Turning arguments into exponentsWhat is -(-2,3,4)?Replace rule also matching complex numbersCollecting terms of even exponentsApplying functions repeatedly on parts of complicated expressionsSplitBy with exponentsTurning table into association for ClassifyCollect with symbolic exponentscoordinate change in a list of pointsTurning a NB into a Repeatable Script?How to Append Sequence of Numbers to List?Expression replacement with variable exponents
Turning arguments into exponents
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Turning arguments into exponents
What is -(-2,3,4)?Replace rule also matching complex numbersCollecting terms of even exponentsApplying functions repeatedly on parts of complicated expressionsSplitBy with exponentsTurning table into association for ClassifyCollect with symbolic exponentscoordinate change in a list of pointsTurning a NB into a Repeatable Script?How to Append Sequence of Numbers to List?Expression replacement with variable exponents
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I'd like to replace every instance of L[stuff], e.g.,
L[-1,-2,-2]
in an expression with x^(-FromDigits[stuff]), e.g.
x^122
.
But I'm running into an issue-- I'd like to do this with the code
L[-1, -2, -2] /. L[a___] -> x^(FromDigits[-a])
but this produces
2/x^6 + x^4 + x^5.
In trouble-shooting, I noticed that while
L[-1, -2, -2] /. L[a___] -> a
produces the output I expected, namely -1,-2,-2, I found that
L[-1, -2, -2] /. L[a___] -> -a
just produces 4 rather than 1,2,2. I must be misunderstanding how Sequence behaves when it shows up in the right side of a Rule since when I try just inputting -Sequence[-1,-2,-2] I get the expected result 1,2,2. Does anyone see where I'm going wrong? Thanks!
list-manipulation replacement symbolic polynomials sequence
asked 8 hours ago
DiffycueDiffycue
2521 silver badge6 bronze badges
$endgroup$
add a comment |
$begingroup$
I'd like to replace every instance of L[stuff], e.g.,
L[-1,-2,-2]
in an expression with x^(-FromDigits[stuff]), e.g.
x^122
.
But I'm running into an issue-- I'd like to do this with the code
L[-1, -2, -2] /. L[a___] -> x^(FromDigits[-a])
but this produces
2/x^6 + x^4 + x^5.
In trouble-shooting, I noticed that while
L[-1, -2, -2] /. L[a___] -> a
produces the output I expected, namely -1,-2,-2, I found that
L[-1, -2, -2] /. L[a___] -> -a
just produces 4 rather than 1,2,2. I must be misunderstanding how Sequence behaves when it shows up in the right side of a Rule since when I try just inputting -Sequence[-1,-2,-2] I get the expected result 1,2,2. Does anyone see where I'm going wrong? Thanks!
list-manipulation replacement symbolic polynomials sequence
asked 8 hours ago
DiffycueDiffycue
2521 silver badge6 bronze badges
$endgroup$
2
$begingroup$
funny, I asked a very similar question last week, I wonder if its the same issue. Anyway, here is a possible workaround:L[-1, -2, -2] /. L[a___] :> x^(FromDigits[-a]).
$endgroup$
– AccidentalFourierTransform
8 hours ago
$begingroup$
@AccidentalFourierTransform yup it would appear to be exactly the same issue and my reaction toFullForm[-b]was exactly the same as yours lol
$endgroup$
– Diffycue
8 hours ago
add a comment |
$begingroup$
I'd like to replace every instance of L[stuff], e.g.,
L[-1,-2,-2]
in an expression with x^(-FromDigits[stuff]), e.g.
x^122
.
But I'm running into an issue-- I'd like to do this with the code
L[-1, -2, -2] /. L[a___] -> x^(FromDigits[-a])
but this produces
2/x^6 + x^4 + x^5.
In trouble-shooting, I noticed that while
L[-1, -2, -2] /. L[a___] -> a
produces the output I expected, namely -1,-2,-2, I found that
L[-1, -2, -2] /. L[a___] -> -a
just produces 4 rather than 1,2,2. I must be misunderstanding how Sequence behaves when it shows up in the right side of a Rule since when I try just inputting -Sequence[-1,-2,-2] I get the expected result 1,2,2. Does anyone see where I'm going wrong? Thanks!
list-manipulation replacement symbolic polynomials sequence
asked 8 hours ago
DiffycueDiffycue
2521 silver badge6 bronze badges
$endgroup$
I'd like to replace every instance of L[stuff], e.g.,
L[-1,-2,-2]
in an expression with x^(-FromDigits[stuff]), e.g.
x^122
.
But I'm running into an issue-- I'd like to do this with the code
L[-1, -2, -2] /. L[a___] -> x^(FromDigits[-a])
but this produces
2/x^6 + x^4 + x^5.
In trouble-shooting, I noticed that while
L[-1, -2, -2] /. L[a___] -> a
produces the output I expected, namely -1,-2,-2, I found that
L[-1, -2, -2] /. L[a___] -> -a
just produces 4 rather than 1,2,2. I must be misunderstanding how Sequence behaves when it shows up in the right side of a Rule since when I try just inputting -Sequence[-1,-2,-2] I get the expected result 1,2,2. Does anyone see where I'm going wrong? Thanks!
list-manipulation replacement symbolic polynomials sequence
list-manipulation replacement symbolic polynomials sequence
asked 8 hours ago
DiffycueDiffycue
2521 silver badge6 bronze badges
asked 8 hours ago
DiffycueDiffycue
2521 silver badge6 bronze badges
asked 8 hours ago
DiffycueDiffycue
2521 silver badge6 bronze badges
asked 8 hours ago
DiffycueDiffycue
2521 silver badge6 bronze badges
asked 8 hours ago
DiffycueDiffycue
2521 silver badge6 bronze badges
2521 silver badge6 bronze badges
2
$begingroup$
funny, I asked a very similar question last week, I wonder if its the same issue. Anyway, here is a possible workaround:L[-1, -2, -2] /. L[a___] :> x^(FromDigits[-a]).
$endgroup$
– AccidentalFourierTransform
8 hours ago
$begingroup$
@AccidentalFourierTransform yup it would appear to be exactly the same issue and my reaction toFullForm[-b]was exactly the same as yours lol
$endgroup$
– Diffycue
8 hours ago
add a comment |
2
$begingroup$
funny, I asked a very similar question last week, I wonder if its the same issue. Anyway, here is a possible workaround:L[-1, -2, -2] /. L[a___] :> x^(FromDigits[-a]).
$endgroup$
– AccidentalFourierTransform
8 hours ago
$begingroup$
@AccidentalFourierTransform yup it would appear to be exactly the same issue and my reaction toFullForm[-b]was exactly the same as yours lol
$endgroup$
– Diffycue
8 hours ago
2
2
$begingroup$
funny, I asked a very similar question last week, I wonder if its the same issue. Anyway, here is a possible workaround:
L[-1, -2, -2] /. L[a___] :> x^(FromDigits[-a]).$endgroup$
– AccidentalFourierTransform
8 hours ago
$begingroup$
funny, I asked a very similar question last week, I wonder if its the same issue. Anyway, here is a possible workaround:
L[-1, -2, -2] /. L[a___] :> x^(FromDigits[-a]).$endgroup$
– AccidentalFourierTransform
8 hours ago
$begingroup$
@AccidentalFourierTransform yup it would appear to be exactly the same issue and my reaction to
FullForm[-b] was exactly the same as yours lol$endgroup$
– Diffycue
8 hours ago
$begingroup$
@AccidentalFourierTransform yup it would appear to be exactly the same issue and my reaction to
FullForm[-b] was exactly the same as yours lol$endgroup$
– Diffycue
8 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Here's an explanation of why you get 4. As mentioned above, your problem is the use of Rule instead of RuleDelayed, which means the RHS of the rule evaluates before the rule is used. So, let's see what the RHS of the rule evaluates to:
-a //FullForm
List[Times[-1,a]]
So, when evaluated, the rule becomes:
L[a___] -> Times[-1, a]
Now, it should be clear what happens. When the rule is applied, a gets bound to the sequence -1, -2, -2, and so the output (before evaluation) is:
Times[-1, -1, -2, -2]
which evaluates to:
4
answered 7 hours ago
Carl WollCarl Woll
86.7k3 gold badges114 silver badges221 bronze badges
$endgroup$
add a comment |
$begingroup$
It's just a matter of using RuleDelayed, or rather :> instead of ->.
L[-1, -2, -2] /. L[a___] :> x^(FromDigits[-a])
x^122
L[-1, -2, -2] /. L[a___] :> -a
1, 2, 2
answered 8 hours ago
That Gravity GuyThat Gravity Guy
2,2311 gold badge6 silver badges15 bronze badges
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here's an explanation of why you get 4. As mentioned above, your problem is the use of Rule instead of RuleDelayed, which means the RHS of the rule evaluates before the rule is used. So, let's see what the RHS of the rule evaluates to:
-a //FullForm
List[Times[-1,a]]
So, when evaluated, the rule becomes:
L[a___] -> Times[-1, a]
Now, it should be clear what happens. When the rule is applied, a gets bound to the sequence -1, -2, -2, and so the output (before evaluation) is:
Times[-1, -1, -2, -2]
which evaluates to:
4
answered 7 hours ago
Carl WollCarl Woll
86.7k3 gold badges114 silver badges221 bronze badges
$endgroup$
add a comment |
$begingroup$
Here's an explanation of why you get 4. As mentioned above, your problem is the use of Rule instead of RuleDelayed, which means the RHS of the rule evaluates before the rule is used. So, let's see what the RHS of the rule evaluates to:
-a //FullForm
List[Times[-1,a]]
So, when evaluated, the rule becomes:
L[a___] -> Times[-1, a]
Now, it should be clear what happens. When the rule is applied, a gets bound to the sequence -1, -2, -2, and so the output (before evaluation) is:
Times[-1, -1, -2, -2]
which evaluates to:
4
answered 7 hours ago
Carl WollCarl Woll
86.7k3 gold badges114 silver badges221 bronze badges
$endgroup$
add a comment |
$begingroup$
Here's an explanation of why you get 4. As mentioned above, your problem is the use of Rule instead of RuleDelayed, which means the RHS of the rule evaluates before the rule is used. So, let's see what the RHS of the rule evaluates to:
-a //FullForm
List[Times[-1,a]]
So, when evaluated, the rule becomes:
L[a___] -> Times[-1, a]
Now, it should be clear what happens. When the rule is applied, a gets bound to the sequence -1, -2, -2, and so the output (before evaluation) is:
Times[-1, -1, -2, -2]
which evaluates to:
4
answered 7 hours ago
Carl WollCarl Woll
86.7k3 gold badges114 silver badges221 bronze badges
$endgroup$
Here's an explanation of why you get 4. As mentioned above, your problem is the use of Rule instead of RuleDelayed, which means the RHS of the rule evaluates before the rule is used. So, let's see what the RHS of the rule evaluates to:
-a //FullForm
List[Times[-1,a]]
So, when evaluated, the rule becomes:
L[a___] -> Times[-1, a]
Now, it should be clear what happens. When the rule is applied, a gets bound to the sequence -1, -2, -2, and so the output (before evaluation) is:
Times[-1, -1, -2, -2]
which evaluates to:
4
answered 7 hours ago
Carl WollCarl Woll
86.7k3 gold badges114 silver badges221 bronze badges
answered 7 hours ago
Carl WollCarl Woll
86.7k3 gold badges114 silver badges221 bronze badges
answered 7 hours ago
Carl WollCarl Woll
86.7k3 gold badges114 silver badges221 bronze badges
answered 7 hours ago
Carl WollCarl Woll
86.7k3 gold badges114 silver badges221 bronze badges
86.7k3 gold badges114 silver badges221 bronze badges
add a comment |
add a comment |
$begingroup$
It's just a matter of using RuleDelayed, or rather :> instead of ->.
L[-1, -2, -2] /. L[a___] :> x^(FromDigits[-a])
x^122
L[-1, -2, -2] /. L[a___] :> -a
1, 2, 2
answered 8 hours ago
That Gravity GuyThat Gravity Guy
2,2311 gold badge6 silver badges15 bronze badges
$endgroup$
add a comment |
$begingroup$
It's just a matter of using RuleDelayed, or rather :> instead of ->.
L[-1, -2, -2] /. L[a___] :> x^(FromDigits[-a])
x^122
L[-1, -2, -2] /. L[a___] :> -a
1, 2, 2
answered 8 hours ago
That Gravity GuyThat Gravity Guy
2,2311 gold badge6 silver badges15 bronze badges
$endgroup$
add a comment |
$begingroup$
It's just a matter of using RuleDelayed, or rather :> instead of ->.
L[-1, -2, -2] /. L[a___] :> x^(FromDigits[-a])
x^122
L[-1, -2, -2] /. L[a___] :> -a
1, 2, 2
answered 8 hours ago
That Gravity GuyThat Gravity Guy
2,2311 gold badge6 silver badges15 bronze badges
$endgroup$
It's just a matter of using RuleDelayed, or rather :> instead of ->.
L[-1, -2, -2] /. L[a___] :> x^(FromDigits[-a])
x^122
L[-1, -2, -2] /. L[a___] :> -a
1, 2, 2
answered 8 hours ago
That Gravity GuyThat Gravity Guy
2,2311 gold badge6 silver badges15 bronze badges
answered 8 hours ago
That Gravity GuyThat Gravity Guy
2,2311 gold badge6 silver badges15 bronze badges
answered 8 hours ago
That Gravity GuyThat Gravity Guy
2,2311 gold badge6 silver badges15 bronze badges
answered 8 hours ago
That Gravity GuyThat Gravity Guy
2,2311 gold badge6 silver badges15 bronze badges
2,2311 gold badge6 silver badges15 bronze badges
add a comment |
add a comment |
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2
$begingroup$
funny, I asked a very similar question last week, I wonder if its the same issue. Anyway, here is a possible workaround:
L[-1, -2, -2] /. L[a___] :> x^(FromDigits[-a]).$endgroup$
– AccidentalFourierTransform
8 hours ago
$begingroup$
@AccidentalFourierTransform yup it would appear to be exactly the same issue and my reaction to
FullForm[-b]was exactly the same as yours lol$endgroup$
– Diffycue
8 hours ago