How to bring out all factors recursively except one particular term?












4














I want to write a function 



keepOnly[expr_, keep_]


Such that 



keepOnly[f[f2*g[g2*h[h2*keep, h1], g1], f1], keep]


becomes 



f2*h2*g2*f[g[h[keep, h1], g1], f1]


In other words, we take all the factors out except for the term keep.






pattern-matching expression-manipulation







share|improve this question




















  • 1




    Are you sure that g1 in the list of multiplicative factors is not an error? The pattern suggests that it should be g2.
    – Shredderroy
    Dec 7 at 21:30










  • @Shredderroy Fixed. Thanks!
    – ablmf
    Dec 7 at 22:03
















4














I want to write a function 



keepOnly[expr_, keep_]


Such that 



keepOnly[f[f2*g[g2*h[h2*keep, h1], g1], f1], keep]


becomes 



f2*h2*g2*f[g[h[keep, h1], g1], f1]


In other words, we take all the factors out except for the term keep.






pattern-matching expression-manipulation







share|improve this question




















  • 1




    Are you sure that g1 in the list of multiplicative factors is not an error? The pattern suggests that it should be g2.
    – Shredderroy
    Dec 7 at 21:30










  • @Shredderroy Fixed. Thanks!
    – ablmf
    Dec 7 at 22:03














4












4








4


1





I want to write a function 



keepOnly[expr_, keep_]


Such that 



keepOnly[f[f2*g[g2*h[h2*keep, h1], g1], f1], keep]


becomes 



f2*h2*g2*f[g[h[keep, h1], g1], f1]


In other words, we take all the factors out except for the term keep.






pattern-matching expression-manipulation







share|improve this question















I want to write a function 



keepOnly[expr_, keep_]


Such that 



keepOnly[f[f2*g[g2*h[h2*keep, h1], g1], f1], keep]


becomes 



f2*h2*g2*f[g[h[keep, h1], g1], f1]


In other words, we take all the factors out except for the term keep.






pattern-matching expression-manipulation




pattern-matching expression-manipulation






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Dec 7 at 22:03

























asked Dec 7 at 20:27









ablmf

24617




24617








  • 1




    Are you sure that g1 in the list of multiplicative factors is not an error? The pattern suggests that it should be g2.
    – Shredderroy
    Dec 7 at 21:30










  • @Shredderroy Fixed. Thanks!
    – ablmf
    Dec 7 at 22:03














  • 1




    Are you sure that g1 in the list of multiplicative factors is not an error? The pattern suggests that it should be g2.
    – Shredderroy
    Dec 7 at 21:30










  • @Shredderroy Fixed. Thanks!
    – ablmf
    Dec 7 at 22:03








1




1




Are you sure that g1 in the list of multiplicative factors is not an error? The pattern suggests that it should be g2.
– Shredderroy
Dec 7 at 21:30




Are you sure that g1 in the list of multiplicative factors is not an error? The pattern suggests that it should be g2.
– Shredderroy
Dec 7 at 21:30












@Shredderroy Fixed. Thanks!
– ablmf
Dec 7 at 22:03




@Shredderroy Fixed. Thanks!
– ablmf
Dec 7 at 22:03










1 Answer
1






active

oldest

votes


















6














exp = f[f2*g[g2*h[h2*keep, h1], g1], f1];



FixedPoint[Replace[#, a_[b_. c_, d___] /; Not[FreeQ[c, keep]] :> b a[c, d], {0, ∞}] &, exp]



f2 g2 h2 f[g[h[keep, h1], g1], f1]




Alternatively,



FixedPoint[Replace[#, a_[b_. c_?(Not@*FreeQ[keep]), d___] :> b a[c, d], {0, ∞}] &, exp]



f2 g2 h2 f[g[h[keep, h1], g1], f1]







share|improve this answer























  • Doesn't ReplaceRepeated with pretty much the same patterns also do it? ReplaceRepeated[f[f2 * g[g2 * h[h2 * keep, h1], g1], f1], a_[b_ * c_?(Not@*FreeQ[keep]), d_] :> b * a[c, d]]
    – Shredderroy
    Dec 7 at 21:27












  • I was a bit confused at first because the original question had g1 as one of the multiplicative factors, not g2.
    – Shredderroy
    Dec 7 at 21:28










  • @Shredderroy, yes ReplaceRepeated also works an is simpler. It doesn't work with the pattern I had tried (a_[b_. c_, d___]), so I used Replace FixedPoint combination.
    – kglr
    Dec 7 at 21:57











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1 Answer
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6














exp = f[f2*g[g2*h[h2*keep, h1], g1], f1];



FixedPoint[Replace[#, a_[b_. c_, d___] /; Not[FreeQ[c, keep]] :> b a[c, d], {0, ∞}] &, exp]



f2 g2 h2 f[g[h[keep, h1], g1], f1]




Alternatively,



FixedPoint[Replace[#, a_[b_. c_?(Not@*FreeQ[keep]), d___] :> b a[c, d], {0, ∞}] &, exp]



f2 g2 h2 f[g[h[keep, h1], g1], f1]







share|improve this answer























  • Doesn't ReplaceRepeated with pretty much the same patterns also do it? ReplaceRepeated[f[f2 * g[g2 * h[h2 * keep, h1], g1], f1], a_[b_ * c_?(Not@*FreeQ[keep]), d_] :> b * a[c, d]]
    – Shredderroy
    Dec 7 at 21:27












  • I was a bit confused at first because the original question had g1 as one of the multiplicative factors, not g2.
    – Shredderroy
    Dec 7 at 21:28










  • @Shredderroy, yes ReplaceRepeated also works an is simpler. It doesn't work with the pattern I had tried (a_[b_. c_, d___]), so I used Replace FixedPoint combination.
    – kglr
    Dec 7 at 21:57
















6














exp = f[f2*g[g2*h[h2*keep, h1], g1], f1];



FixedPoint[Replace[#, a_[b_. c_, d___] /; Not[FreeQ[c, keep]] :> b a[c, d], {0, ∞}] &, exp]



f2 g2 h2 f[g[h[keep, h1], g1], f1]




Alternatively,



FixedPoint[Replace[#, a_[b_. c_?(Not@*FreeQ[keep]), d___] :> b a[c, d], {0, ∞}] &, exp]



f2 g2 h2 f[g[h[keep, h1], g1], f1]







share|improve this answer























  • Doesn't ReplaceRepeated with pretty much the same patterns also do it? ReplaceRepeated[f[f2 * g[g2 * h[h2 * keep, h1], g1], f1], a_[b_ * c_?(Not@*FreeQ[keep]), d_] :> b * a[c, d]]
    – Shredderroy
    Dec 7 at 21:27












  • I was a bit confused at first because the original question had g1 as one of the multiplicative factors, not g2.
    – Shredderroy
    Dec 7 at 21:28










  • @Shredderroy, yes ReplaceRepeated also works an is simpler. It doesn't work with the pattern I had tried (a_[b_. c_, d___]), so I used Replace FixedPoint combination.
    – kglr
    Dec 7 at 21:57














6












6








6






exp = f[f2*g[g2*h[h2*keep, h1], g1], f1];



FixedPoint[Replace[#, a_[b_. c_, d___] /; Not[FreeQ[c, keep]] :> b a[c, d], {0, ∞}] &, exp]



f2 g2 h2 f[g[h[keep, h1], g1], f1]




Alternatively,



FixedPoint[Replace[#, a_[b_. c_?(Not@*FreeQ[keep]), d___] :> b a[c, d], {0, ∞}] &, exp]



f2 g2 h2 f[g[h[keep, h1], g1], f1]







share|improve this answer














exp = f[f2*g[g2*h[h2*keep, h1], g1], f1];



FixedPoint[Replace[#, a_[b_. c_, d___] /; Not[FreeQ[c, keep]] :> b a[c, d], {0, ∞}] &, exp]



f2 g2 h2 f[g[h[keep, h1], g1], f1]




Alternatively,



FixedPoint[Replace[#, a_[b_. c_?(Not@*FreeQ[keep]), d___] :> b a[c, d], {0, ∞}] &, exp]



f2 g2 h2 f[g[h[keep, h1], g1], f1]








share|improve this answer














share|improve this answer



share|improve this answer








edited Dec 7 at 21:23

























answered Dec 7 at 20:55









kglr

176k9197402




176k9197402












  • Doesn't ReplaceRepeated with pretty much the same patterns also do it? ReplaceRepeated[f[f2 * g[g2 * h[h2 * keep, h1], g1], f1], a_[b_ * c_?(Not@*FreeQ[keep]), d_] :> b * a[c, d]]
    – Shredderroy
    Dec 7 at 21:27












  • I was a bit confused at first because the original question had g1 as one of the multiplicative factors, not g2.
    – Shredderroy
    Dec 7 at 21:28










  • @Shredderroy, yes ReplaceRepeated also works an is simpler. It doesn't work with the pattern I had tried (a_[b_. c_, d___]), so I used Replace FixedPoint combination.
    – kglr
    Dec 7 at 21:57


















  • Doesn't ReplaceRepeated with pretty much the same patterns also do it? ReplaceRepeated[f[f2 * g[g2 * h[h2 * keep, h1], g1], f1], a_[b_ * c_?(Not@*FreeQ[keep]), d_] :> b * a[c, d]]
    – Shredderroy
    Dec 7 at 21:27












  • I was a bit confused at first because the original question had g1 as one of the multiplicative factors, not g2.
    – Shredderroy
    Dec 7 at 21:28










  • @Shredderroy, yes ReplaceRepeated also works an is simpler. It doesn't work with the pattern I had tried (a_[b_. c_, d___]), so I used Replace FixedPoint combination.
    – kglr
    Dec 7 at 21:57
















Doesn't ReplaceRepeated with pretty much the same patterns also do it? ReplaceRepeated[f[f2 * g[g2 * h[h2 * keep, h1], g1], f1], a_[b_ * c_?(Not@*FreeQ[keep]), d_] :> b * a[c, d]]
– Shredderroy
Dec 7 at 21:27






Doesn't ReplaceRepeated with pretty much the same patterns also do it? ReplaceRepeated[f[f2 * g[g2 * h[h2 * keep, h1], g1], f1], a_[b_ * c_?(Not@*FreeQ[keep]), d_] :> b * a[c, d]]
– Shredderroy
Dec 7 at 21:27














I was a bit confused at first because the original question had g1 as one of the multiplicative factors, not g2.
– Shredderroy
Dec 7 at 21:28




I was a bit confused at first because the original question had g1 as one of the multiplicative factors, not g2.
– Shredderroy
Dec 7 at 21:28












@Shredderroy, yes ReplaceRepeated also works an is simpler. It doesn't work with the pattern I had tried (a_[b_. c_, d___]), so I used Replace FixedPoint combination.
– kglr
Dec 7 at 21:57




@Shredderroy, yes ReplaceRepeated also works an is simpler. It doesn't work with the pattern I had tried (a_[b_. c_, d___]), so I used Replace FixedPoint combination.
– kglr
Dec 7 at 21:57


















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