Assemble separate sentences connected by dictionary attributes-CodePudding

I have the following list of dictionaries and I'm trying to come up with a single connected sentence based on whether the fact that the dictionary of the child sentences have the "ref" attribute connecting it to the father sentence.

clause_list = [
    {"id": "T1", "text": "hi"},
    {"id": "T2", "text": "I'm", "ref": "T1"},
    {"id": "T3", "text": "Simone", "ref": "T2"},
]

Expected output is "hi I'm Simone" but avoiding sentences like "hi I'm" or "I'm Simone"

What I tried so far is the following, but no matter how I flip it, the undesired sentences always get printed.

for c in clause_list:
  for child in clause_list:
    try:
      if c["id"] == child["ref"] and "ref" not in c.keys():
        print(c["text"], child["text"])
      elif c["id"] == child["ref"] and "ref" in c.keys():
        for father in clause_list:
          if father["id"] == c["ref"] and "ref" not in father.keys():
            print(father["text"], c["text"], child["text"])
    except KeyError:
      pass

CodePudding user response：

So right now you have your data set up as a list of dictionaries. I think it will be more helpful if you set it up as a linked list instead and use a bit of recursion. Maybe something like below. As a note, you'll need something to indicate the start of the sentence (I'm going to use "parent=True"):

clause_list = {
  # ID's are the keys
  'T1':{'text':"hi ", 'child':'T2','parent':True}, 
  'T2':{'text':"I'm ", 'child':'T3'},
  'T3':{'text':"Simone", 'child':None},
}

Then your code can look like this:

# Recursive function
def print_child(child_id, clause_list):
  print(clause_list[child_id]['text'])
  if 'child' in clause_list[child_id]:
    print_child(clause_list[child_id]['child'], clause_list)

# Main function
for id in clause_list:
  if 'parent' in clause_list[id]:
    print(clause_list[id]['text'])
    # Recurse through the parent to find others
    if 'child' in clause_list[id]:
      print_child(clause_list[id]['child'], clause_list)

Cloud definitely be cleaned up a bit and optimized, but this is the general gist. Hope it helps!

CodePudding user response：

probablly is best use a class and not a dict, but you could convert to list of dict to list of classes easy. This work. Convert the list of dict to list of Clauses. and the clauses has the method search_ref that print the partial text if there are no referenced object, or add the referenced object and continue if there are. if you have 2 objects i don't know exactly what you want

clause_list = [
    {"id": "T1", "text": "hi"},
    {"id": "T2", "text": "I'm", "ref": "T1"},
    {"id": "T3", "text": "Simone", "ref": "T2"},
]
class Clause:
    def __init__(self, id, text, ref:None):
        self.id = id
        self.text = text
        self.ref = ref
    
    def search_ref(self, Clauses, text=''):
        parcialText = text   ' '   self.text
        for clause in Clauses:
            if clause.ref == self.id:
                return clause.search_ref(Clauses, parcialText)
        print(parcialText)
    
Clauses = [Clause(id=c['id'], text=c['text'], ref=c.get('ref')) for c in clause_list]

for c in Clauses:
    if c.ref is None:
        c.search_ref(Clauses)

CodePudding user response：

Another solution using a recursion:

clause_list = [
    {"id": "T1", "text": "hi"},
    {"id": "T2", "text": "I'm", "ref": "T1"},
    {"id": "T3", "text": "Simone", "ref": "T2"},
]


inv_dict = {d.get("ref", ""): (d["id"], d["text"]) for d in clause_list}


def get_sentence(inv_dict, key=""):
    if key in inv_dict:
        id_, text = inv_dict[key]
        return [text]   get_sentence(inv_dict, id_)
    return []


print(" ".join(get_sentence(inv_dict)))

Prints:

hi I'm Simone