Removing fractions without a particular variableFilter out all terms not involving a given variableAdd...
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Removing fractions without a particular variable
Filter out all terms not involving a given variableAdd Pattern to List Removing DuplicatesHow to make this particular replacement involving expAutomatic substitution of fractionsHow to select the value of a substitution rule for a particular variable?Why is `Block` significantly faster than `ReplaceAll` for inputting variable valuesApply tranformation to particular list elementsMake substitutions without specifying the objectHow to replace Variable With Variableusing DeleteCases to delete vectors with particular index as 0Relative Chop for Fractions with Trigonometric Terms
$begingroup$
As a very minimal example, consider a sum of fractions such as:
$A=frac{a}{s_{123}bc}+frac{d}{e}$
In practice, I have many hundreds of thousands of these fractions generated from a recursion relation. My desired result comes around from the prescription in my theory, where I am required to multiply through by $s_{123}$ and then take $s_{123}to 0$. Upon replicating in MMA, I use a code similar to:
A*s[1,2,3]//Expand
%/.{s[1,2,3]->0}
This Expand is necessary since by simply multiplying A by the relevant factor, MMA will first interpret it as $s_{123} big (frac{a}{s_{123}bc}+frac{d}{e}big)$. I think this is what is slowing my code down substantially, and crashes MMA due to a lack of memory.
To get around this, I have had a few ideas;
- Use something like
DeleteCaseas a simple method (or something perhaps similar to Filter out all terms not involving a given variable which might be a bit overkill for my needs) to delete all terms not involving $s_{123}$, and then set the remaining factors of it in the leftover terms to 1. - Instead of using
Expand, define a new function which brings common factors inside to each fraction.
Which of these, alongside any other methods, would you suggest? How could I go about implementing them? I think 2 would be easier to do, but at this point, I am needing efficiency in my code on such a large scale.
performance-tuning replacement deletecases
asked 6 hours ago
BradBrad
497
$endgroup$
add a comment |
$begingroup$
As a very minimal example, consider a sum of fractions such as:
$A=frac{a}{s_{123}bc}+frac{d}{e}$
In practice, I have many hundreds of thousands of these fractions generated from a recursion relation. My desired result comes around from the prescription in my theory, where I am required to multiply through by $s_{123}$ and then take $s_{123}to 0$. Upon replicating in MMA, I use a code similar to:
A*s[1,2,3]//Expand
%/.{s[1,2,3]->0}
This Expand is necessary since by simply multiplying A by the relevant factor, MMA will first interpret it as $s_{123} big (frac{a}{s_{123}bc}+frac{d}{e}big)$. I think this is what is slowing my code down substantially, and crashes MMA due to a lack of memory.
To get around this, I have had a few ideas;
- Use something like
DeleteCaseas a simple method (or something perhaps similar to Filter out all terms not involving a given variable which might be a bit overkill for my needs) to delete all terms not involving $s_{123}$, and then set the remaining factors of it in the leftover terms to 1. - Instead of using
Expand, define a new function which brings common factors inside to each fraction.
Which of these, alongside any other methods, would you suggest? How could I go about implementing them? I think 2 would be easier to do, but at this point, I am needing efficiency in my code on such a large scale.
performance-tuning replacement deletecases
asked 6 hours ago
BradBrad
497
$endgroup$
$begingroup$
You might find that working with aListof terms is better (e.g. effectivelyList@@A) than working with a sum of terms. PresumablyExpandis comparing a large number of terms, looking for common subexpressions.
$endgroup$
– mikado
3 hours ago
add a comment |
$begingroup$
As a very minimal example, consider a sum of fractions such as:
$A=frac{a}{s_{123}bc}+frac{d}{e}$
In practice, I have many hundreds of thousands of these fractions generated from a recursion relation. My desired result comes around from the prescription in my theory, where I am required to multiply through by $s_{123}$ and then take $s_{123}to 0$. Upon replicating in MMA, I use a code similar to:
A*s[1,2,3]//Expand
%/.{s[1,2,3]->0}
This Expand is necessary since by simply multiplying A by the relevant factor, MMA will first interpret it as $s_{123} big (frac{a}{s_{123}bc}+frac{d}{e}big)$. I think this is what is slowing my code down substantially, and crashes MMA due to a lack of memory.
To get around this, I have had a few ideas;
- Use something like
DeleteCaseas a simple method (or something perhaps similar to Filter out all terms not involving a given variable which might be a bit overkill for my needs) to delete all terms not involving $s_{123}$, and then set the remaining factors of it in the leftover terms to 1. - Instead of using
Expand, define a new function which brings common factors inside to each fraction.
Which of these, alongside any other methods, would you suggest? How could I go about implementing them? I think 2 would be easier to do, but at this point, I am needing efficiency in my code on such a large scale.
performance-tuning replacement deletecases
asked 6 hours ago
BradBrad
497
$endgroup$
As a very minimal example, consider a sum of fractions such as:
$A=frac{a}{s_{123}bc}+frac{d}{e}$
In practice, I have many hundreds of thousands of these fractions generated from a recursion relation. My desired result comes around from the prescription in my theory, where I am required to multiply through by $s_{123}$ and then take $s_{123}to 0$. Upon replicating in MMA, I use a code similar to:
A*s[1,2,3]//Expand
%/.{s[1,2,3]->0}
This Expand is necessary since by simply multiplying A by the relevant factor, MMA will first interpret it as $s_{123} big (frac{a}{s_{123}bc}+frac{d}{e}big)$. I think this is what is slowing my code down substantially, and crashes MMA due to a lack of memory.
To get around this, I have had a few ideas;
- Use something like
DeleteCaseas a simple method (or something perhaps similar to Filter out all terms not involving a given variable which might be a bit overkill for my needs) to delete all terms not involving $s_{123}$, and then set the remaining factors of it in the leftover terms to 1. - Instead of using
Expand, define a new function which brings common factors inside to each fraction.
Which of these, alongside any other methods, would you suggest? How could I go about implementing them? I think 2 would be easier to do, but at this point, I am needing efficiency in my code on such a large scale.
performance-tuning replacement deletecases
performance-tuning replacement deletecases
asked 6 hours ago
BradBrad
497
asked 6 hours ago
BradBrad
497
edited 4 hours ago
Brad
asked 6 hours ago
BradBrad
497
asked 6 hours ago
BradBrad
497
asked 6 hours ago
BradBrad
497
497
$begingroup$
You might find that working with aListof terms is better (e.g. effectivelyList@@A) than working with a sum of terms. PresumablyExpandis comparing a large number of terms, looking for common subexpressions.
$endgroup$
– mikado
3 hours ago
add a comment |
$begingroup$
You might find that working with aListof terms is better (e.g. effectivelyList@@A) than working with a sum of terms. PresumablyExpandis comparing a large number of terms, looking for common subexpressions.
$endgroup$
– mikado
3 hours ago
$begingroup$
You might find that working with a
List of terms is better (e.g. effectively List@@A) than working with a sum of terms. Presumably Expand is comparing a large number of terms, looking for common subexpressions.$endgroup$
– mikado
3 hours ago
$begingroup$
You might find that working with a
List of terms is better (e.g. effectively List@@A) than working with a sum of terms. Presumably Expand is comparing a large number of terms, looking for common subexpressions.$endgroup$
– mikado
3 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Pick out only those terms in A that contain exactly one s123 in the denominator:
A = a/(s123 b c) + d/e;
Select[A, Exponent[#, s123] == -1 &] /. s123 -> 1
a/(b c)
or with pattern matching:
Total[Cases[A, _/s123 | 1/s123, {1}]] /. s123 -> 1
a/(b c)
You'll have to check what's fastest in your case. Also, for added speed you may leave out the special pattern 1/s123 (*) in the second solution if you're sure it never occurs.
(*) We need a special pattern for 1/s123 because it does not match the pattern _/s123. Why not? The full form of x/s123 (with anything for x) has head Times:
x/s123 //FullForm
(* Times[Power[s123, -1], x] *)
The full form of 1/s123, on the other hand, has head Power:
1/s123 //FullForm
(* Power[s123, -1] *)
answered 2 hours ago
RomanRoman
2,249716
$endgroup$
$begingroup$
Thank you for your very helpful and informative comment. Only off the top of your head, can you think of any other alternative methods that may be faster which I have not considered before?
$endgroup$
– Brad
44 mins ago
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
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votes
$begingroup$
Pick out only those terms in A that contain exactly one s123 in the denominator:
A = a/(s123 b c) + d/e;
Select[A, Exponent[#, s123] == -1 &] /. s123 -> 1
a/(b c)
or with pattern matching:
Total[Cases[A, _/s123 | 1/s123, {1}]] /. s123 -> 1
a/(b c)
You'll have to check what's fastest in your case. Also, for added speed you may leave out the special pattern 1/s123 (*) in the second solution if you're sure it never occurs.
(*) We need a special pattern for 1/s123 because it does not match the pattern _/s123. Why not? The full form of x/s123 (with anything for x) has head Times:
x/s123 //FullForm
(* Times[Power[s123, -1], x] *)
The full form of 1/s123, on the other hand, has head Power:
1/s123 //FullForm
(* Power[s123, -1] *)
answered 2 hours ago
RomanRoman
2,249716
$endgroup$
$begingroup$
Thank you for your very helpful and informative comment. Only off the top of your head, can you think of any other alternative methods that may be faster which I have not considered before?
$endgroup$
– Brad
44 mins ago
add a comment |
$begingroup$
Pick out only those terms in A that contain exactly one s123 in the denominator:
A = a/(s123 b c) + d/e;
Select[A, Exponent[#, s123] == -1 &] /. s123 -> 1
a/(b c)
or with pattern matching:
Total[Cases[A, _/s123 | 1/s123, {1}]] /. s123 -> 1
a/(b c)
You'll have to check what's fastest in your case. Also, for added speed you may leave out the special pattern 1/s123 (*) in the second solution if you're sure it never occurs.
(*) We need a special pattern for 1/s123 because it does not match the pattern _/s123. Why not? The full form of x/s123 (with anything for x) has head Times:
x/s123 //FullForm
(* Times[Power[s123, -1], x] *)
The full form of 1/s123, on the other hand, has head Power:
1/s123 //FullForm
(* Power[s123, -1] *)
answered 2 hours ago
RomanRoman
2,249716
$endgroup$
$begingroup$
Thank you for your very helpful and informative comment. Only off the top of your head, can you think of any other alternative methods that may be faster which I have not considered before?
$endgroup$
– Brad
44 mins ago
add a comment |
$begingroup$
Pick out only those terms in A that contain exactly one s123 in the denominator:
A = a/(s123 b c) + d/e;
Select[A, Exponent[#, s123] == -1 &] /. s123 -> 1
a/(b c)
or with pattern matching:
Total[Cases[A, _/s123 | 1/s123, {1}]] /. s123 -> 1
a/(b c)
You'll have to check what's fastest in your case. Also, for added speed you may leave out the special pattern 1/s123 (*) in the second solution if you're sure it never occurs.
(*) We need a special pattern for 1/s123 because it does not match the pattern _/s123. Why not? The full form of x/s123 (with anything for x) has head Times:
x/s123 //FullForm
(* Times[Power[s123, -1], x] *)
The full form of 1/s123, on the other hand, has head Power:
1/s123 //FullForm
(* Power[s123, -1] *)
answered 2 hours ago
RomanRoman
2,249716
$endgroup$
Pick out only those terms in A that contain exactly one s123 in the denominator:
A = a/(s123 b c) + d/e;
Select[A, Exponent[#, s123] == -1 &] /. s123 -> 1
a/(b c)
or with pattern matching:
Total[Cases[A, _/s123 | 1/s123, {1}]] /. s123 -> 1
a/(b c)
You'll have to check what's fastest in your case. Also, for added speed you may leave out the special pattern 1/s123 (*) in the second solution if you're sure it never occurs.
(*) We need a special pattern for 1/s123 because it does not match the pattern _/s123. Why not? The full form of x/s123 (with anything for x) has head Times:
x/s123 //FullForm
(* Times[Power[s123, -1], x] *)
The full form of 1/s123, on the other hand, has head Power:
1/s123 //FullForm
(* Power[s123, -1] *)
answered 2 hours ago
RomanRoman
2,249716
edited 49 mins ago
answered 2 hours ago
RomanRoman
2,249716
answered 2 hours ago
RomanRoman
2,249716
answered 2 hours ago
RomanRoman
2,249716
2,249716
$begingroup$
Thank you for your very helpful and informative comment. Only off the top of your head, can you think of any other alternative methods that may be faster which I have not considered before?
$endgroup$
– Brad
44 mins ago
add a comment |
$begingroup$
Thank you for your very helpful and informative comment. Only off the top of your head, can you think of any other alternative methods that may be faster which I have not considered before?
$endgroup$
– Brad
44 mins ago
$begingroup$
Thank you for your very helpful and informative comment. Only off the top of your head, can you think of any other alternative methods that may be faster which I have not considered before?
$endgroup$
– Brad
44 mins ago
$begingroup$
Thank you for your very helpful and informative comment. Only off the top of your head, can you think of any other alternative methods that may be faster which I have not considered before?
$endgroup$
– Brad
44 mins ago
add a comment |
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$begingroup$
You might find that working with a
Listof terms is better (e.g. effectivelyList@@A) than working with a sum of terms. PresumablyExpandis comparing a large number of terms, looking for common subexpressions.$endgroup$
– mikado
3 hours ago