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











up vote
4
down vote

favorite
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




















  • 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
    yesterday










  • @Shredderroy Fixed. Thanks!
    – ablmf
    yesterday















up vote
4
down vote

favorite
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




















  • 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
    yesterday










  • @Shredderroy Fixed. Thanks!
    – ablmf
    yesterday













up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






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 yesterday

























asked yesterday









ablmf

23917




23917








  • 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
    yesterday










  • @Shredderroy Fixed. Thanks!
    – ablmf
    yesterday














  • 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
    yesterday










  • @Shredderroy Fixed. Thanks!
    – ablmf
    yesterday








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
yesterday




Are you sure that g1 in the list of multiplicative factors is not an error? The pattern suggests that it should be g2.
– Shredderroy
yesterday












@Shredderroy Fixed. Thanks!
– ablmf
yesterday




@Shredderroy Fixed. Thanks!
– ablmf
yesterday










1 Answer
1






active

oldest

votes

















up vote
6
down vote



accepted










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
    yesterday












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










  • @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
    yesterday











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1 Answer
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1 Answer
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oldest

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active

oldest

votes








up vote
6
down vote



accepted










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
    yesterday












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










  • @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
    yesterday















up vote
6
down vote



accepted










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
    yesterday












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










  • @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
    yesterday













up vote
6
down vote



accepted







up vote
6
down vote



accepted






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 yesterday

























answered yesterday









kglr

175k9197402




175k9197402












  • 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
    yesterday












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










  • @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
    yesterday


















  • 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
    yesterday












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










  • @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
    yesterday
















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
yesterday






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
yesterday














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




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












@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
yesterday




@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
yesterday


















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