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
asked Dec 7 at 20:27
ablmf
24417
add a comment |
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
asked Dec 7 at 20:27
ablmf
24417
1
Are you sure thatg1in the list of multiplicative factors is not an error? The pattern suggests that it should beg2.
– Shredderroy
Dec 7 at 21:30
@Shredderroy Fixed. Thanks!
– ablmf
Dec 7 at 22:03
add a comment |
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
asked Dec 7 at 20:27
ablmf
24417
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
asked Dec 7 at 20:27
ablmf
24417
asked Dec 7 at 20:27
ablmf
24417
edited Dec 7 at 22:03
asked Dec 7 at 20:27
ablmf
24417
asked Dec 7 at 20:27
ablmf
24417
asked Dec 7 at 20:27
ablmf
24417
24417
1
Are you sure thatg1in the list of multiplicative factors is not an error? The pattern suggests that it should beg2.
– Shredderroy
Dec 7 at 21:30
@Shredderroy Fixed. Thanks!
– ablmf
Dec 7 at 22:03
add a comment |
1
Are you sure thatg1in the list of multiplicative factors is not an error? The pattern suggests that it should beg2.
– 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
add a comment |
1 Answer
1
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oldest
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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]
answered Dec 7 at 20:55
kglr
175k9197402
Doesn'tReplaceRepeatedwith 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 hadg1as one of the multiplicative factors, notg2.
– Shredderroy
Dec 7 at 21:28
@Shredderroy, yesReplaceRepeatedalso works an is simpler. It doesn't work with the pattern I had tried (a_[b_. c_, d___]), so I usedReplaceFixedPointcombination.
– kglr
Dec 7 at 21:57
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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]
answered Dec 7 at 20:55
kglr
175k9197402
Doesn'tReplaceRepeatedwith 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 hadg1as one of the multiplicative factors, notg2.
– Shredderroy
Dec 7 at 21:28
@Shredderroy, yesReplaceRepeatedalso works an is simpler. It doesn't work with the pattern I had tried (a_[b_. c_, d___]), so I usedReplaceFixedPointcombination.
– kglr
Dec 7 at 21:57
add a comment |
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]
answered Dec 7 at 20:55
kglr
175k9197402
Doesn'tReplaceRepeatedwith 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 hadg1as one of the multiplicative factors, notg2.
– Shredderroy
Dec 7 at 21:28
@Shredderroy, yesReplaceRepeatedalso works an is simpler. It doesn't work with the pattern I had tried (a_[b_. c_, d___]), so I usedReplaceFixedPointcombination.
– kglr
Dec 7 at 21:57
add a comment |
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]
answered Dec 7 at 20:55
kglr
175k9197402
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]
answered Dec 7 at 20:55
kglr
175k9197402
edited Dec 7 at 21:23
answered Dec 7 at 20:55
kglr
175k9197402
answered Dec 7 at 20:55
kglr
175k9197402
answered Dec 7 at 20:55
kglr
175k9197402
175k9197402
Doesn'tReplaceRepeatedwith 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 hadg1as one of the multiplicative factors, notg2.
– Shredderroy
Dec 7 at 21:28
@Shredderroy, yesReplaceRepeatedalso works an is simpler. It doesn't work with the pattern I had tried (a_[b_. c_, d___]), so I usedReplaceFixedPointcombination.
– kglr
Dec 7 at 21:57
add a comment |
Doesn'tReplaceRepeatedwith 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 hadg1as one of the multiplicative factors, notg2.
– Shredderroy
Dec 7 at 21:28
@Shredderroy, yesReplaceRepeatedalso works an is simpler. It doesn't work with the pattern I had tried (a_[b_. c_, d___]), so I usedReplaceFixedPointcombination.
– 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
add a comment |
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1
Are you sure that
g1in the list of multiplicative factors is not an error? The pattern suggests that it should beg2.– Shredderroy
Dec 7 at 21:30
@Shredderroy Fixed. Thanks!
– ablmf
Dec 7 at 22:03