], om, n]
n[f1[om[1]] f2[om[2]] f3[om[3]]]
answered 36 mins ago
m_goldbergm_goldberg
84.7k872196
$endgroup$
$begingroup$
Thank you for another possible answer. I think in the current form it doesn't cover the case of equal functions.
$endgroup$
– Display Name
22 mins ago
add a comment |
0
0
0
$begingroup$
A simple function can do it.
changeArg[p : Times[__], var_Symbol, head_Symbol] :=
head[MapIndexed[#1[var[#2[[1]]]] &, Head /@ p]]
changeArg[f1[t] f2[t] f3[t], om, n]
n[f1[om[1]] f2[om[2]] f3[om[3]]]
answered 36 mins ago
m_goldbergm_goldberg
84.7k872196
$endgroup$
A simple function can do it.
changeArg[p : Times[__], var_Symbol, head_Symbol] :=
head[MapIndexed[#1[var[#2[[1]]]] &, Head /@ p]]
changeArg[f1[t] f2[t] f3[t], om, n]
n[f1[om[1]] f2[om[2]] f3[om[3]]]
answered 36 mins ago
m_goldbergm_goldberg
84.7k872196
answered 36 mins ago
m_goldbergm_goldberg
84.7k872196
answered 36 mins ago
m_goldbergm_goldberg
84.7k872196
answered 36 mins ago
m_goldbergm_goldberg
84.7k872196
84.7k872196
$begingroup$
Thank you for another possible answer. I think in the current form it doesn't cover the case of equal functions.
$endgroup$
– Display Name
22 mins ago
add a comment |
$begingroup$
Thank you for another possible answer. I think in the current form it doesn't cover the case of equal functions.
$endgroup$
– Display Name
22 mins ago
$begingroup$
Thank you for another possible answer. I think in the current form it doesn't cover the case of equal functions.
$endgroup$
– Display Name
22 mins ago
$begingroup$
Thank you for another possible answer. I think in the current form it doesn't cover the case of equal functions.
$endgroup$
– Display Name
22 mins ago
add a comment |
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