How can I update an association with a function?

up vote
3
down vote

favorite

I'm not sure if this is a bug, or if I just am not understanding, properly, how to create a function.

I am able to use a list of rules to update an association like so:

AssociateTo[pools, asocs]

Where pools is:

<|p10->-0.271069,p11->-0.748554,p12->-0.772571,p13->0.801179,p14->0.823549,p15->-0.617603,p16->-0.171479,p17->0.860391,p18->-0.0937823,p19->0.142897|>

And asocs is:

p12 -> -1.0208, p11 -> -1.09988*10^7, p15 -> -0.0190685, 
 p19 -> 0.075402, p10 -> -0.0904602, p16 -> -0.147492, 
 p17 -> 0.353486, p13 -> 869.624

However, when I try to use this function:

updatePoolwithValues[a_, newVals_] := AssociateTo[a, newVals];

I get the following error message:

<|p10->-0.271069,p11->-0.748554,p12->-0.772571,p13->0.801179,p14->0.823549,p15->-0.617603,p16->-0.171479,p17->0.860391,p18->-0.0937823,p19->0.142897|> is not a variable with a value, so its value cannot be changed.

Is this an error, or is there something I am not understanding?

Thanks!

asked 4 hours ago

Jmeeks29ig

34818

add a commentÂ |Â

up vote
3
down vote

favorite

I'm not sure if this is a bug, or if I just am not understanding, properly, how to create a function.

I am able to use a list of rules to update an association like so:

AssociateTo[pools, asocs]

Where pools is:

<|p10->-0.271069,p11->-0.748554,p12->-0.772571,p13->0.801179,p14->0.823549,p15->-0.617603,p16->-0.171479,p17->0.860391,p18->-0.0937823,p19->0.142897|>

And asocs is:

p12 -> -1.0208, p11 -> -1.09988*10^7, p15 -> -0.0190685, 
 p19 -> 0.075402, p10 -> -0.0904602, p16 -> -0.147492, 
 p17 -> 0.353486, p13 -> 869.624

However, when I try to use this function:

updatePoolwithValues[a_, newVals_] := AssociateTo[a, newVals];

I get the following error message:

<|p10->-0.271069,p11->-0.748554,p12->-0.772571,p13->0.801179,p14->0.823549,p15->-0.617603,p16->-0.171479,p17->0.860391,p18->-0.0937823,p19->0.142897|> is not a variable with a value, so its value cannot be changed.

Is this an error, or is there something I am not understanding?

Thanks!

asked 4 hours ago

Jmeeks29ig

34818

add a commentÂ |Â

up vote
3
down vote

favorite

I'm not sure if this is a bug, or if I just am not understanding, properly, how to create a function.

I am able to use a list of rules to update an association like so:

AssociateTo[pools, asocs]

Where pools is:

<|p10->-0.271069,p11->-0.748554,p12->-0.772571,p13->0.801179,p14->0.823549,p15->-0.617603,p16->-0.171479,p17->0.860391,p18->-0.0937823,p19->0.142897|>

And asocs is:

p12 -> -1.0208, p11 -> -1.09988*10^7, p15 -> -0.0190685, 
 p19 -> 0.075402, p10 -> -0.0904602, p16 -> -0.147492, 
 p17 -> 0.353486, p13 -> 869.624

However, when I try to use this function:

updatePoolwithValues[a_, newVals_] := AssociateTo[a, newVals];

I get the following error message:

<|p10->-0.271069,p11->-0.748554,p12->-0.772571,p13->0.801179,p14->0.823549,p15->-0.617603,p16->-0.171479,p17->0.860391,p18->-0.0937823,p19->0.142897|> is not a variable with a value, so its value cannot be changed.

Is this an error, or is there something I am not understanding?

Thanks!

asked 4 hours ago

Jmeeks29ig

34818

I'm not sure if this is a bug, or if I just am not understanding, properly, how to create a function.

I am able to use a list of rules to update an association like so:

AssociateTo[pools, asocs]

Where pools is:

<|p10->-0.271069,p11->-0.748554,p12->-0.772571,p13->0.801179,p14->0.823549,p15->-0.617603,p16->-0.171479,p17->0.860391,p18->-0.0937823,p19->0.142897|>

And asocs is:

p12 -> -1.0208, p11 -> -1.09988*10^7, p15 -> -0.0190685, 
 p19 -> 0.075402, p10 -> -0.0904602, p16 -> -0.147492, 
 p17 -> 0.353486, p13 -> 869.624

However, when I try to use this function:

updatePoolwithValues[a_, newVals_] := AssociateTo[a, newVals];

I get the following error message:

<|p10->-0.271069,p11->-0.748554,p12->-0.772571,p13->0.801179,p14->0.823549,p15->-0.617603,p16->-0.171479,p17->0.860391,p18->-0.0937823,p19->0.142897|> is not a variable with a value, so its value cannot be changed.

Is this an error, or is there something I am not understanding?

Thanks!

functions associations

asked 4 hours ago

Jmeeks29ig

34818

asked 4 hours ago

Jmeeks29ig

34818

asked 4 hours ago

Jmeeks29ig

34818

asked 4 hours ago

Jmeeks29ig

34818

asked 4 hours ago

Jmeeks29ig

34818

add a commentÂ |Â

1 Answer
1

active

oldest

votes

up vote
4
down vote

accepted

I am going to use simpler numbers and a shorter set of rules and lists to make it easier to focus on the problem.

pool = <|p10 -> 10, p11 -> 11, p12 -> 12, p13 -> 13, p14 -> 14, 
 p15 -> 15|>;

assoc = p12 -> -12, p11 -> -11, p16 -> -16;

AssociateTo[pool, assoc]

(* <|p10 -> 10, p13 -> 13, p14 -> 14, p15 -> 15, p12 -> -12, 
 p11 -> -11, p16 -> -16|> *)

After execution check that the symbol pool has been changed.

pool

(* <|p10 -> 10, p13 -> 13, p14 -> 14, p15 -> 15, p12 -> -12, 
 p11 -> -11, p16 -> -16|> *)

AssociateTo has the attribute HoldFirst

Attributes[AssociateTo]

(* HoldFirst, Protected *)

Now let's define your function. I am going to use a slightly different name to clarify what we are attempting to do:

myAssociateTo[a_, newVals_] := AssociateTo[a, newVals]

If you apply this to pool and assoc you will indeed get an error.

myAssociateTo[pool, assoc]

During evaluation of AssociateTo::rvalue: <|p10->10,p11->11,p12->12,
p13->13,p14->14,p15->15|> is not a variable with a value, so its value
cannot be changed.

Let's give myAssociateTo the same attribute as AssociateTo

SetAttributes[myAssociateTo, HoldFirst]

Now we reset pool and try it.

pool = <|p10 -> 10, p11 -> 11, p12 -> 12, p13 -> 13, p14 -> 14, 
 p15 -> 15|>;

myAssociateTo[pool, assoc]

(* <|p10 -> 10, p11 -> -11, p12 -> -12, p13 -> 13, p14 -> 14, 
 p15 -> 15, p16 -> -16|> *)

The desired result has been achieved.

Note further that the symbol pool has been equated to the new value.

pool

(* <|p10 -> 10, p11 -> -11, p12 -> -12, p13 -> 13, p14 -> 14, 
 p15 -> 15, p16 -> -16|> *)

edited 3 hours ago

answered 4 hours ago

Jack LaVigne

11.2k21329

Thank you Carl. Corrected the typo.
â€“Â Jack LaVigne
3 hours ago

I am wondering if I can ask about the necessity to rename a built-in function.
â€“Â ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³
2 hours ago

@ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³ because you want to change its Attributes ?
â€“Â chris
29 mins ago

@Jack LaVigne, Thanks, I really appreciate that! It makes a lot more sense to me now
â€“Â Jmeeks29ig
13 mins ago

add a commentÂ |Â

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
4
down vote

accepted

I am going to use simpler numbers and a shorter set of rules and lists to make it easier to focus on the problem.

pool = <|p10 -> 10, p11 -> 11, p12 -> 12, p13 -> 13, p14 -> 14, 
 p15 -> 15|>;

assoc = p12 -> -12, p11 -> -11, p16 -> -16;

AssociateTo[pool, assoc]

(* <|p10 -> 10, p13 -> 13, p14 -> 14, p15 -> 15, p12 -> -12, 
 p11 -> -11, p16 -> -16|> *)

After execution check that the symbol pool has been changed.

pool

(* <|p10 -> 10, p13 -> 13, p14 -> 14, p15 -> 15, p12 -> -12, 
 p11 -> -11, p16 -> -16|> *)

AssociateTo has the attribute HoldFirst

Attributes[AssociateTo]

(* HoldFirst, Protected *)

Now let's define your function. I am going to use a slightly different name to clarify what we are attempting to do:

myAssociateTo[a_, newVals_] := AssociateTo[a, newVals]

If you apply this to pool and assoc you will indeed get an error.

myAssociateTo[pool, assoc]

During evaluation of AssociateTo::rvalue: <|p10->10,p11->11,p12->12,
p13->13,p14->14,p15->15|> is not a variable with a value, so its value
cannot be changed.

Let's give myAssociateTo the same attribute as AssociateTo

SetAttributes[myAssociateTo, HoldFirst]

Now we reset pool and try it.

pool = <|p10 -> 10, p11 -> 11, p12 -> 12, p13 -> 13, p14 -> 14, 
 p15 -> 15|>;

myAssociateTo[pool, assoc]

(* <|p10 -> 10, p11 -> -11, p12 -> -12, p13 -> 13, p14 -> 14, 
 p15 -> 15, p16 -> -16|> *)

The desired result has been achieved.

Note further that the symbol pool has been equated to the new value.

pool

(* <|p10 -> 10, p11 -> -11, p12 -> -12, p13 -> 13, p14 -> 14, 
 p15 -> 15, p16 -> -16|> *)

edited 3 hours ago

answered 4 hours ago

Jack LaVigne

11.2k21329

Thank you Carl. Corrected the typo.
â€“Â Jack LaVigne
3 hours ago

I am wondering if I can ask about the necessity to rename a built-in function.
â€“Â ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³
2 hours ago

@ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³ because you want to change its Attributes ?
â€“Â chris
29 mins ago

@Jack LaVigne, Thanks, I really appreciate that! It makes a lot more sense to me now
â€“Â Jmeeks29ig
13 mins ago

add a commentÂ |Â

up vote
4
down vote

accepted

I am going to use simpler numbers and a shorter set of rules and lists to make it easier to focus on the problem.

pool = <|p10 -> 10, p11 -> 11, p12 -> 12, p13 -> 13, p14 -> 14, 
 p15 -> 15|>;

assoc = p12 -> -12, p11 -> -11, p16 -> -16;

AssociateTo[pool, assoc]

(* <|p10 -> 10, p13 -> 13, p14 -> 14, p15 -> 15, p12 -> -12, 
 p11 -> -11, p16 -> -16|> *)

After execution check that the symbol pool has been changed.

pool

(* <|p10 -> 10, p13 -> 13, p14 -> 14, p15 -> 15, p12 -> -12, 
 p11 -> -11, p16 -> -16|> *)

AssociateTo has the attribute HoldFirst

Attributes[AssociateTo]

(* HoldFirst, Protected *)

Now let's define your function. I am going to use a slightly different name to clarify what we are attempting to do:

myAssociateTo[a_, newVals_] := AssociateTo[a, newVals]

If you apply this to pool and assoc you will indeed get an error.

myAssociateTo[pool, assoc]

During evaluation of AssociateTo::rvalue: <|p10->10,p11->11,p12->12,
p13->13,p14->14,p15->15|> is not a variable with a value, so its value
cannot be changed.

Let's give myAssociateTo the same attribute as AssociateTo

SetAttributes[myAssociateTo, HoldFirst]

Now we reset pool and try it.

pool = <|p10 -> 10, p11 -> 11, p12 -> 12, p13 -> 13, p14 -> 14, 
 p15 -> 15|>;

myAssociateTo[pool, assoc]

(* <|p10 -> 10, p11 -> -11, p12 -> -12, p13 -> 13, p14 -> 14, 
 p15 -> 15, p16 -> -16|> *)

The desired result has been achieved.

Note further that the symbol pool has been equated to the new value.

pool

(* <|p10 -> 10, p11 -> -11, p12 -> -12, p13 -> 13, p14 -> 14, 
 p15 -> 15, p16 -> -16|> *)

edited 3 hours ago

answered 4 hours ago

Jack LaVigne

11.2k21329

Thank you Carl. Corrected the typo.
â€“Â Jack LaVigne
3 hours ago

I am wondering if I can ask about the necessity to rename a built-in function.
â€“Â ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³
2 hours ago

@ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³ because you want to change its Attributes ?
â€“Â chris
29 mins ago

@Jack LaVigne, Thanks, I really appreciate that! It makes a lot more sense to me now
â€“Â Jmeeks29ig
13 mins ago

add a commentÂ |Â

up vote
4
down vote

accepted

I am going to use simpler numbers and a shorter set of rules and lists to make it easier to focus on the problem.

pool = <|p10 -> 10, p11 -> 11, p12 -> 12, p13 -> 13, p14 -> 14, 
 p15 -> 15|>;

assoc = p12 -> -12, p11 -> -11, p16 -> -16;

AssociateTo[pool, assoc]

(* <|p10 -> 10, p13 -> 13, p14 -> 14, p15 -> 15, p12 -> -12, 
 p11 -> -11, p16 -> -16|> *)

After execution check that the symbol pool has been changed.

pool

(* <|p10 -> 10, p13 -> 13, p14 -> 14, p15 -> 15, p12 -> -12, 
 p11 -> -11, p16 -> -16|> *)

AssociateTo has the attribute HoldFirst

Attributes[AssociateTo]

(* HoldFirst, Protected *)

Now let's define your function. I am going to use a slightly different name to clarify what we are attempting to do:

myAssociateTo[a_, newVals_] := AssociateTo[a, newVals]

If you apply this to pool and assoc you will indeed get an error.

myAssociateTo[pool, assoc]

During evaluation of AssociateTo::rvalue: <|p10->10,p11->11,p12->12,
p13->13,p14->14,p15->15|> is not a variable with a value, so its value
cannot be changed.

Let's give myAssociateTo the same attribute as AssociateTo

SetAttributes[myAssociateTo, HoldFirst]

Now we reset pool and try it.

pool = <|p10 -> 10, p11 -> 11, p12 -> 12, p13 -> 13, p14 -> 14, 
 p15 -> 15|>;

myAssociateTo[pool, assoc]

(* <|p10 -> 10, p11 -> -11, p12 -> -12, p13 -> 13, p14 -> 14, 
 p15 -> 15, p16 -> -16|> *)

The desired result has been achieved.

Note further that the symbol pool has been equated to the new value.

pool

(* <|p10 -> 10, p11 -> -11, p12 -> -12, p13 -> 13, p14 -> 14, 
 p15 -> 15, p16 -> -16|> *)

edited 3 hours ago

answered 4 hours ago

Jack LaVigne

11.2k21329

I am going to use simpler numbers and a shorter set of rules and lists to make it easier to focus on the problem.

pool = <|p10 -> 10, p11 -> 11, p12 -> 12, p13 -> 13, p14 -> 14, 
 p15 -> 15|>;

assoc = p12 -> -12, p11 -> -11, p16 -> -16;

AssociateTo[pool, assoc]

(* <|p10 -> 10, p13 -> 13, p14 -> 14, p15 -> 15, p12 -> -12, 
 p11 -> -11, p16 -> -16|> *)

After execution check that the symbol pool has been changed.

pool

(* <|p10 -> 10, p13 -> 13, p14 -> 14, p15 -> 15, p12 -> -12, 
 p11 -> -11, p16 -> -16|> *)

AssociateTo has the attribute HoldFirst

Attributes[AssociateTo]

(* HoldFirst, Protected *)

Now let's define your function. I am going to use a slightly different name to clarify what we are attempting to do:

myAssociateTo[a_, newVals_] := AssociateTo[a, newVals]

If you apply this to pool and assoc you will indeed get an error.

myAssociateTo[pool, assoc]

During evaluation of AssociateTo::rvalue: <|p10->10,p11->11,p12->12,
p13->13,p14->14,p15->15|> is not a variable with a value, so its value
cannot be changed.

Let's give myAssociateTo the same attribute as AssociateTo

SetAttributes[myAssociateTo, HoldFirst]

Now we reset pool and try it.

pool = <|p10 -> 10, p11 -> 11, p12 -> 12, p13 -> 13, p14 -> 14, 
 p15 -> 15|>;

myAssociateTo[pool, assoc]

(* <|p10 -> 10, p11 -> -11, p12 -> -12, p13 -> 13, p14 -> 14, 
 p15 -> 15, p16 -> -16|> *)

The desired result has been achieved.

Note further that the symbol pool has been equated to the new value.

pool

(* <|p10 -> 10, p11 -> -11, p12 -> -12, p13 -> 13, p14 -> 14, 
 p15 -> 15, p16 -> -16|> *)

edited 3 hours ago

answered 4 hours ago

Jack LaVigne

11.2k21329

edited 3 hours ago

answered 4 hours ago

Jack LaVigne

11.2k21329

answered 4 hours ago

Jack LaVigne

11.2k21329

answered 4 hours ago

Jack LaVigne

11.2k21329

Thank you Carl. Corrected the typo.
â€“Â Jack LaVigne
3 hours ago

I am wondering if I can ask about the necessity to rename a built-in function.
â€“Â ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³
2 hours ago

@ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³ because you want to change its Attributes ?
â€“Â chris
29 mins ago

@Jack LaVigne, Thanks, I really appreciate that! It makes a lot more sense to me now
â€“Â Jmeeks29ig
13 mins ago

add a commentÂ |Â

Thank you Carl. Corrected the typo.
â€“Â Jack LaVigne
3 hours ago

I am wondering if I can ask about the necessity to rename a built-in function.
â€“Â ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³
2 hours ago

@ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³ because you want to change its Attributes ?
â€“Â chris
29 mins ago

@Jack LaVigne, Thanks, I really appreciate that! It makes a lot more sense to me now
â€“Â Jmeeks29ig
13 mins ago

Thank you Carl. Corrected the typo.
â€“Â Jack LaVigne
3 hours ago

I am wondering if I can ask about the necessity to rename a built-in function.
â€“Â ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³
2 hours ago

@ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³ because you want to change its Attributes ?
â€“Â chris
29 mins ago

@Jack LaVigne, Thanks, I really appreciate that! It makes a lot more sense to me now
â€“Â Jmeeks29ig
13 mins ago

add a commentÂ |Â

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