JupiterIO1
Ultimately I’d like to update the values of keys in myMap based on whether the key is in myList and in myOtherList:
myList = ["a", "b", "c"]
myOtherList = ["b"]
myMap = %{"a" => "no", "b" => "no", "c" => "no"}
Enum.each(myList, fn e ->
case Enum.member?(myOtherList, e) do
true -> Map.replace(myMap, e, "yes")
false -> Map.replace(myMap, e, "no")
end
end)
myMap doesn’t update.
I believe the problem would be because of immutability, but I still cannot arrive to a solution. I’m open to other Map function suggestions.
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jgonet
I suggest using if else for this. Also, to mutate your map you have to use explicit assignment, so
my_map = if (...).Oh, and in Elixir
snake_caseis the most widely used convention.sanswork
Check out Enum.reduce
peerreynders
NobbZ
Will
myListalways be identical toMap.keys(myMap)? Or will they diverge?Will
Map.values(myMap)always be identical to a list of"no"? Or might there be other values?In the following solution, I will assume, that
myListandMap.keys(myMap)might be different, while the values all have to be"no".The reason why your original code did not work is, because the inner binding in the
fnyou pass toEnum.each/2does not change the outer binding ofmyMap, always remember, elixir is immutable, it does only support shadowing, but shadowing does never leak its scope.peerreynders
To me it seems peculiar that the original code completely ignores all the values that are already in the Map - what is the point of having a Map then?
JupiterIO1
Let me be more specific by redefining the problem:
For each key in
map_A(we don’t care about the values) get another known map (for example sake:map_B) which is referenced by the key atmap_Aand test whether any of the keys inmap_Bmatches some variablex. If match, update value at key inmap_Ato"yes", else"no". And this is the problem.The problem exists, as NobbZ said:
and I assume in this case that any iterator will not allow this. That goes for
Enum.each/2and:I cannot see further than this:
peerreynders
There is no “iterator”.
foris a comprehension. Simply speaking it consumes an enumerable and produces a list (or whatever else:intoidentifies) - it’s an expression like everything else in Elixir, not a statement.JupiterIO1
The solution was solved in another topic:
peerreynders
Stop thinking of “variables”- all values in Elixir are immutable. However identifiers can be rebound to new values. So the only thing that can “vary” is the value the identifier is bound to - not the value itself.
Imperative languages are about PLace-Oriented Programming (PLOP) - in functional programming you are programming with values (The Value of Values).
Furthermore you don’t “loop” in Elixir - you recurse.
JupiterIO1
You are right. As much as I have learnt and understood this concept, I should be more specific with my language. ty