Map object types ?

[jrhoward]

I am trying iterate over a list ( list-a ) which is ordered and in the context of iterator object retrieve the matching object from another list (list-b) to create another list (list-c) which is ordered the same as list -a. It would be much easier if I could have an Map type object for list-b and retrieve by key but I cannot find any references to map objects. I have to iterate over list-b for every item in list-a which is very inefficient.

It's likely it just because I am not well versed in functional programming. I am wondering if I am missing something and if there is a functional pattern to achieve the above efficiently.

[jrhoward]

ok found association lists in liquidsoap book.

[vitoyucepi]

Hi @jrhoward,
Do you still have a problem?

[jrhoward]

Hey @vitoyucepi
Yes, association list is not what I thought it was.
I have a solution but it seems very inefficient running list.iter inside list.iter

[vitoyucepi]

Did you find list.assoc.mem and list.assoc?
list.assoc.mem checks if a key is among the keys of the associated list.
list.assoc gets value for the given key.

[jrhoward]

I did investigate that. The example in the book uses:

m = [("artist", "Frank Sinatra"), ("title", "Fly me to the moon")]
list.assoc("title", m)

I'm looking for something more like:

m = [ { filename="...", title="...", artist="..." }, { filename="...", title="...", artist="..." } , ... ]

list.iter(  fun(filename) -> get object from m where m.filename == filename, playlist.remaining_files() )

where the new list is ordered like the remaining songs in the playlist
With a map I could do this easily because I could associate the value of filename with a key.

[toots]

Should this be a fold?

m = [ { filename="...", title="...", artist="..." }, { filename="...", title="...", artist="..." } , ... ]

filtered_m = list.fold(fun (filtered_m, el) ->
  if (m.filename == "some value") then
    [...filtered_m, el]
  else
    filtered_m
  end, [], m)

[jrhoward]

Thanks toots I have not investigated fold. Will have a look at that.

vitoyucepi, yes I have a few programming languages under my belt but only want to do that if it's not achievable in liquidsoap.
If I were to it implement outside of liquidsoap I would probably use golang.

I appreciate your help !!

[jrhoward]

regarding the playlog question, I am rendering what is left in a playlist in order using remaining_files() but accessing the files to extract metadata is quite expensive.

I run a function on_reload in a thread for the playlist to create a list of song objects. I am reusing this with remaining_files to access song information rather than reloading metadata from the file system.

this code works but there is an iterator inside the iterator
The root iterator needs to be remaining_files() to preserve order:

// song_list is the list containing all songs generated on_reload

songs = ref([])

def add_song( filename )
    list.iter( fun( song )-> ignore( string.contains(suffix=song.filename, filename )?songs:=list.add( song, !songs ): null() ),!song_list )
end

list.iter( add_song , default_playlist.remaining_files()  )

[vitoyucepi]

yes I have a few programming languages under my belt but only want to do that if it's not achievable in liquidsoap

It will be easier for me to translate a solution from another programming language than from a natural language.
For example

m = [{"filename": "/tmp/test/3.mp3", "title": "3", "artist": "3"},
     {"filename": "/tmp/test/2.mp3", "title": "2", "artist": "2"}]

playlist_remaining_files = ["/tmp/test/1.mp3", "/tmp/test/2.mp3", "/tmp/test/3.mp3", "/tmp/test/4.mp3"]

for filename in playlist_remaining_files:
    for item in m:
        if item["filename"] == filename:
            print(item)

will translate into

m = [{filename = "/tmp/test/3.mp3", title = "3", artist = "3"},
     {filename = "/tmp/test/2.mp3", title = "2", artist = "2"}]
p = playlist("/tmp/test", mode="normal")

for filename = list.iterator(p.remaining_files()) do
  for item = list.iterator(m) do
    if item.filename == filename then
      print(item)
    end
  end
end

For me, it's not clear what are you planning to do.
Maybe it'll be better to store m as

m = [
  ("/tmp/test/3.mp3", {filename = "/tmp/test/3.mp3", title = "3", artist = "3"}),
  ("/tmp/test/2.mp3", {filename = "/tmp/test/2.mp3", title = "2", artist = "2"}),
]

And then

if list.assoc.mem(filename, m) then
  list.assoc(filename, m)
end

[jrhoward]

thanks @vitoyucepi, that will work.
This structure

m = [
  ("/tmp/test/3.mp3", {filename = "/tmp/test/3.mp3", title = "3", artist = "3"}),
  ("/tmp/test/2.mp3", {filename = "/tmp/test/2.mp3", title = "2", artist = "2"}),
]

I was under the impression that list.assoc only worked with (string,string)