this doesn’t do what you would think, though:

iex(4)> match?(a, %{x: 5})
true

My gut pick is this one:

subset? =
  fn (map1, map2) ->
    in_other =
      fn {key, value} ->
        {:ok, value} == Map.fetch(map2, key)
      end
    map1
    |> Map.to_list()
    |> Enum.all?(in_other)
  end

as it simply verifies that all the necessary key/value pairs are present and will stop as soon as that is not the case.

In this case I also prefer to use fetch/2 in order the leverage the underlying map functionality rather than relying on in/2’s macro magic even if it makes the code a bit more verbose (so be it).

One thing to be aware of is that the empty set is a subset of any other set - the same is true for empty maps.

Do not pin here:

iex(1)> a = %{a: 1}
iex(2)> b = %{a: 1, b: 2}
iex(3)> match?(a, b)
true

Now I see my code is actually the same as @peerreynders second suggestion but in only one line.

If you want also, his last suggestion can be written as:

map1 = %{a: 1}
map2 = %{a: 1, b: 2}

map1
|> Map.to_list()
|> Enum.all?(&(&1 in map2))

But I still think MapSet.subset?/2 is clearer, and as you can see in its code, looks like it does pretty much the same thing.

Now that @axelson has a workaround for it, I want to bring a little discussion: shouldn’t his example work? I mean this one:

map1 = %{a: 1}
map2 = %{a: 1, b: 2}

^map1 = map2

I totally expected it to work. Didn’t you?​

I mean, if we think logically: if %{a: 1} = b works and
a = %{a: 1}, then ^a = b should work.

Do any of you know why it doesn’t? Is there a specific reason for it?