Taking the numerator and the denominator
$begingroup$
Is there any way to get the Numerator and the Denominator of an expression split in the same way that Mathematica graphically represents it.
Say I have an expression:
(a^(p1 + p2 - p3) b^(p1 - p2 + p3))/(c d)
When plugged into mathematica it gets represented with the $c$ and $d$ in the denominator but if I were to try to extract the numerator and denominator with respectively Numerator and Denominator I would not get the same split. I understand that a different (probably more sensible) choice of representation is made in Numerator and Denominator. However, is it possible to (automatically and reliably) take the denominator and numerator parts as split in the graphical representation? Mathematica must be able to determine this split since it has to decide how to graphically represent the output.
expression-manipulation
asked yesterday
KvotheKvothe
926317
$endgroup$
add a comment |
$begingroup$
Is there any way to get the Numerator and the Denominator of an expression split in the same way that Mathematica graphically represents it.
Say I have an expression:
(a^(p1 + p2 - p3) b^(p1 - p2 + p3))/(c d)
When plugged into mathematica it gets represented with the $c$ and $d$ in the denominator but if I were to try to extract the numerator and denominator with respectively Numerator and Denominator I would not get the same split. I understand that a different (probably more sensible) choice of representation is made in Numerator and Denominator. However, is it possible to (automatically and reliably) take the denominator and numerator parts as split in the graphical representation? Mathematica must be able to determine this split since it has to decide how to graphically represent the output.
expression-manipulation
asked yesterday
KvotheKvothe
926317
$endgroup$
add a comment |
$begingroup$
Is there any way to get the Numerator and the Denominator of an expression split in the same way that Mathematica graphically represents it.
Say I have an expression:
(a^(p1 + p2 - p3) b^(p1 - p2 + p3))/(c d)
When plugged into mathematica it gets represented with the $c$ and $d$ in the denominator but if I were to try to extract the numerator and denominator with respectively Numerator and Denominator I would not get the same split. I understand that a different (probably more sensible) choice of representation is made in Numerator and Denominator. However, is it possible to (automatically and reliably) take the denominator and numerator parts as split in the graphical representation? Mathematica must be able to determine this split since it has to decide how to graphically represent the output.
expression-manipulation
asked yesterday
KvotheKvothe
926317
$endgroup$
Is there any way to get the Numerator and the Denominator of an expression split in the same way that Mathematica graphically represents it.
Say I have an expression:
(a^(p1 + p2 - p3) b^(p1 - p2 + p3))/(c d)
When plugged into mathematica it gets represented with the $c$ and $d$ in the denominator but if I were to try to extract the numerator and denominator with respectively Numerator and Denominator I would not get the same split. I understand that a different (probably more sensible) choice of representation is made in Numerator and Denominator. However, is it possible to (automatically and reliably) take the denominator and numerator parts as split in the graphical representation? Mathematica must be able to determine this split since it has to decide how to graphically represent the output.
expression-manipulation
expression-manipulation
asked yesterday
KvotheKvothe
926317
asked yesterday
KvotheKvothe
926317
asked yesterday
KvotheKvothe
926317
asked yesterday
KvotheKvothe
926317
asked yesterday
KvotheKvothe
926317
926317
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The problem is that the negative terms in the exponent are being included in the denominator. A way to workaround this is to inactivate Plus, use Numerator and Denominator, and reactivate:
expr = (a^(p1+p2-p3) b^(p1-p2+p3))/(c d);
Activate @* Through @* {Numerator, Denominator} @* ReplaceAll[Plus->Inactive[Plus]] @ expr
{a^(p1 + p2 - p3) b^(p1 - p2 + p3), c d}
answered yesterday
Carl WollCarl Woll
70.9k394184
$endgroup$
add a comment |
$begingroup$
So, for your particular expression (which is a single fraction), the following kluge works:
expr = (a^(p1 + p2 - p3) b^(p1 - p2 + p3))/(c d);
ToExpression /@ List @@ ToBoxes@expr
(* {a^(p1 + p2 - p3) b^(p1 - p2 + p3), c d} *)
This is a little tricky to automat, because you have to inspect the formatting expression that you get.
This uses the visual formatting in the following way. ToBoxes converts the output to front-end formatting:
ToBoxes@expr
(* FractionBox[
RowBox[{SuperscriptBox["a", RowBox[{"p1", "+", "p2", "-", "p3"}]], " ", SuperscriptBox["b", RowBox[{"p1", "-", "p2", "+", "p3"}]]}],
RowBox[{"c", " ", "d"}]
] *)
Then, noting that the numerator and the denominator are the first and second elements of the the FractionBox expression, we replace FractionBox with List using List@@, and then convert the remaining formatting expression to Mathematica input expressions using ToExpression.
answered yesterday
marchmarch
17.4k22769
$endgroup$
add a comment |
$begingroup$
Here is another option using TeXForm.
expr = (a^(p1 + p2 - p3) b^(p1 - p2 + p3))/(c d);
t = ToString[TeXForm[expr]]
(* frac{a^{text{p1}+text{p2}-text{p3}} b^{text{p1}-text{p2}+text{p3}}}{c d} *)
Numerator:
ToExpression[StringCases[t, RegularExpression["frac{(.+)}{(.+)}$"] -> "$1"][[1]], TeXForm]
(* a^(p1 + p2 - p3) b^(p1 - p2 + p3) *)
Denominator:
ToExpression[StringCases[t, RegularExpression["frac{(.+)}{(.+)}$"] -> "$2"][[1]], TeXForm]
(* cd *)
answered yesterday
MelaGoMelaGo
2163
$endgroup$
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 problem is that the negative terms in the exponent are being included in the denominator. A way to workaround this is to inactivate Plus, use Numerator and Denominator, and reactivate:
expr = (a^(p1+p2-p3) b^(p1-p2+p3))/(c d);
Activate @* Through @* {Numerator, Denominator} @* ReplaceAll[Plus->Inactive[Plus]] @ expr
{a^(p1 + p2 - p3) b^(p1 - p2 + p3), c d}
answered yesterday
Carl WollCarl Woll
70.9k394184
$endgroup$
add a comment |
$begingroup$
The problem is that the negative terms in the exponent are being included in the denominator. A way to workaround this is to inactivate Plus, use Numerator and Denominator, and reactivate:
expr = (a^(p1+p2-p3) b^(p1-p2+p3))/(c d);
Activate @* Through @* {Numerator, Denominator} @* ReplaceAll[Plus->Inactive[Plus]] @ expr
{a^(p1 + p2 - p3) b^(p1 - p2 + p3), c d}
answered yesterday
Carl WollCarl Woll
70.9k394184
$endgroup$
add a comment |
$begingroup$
The problem is that the negative terms in the exponent are being included in the denominator. A way to workaround this is to inactivate Plus, use Numerator and Denominator, and reactivate:
expr = (a^(p1+p2-p3) b^(p1-p2+p3))/(c d);
Activate @* Through @* {Numerator, Denominator} @* ReplaceAll[Plus->Inactive[Plus]] @ expr
{a^(p1 + p2 - p3) b^(p1 - p2 + p3), c d}
answered yesterday
Carl WollCarl Woll
70.9k394184
$endgroup$
The problem is that the negative terms in the exponent are being included in the denominator. A way to workaround this is to inactivate Plus, use Numerator and Denominator, and reactivate:
expr = (a^(p1+p2-p3) b^(p1-p2+p3))/(c d);
Activate @* Through @* {Numerator, Denominator} @* ReplaceAll[Plus->Inactive[Plus]] @ expr
{a^(p1 + p2 - p3) b^(p1 - p2 + p3), c d}
answered yesterday
Carl WollCarl Woll
70.9k394184
answered yesterday
Carl WollCarl Woll
70.9k394184
answered yesterday
Carl WollCarl Woll
70.9k394184
answered yesterday
Carl WollCarl Woll
70.9k394184
70.9k394184
add a comment |
add a comment |
$begingroup$
So, for your particular expression (which is a single fraction), the following kluge works:
expr = (a^(p1 + p2 - p3) b^(p1 - p2 + p3))/(c d);
ToExpression /@ List @@ ToBoxes@expr
(* {a^(p1 + p2 - p3) b^(p1 - p2 + p3), c d} *)
This is a little tricky to automat, because you have to inspect the formatting expression that you get.
This uses the visual formatting in the following way. ToBoxes converts the output to front-end formatting:
ToBoxes@expr
(* FractionBox[
RowBox[{SuperscriptBox["a", RowBox[{"p1", "+", "p2", "-", "p3"}]], " ", SuperscriptBox["b", RowBox[{"p1", "-", "p2", "+", "p3"}]]}],
RowBox[{"c", " ", "d"}]
] *)
Then, noting that the numerator and the denominator are the first and second elements of the the FractionBox expression, we replace FractionBox with List using List@@, and then convert the remaining formatting expression to Mathematica input expressions using ToExpression.
answered yesterday
marchmarch
17.4k22769
$endgroup$
add a comment |
$begingroup$
So, for your particular expression (which is a single fraction), the following kluge works:
expr = (a^(p1 + p2 - p3) b^(p1 - p2 + p3))/(c d);
ToExpression /@ List @@ ToBoxes@expr
(* {a^(p1 + p2 - p3) b^(p1 - p2 + p3), c d} *)
This is a little tricky to automat, because you have to inspect the formatting expression that you get.
This uses the visual formatting in the following way. ToBoxes converts the output to front-end formatting:
ToBoxes@expr
(* FractionBox[
RowBox[{SuperscriptBox["a", RowBox[{"p1", "+", "p2", "-", "p3"}]], " ", SuperscriptBox["b", RowBox[{"p1", "-", "p2", "+", "p3"}]]}],
RowBox[{"c", " ", "d"}]
] *)
Then, noting that the numerator and the denominator are the first and second elements of the the FractionBox expression, we replace FractionBox with List using List@@, and then convert the remaining formatting expression to Mathematica input expressions using ToExpression.
answered yesterday
marchmarch
17.4k22769
$endgroup$
add a comment |
$begingroup$
So, for your particular expression (which is a single fraction), the following kluge works:
expr = (a^(p1 + p2 - p3) b^(p1 - p2 + p3))/(c d);
ToExpression /@ List @@ ToBoxes@expr
(* {a^(p1 + p2 - p3) b^(p1 - p2 + p3), c d} *)
This is a little tricky to automat, because you have to inspect the formatting expression that you get.
This uses the visual formatting in the following way. ToBoxes converts the output to front-end formatting:
ToBoxes@expr
(* FractionBox[
RowBox[{SuperscriptBox["a", RowBox[{"p1", "+", "p2", "-", "p3"}]], " ", SuperscriptBox["b", RowBox[{"p1", "-", "p2", "+", "p3"}]]}],
RowBox[{"c", " ", "d"}]
] *)
Then, noting that the numerator and the denominator are the first and second elements of the the FractionBox expression, we replace FractionBox with List using List@@, and then convert the remaining formatting expression to Mathematica input expressions using ToExpression.
answered yesterday
marchmarch
17.4k22769
$endgroup$
So, for your particular expression (which is a single fraction), the following kluge works:
expr = (a^(p1 + p2 - p3) b^(p1 - p2 + p3))/(c d);
ToExpression /@ List @@ ToBoxes@expr
(* {a^(p1 + p2 - p3) b^(p1 - p2 + p3), c d} *)
This is a little tricky to automat, because you have to inspect the formatting expression that you get.
This uses the visual formatting in the following way. ToBoxes converts the output to front-end formatting:
ToBoxes@expr
(* FractionBox[
RowBox[{SuperscriptBox["a", RowBox[{"p1", "+", "p2", "-", "p3"}]], " ", SuperscriptBox["b", RowBox[{"p1", "-", "p2", "+", "p3"}]]}],
RowBox[{"c", " ", "d"}]
] *)
Then, noting that the numerator and the denominator are the first and second elements of the the FractionBox expression, we replace FractionBox with List using List@@, and then convert the remaining formatting expression to Mathematica input expressions using ToExpression.
answered yesterday
marchmarch
17.4k22769
answered yesterday
marchmarch
17.4k22769
answered yesterday
marchmarch
17.4k22769
answered yesterday
marchmarch
17.4k22769
17.4k22769
add a comment |
add a comment |
$begingroup$
Here is another option using TeXForm.
expr = (a^(p1 + p2 - p3) b^(p1 - p2 + p3))/(c d);
t = ToString[TeXForm[expr]]
(* frac{a^{text{p1}+text{p2}-text{p3}} b^{text{p1}-text{p2}+text{p3}}}{c d} *)
Numerator:
ToExpression[StringCases[t, RegularExpression["frac{(.+)}{(.+)}$"] -> "$1"][[1]], TeXForm]
(* a^(p1 + p2 - p3) b^(p1 - p2 + p3) *)
Denominator:
ToExpression[StringCases[t, RegularExpression["frac{(.+)}{(.+)}$"] -> "$2"][[1]], TeXForm]
(* cd *)
answered yesterday
MelaGoMelaGo
2163
$endgroup$
add a comment |
$begingroup$
Here is another option using TeXForm.
expr = (a^(p1 + p2 - p3) b^(p1 - p2 + p3))/(c d);
t = ToString[TeXForm[expr]]
(* frac{a^{text{p1}+text{p2}-text{p3}} b^{text{p1}-text{p2}+text{p3}}}{c d} *)
Numerator:
ToExpression[StringCases[t, RegularExpression["frac{(.+)}{(.+)}$"] -> "$1"][[1]], TeXForm]
(* a^(p1 + p2 - p3) b^(p1 - p2 + p3) *)
Denominator:
ToExpression[StringCases[t, RegularExpression["frac{(.+)}{(.+)}$"] -> "$2"][[1]], TeXForm]
(* cd *)
answered yesterday
MelaGoMelaGo
2163
$endgroup$
add a comment |
$begingroup$
Here is another option using TeXForm.
expr = (a^(p1 + p2 - p3) b^(p1 - p2 + p3))/(c d);
t = ToString[TeXForm[expr]]
(* frac{a^{text{p1}+text{p2}-text{p3}} b^{text{p1}-text{p2}+text{p3}}}{c d} *)
Numerator:
ToExpression[StringCases[t, RegularExpression["frac{(.+)}{(.+)}$"] -> "$1"][[1]], TeXForm]
(* a^(p1 + p2 - p3) b^(p1 - p2 + p3) *)
Denominator:
ToExpression[StringCases[t, RegularExpression["frac{(.+)}{(.+)}$"] -> "$2"][[1]], TeXForm]
(* cd *)
answered yesterday
MelaGoMelaGo
2163
$endgroup$
Here is another option using TeXForm.
expr = (a^(p1 + p2 - p3) b^(p1 - p2 + p3))/(c d);
t = ToString[TeXForm[expr]]
(* frac{a^{text{p1}+text{p2}-text{p3}} b^{text{p1}-text{p2}+text{p3}}}{c d} *)
Numerator:
ToExpression[StringCases[t, RegularExpression["frac{(.+)}{(.+)}$"] -> "$1"][[1]], TeXForm]
(* a^(p1 + p2 - p3) b^(p1 - p2 + p3) *)
Denominator:
ToExpression[StringCases[t, RegularExpression["frac{(.+)}{(.+)}$"] -> "$2"][[1]], TeXForm]
(* cd *)
answered yesterday
MelaGoMelaGo
2163
answered yesterday
MelaGoMelaGo
2163
answered yesterday
MelaGoMelaGo
2163
answered yesterday
MelaGoMelaGo
2163
2163
add a comment |
add a comment |
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