Thank you Arnoud. Your answer is very informative. Since (awkwardly) posting my question, I had looked more at the documentation regarding Associations and had come up with a better understanding of how association assignment worked. Your explanation cemented for me the understanding I had gleaned from reading the documentation. I also commend you on being able to answer an admittedly poorly asked question,

Hi Henrick,

I am going to take a stab at this, but your question is written in a very complicated way (try to ask a smaller simpler question, if at all possible; that way people may be able to help you quicker).

Let's start with an association (note that you can type four spaces at the beginning of a line to make it look like code):

x = <| "a" -> 1, "b" -> 2, "c" -> 3 |>

And a function which gets called with an association:

f[y_Association] := y

Now I can call f with x:

f[x]

And this returns the association:

<| "a" -> 1, "b" -> 2, "c" -> 3 |>

OK, so far, so good. Not let's say you want to return an association but with a modified key or value. First, let's do it the wrong way:

f[y_Association] := ( y["a"] = 5; y)

If you do this and call it with:

f[x]

you get an error:

(Association::setps) <|a->3,b->2,c->3|> in the part assignment is not a symbol.

The reason for that is that y is not a variable but the expression that you passed in. So this:

f[x]

becomes:

f[<| "a" -> 1, "b" -> 2, "c" -> 3 |>]

and the code on the right hand side becomes:

( <| "a" -> 1, "b" -> 2, "c" -> 3 |>["a"] = 5; <| "a" -> 1, "b" -> 2, "c" -> 3 |>)

which leads to the error because you can not assign to an association.

To fix this you need to use a Module with a temporary variable to which you assign your association:

f[y_Association] := Module[ {z=y}, z["a"]=5; z ]

Now if you call it with x:

f[x]

you get no error and a plausible useful result:

<|"a" -> 5, "b" -> 2, "c" -> 3|>

If you want to add a key, you can do something like this:

f[y_Association] := Module[{z = y}, z["d"] = 5; z]

If you want to drop a key you can use either:

f[y_Association] := KeyDrop[y, "a"]

or:

f[y_Association] := Module[{z = y}, KeyDrop[z, "a"]]

(Module is not needed here, since you're not trying to assign to an expression in this case).

Finally, if you want your original x to be modified by f you simply assign to it:

x = f[x]

So here, f is called with x, uses a temporary variable to modify the expression, returns the result, and assigns that result to x.

I think the last bit might solve the second part of your question.

Hope this helps.

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