Python建模树

Question

I have rethought my entire problem. I have simplified it down, and now I understand how I can explain it.

I need to express search terms along with a metric in a tree.

As an example here is some input

phrases = (
    ("removals lisbon", 1),
    ("moving to india", 3),
    ("moving to indonesia", 1),
    ("removals dublin", 3),
    ("moving to malta", 45),
    ("move to brazil", 2),
    ("moving chicago", 1),
    ("moving to california", 29),
    ("moving to brussels", 4),
    ("moving to bangladesh", 1),
    ("removals california", 2),
    ("moving from spain", 4),
    ("moving to russia", 3),
    ("move to los angeles", 2),
    ("move to germany", 1),
    ("moving to poalnd", 1),
    ("removals stockholm", 1),
    ("removal to poland", 1),
    ("moves uk", 7),
    ("moving hamburger", 1),
    ("move to malta", 8),
    ("move to london", 1),
    ("moving from cyprus", 1),
    ("move to japan", 5)
)

Starting with the most common word, the most common word is to , taking this word, we can build a tree of all the phrases containing to , we do this by finding all the phrases that contain to , then (ignoring to itself) we simply add a child which is all the words that make up those phrases and their associated score. We then pick the most popular word of the children and repeat then we end up with another set of children not containing to or from , we then keep going down until we run out of depth , we then climb back up and go down another branch.

This will give a structure looking something like

to
├── move
│   └── city
├── moves
│   └── city
├── moving
│   └── city
└── removals
    └── city

And so on.

Once I have this tree I can display it and its all fine.

I started work on this with the following modified code

def count(phrases, ignore=()):
    counter = Counter()
    for phrase, _ in phrases:
        for word in phrase.split(" "):
            if word not in ignore:
                counter[word] += 1
    return counter


def filter_word(word, phrases):
    for phrase, count in phrases:
        if word in phrase.split(" "):
            yield phrase, count


class Node(object):
    def __init__(self, word, clicks):
        self.word = word
        self.clicks = clicks

        self.children = []
        self.unprocessed_children = []


def build_tree(pair, phrases, depth, lower_score):
    word, clicks = pair
    root = current = Node(*pair)
    visited = [word]

    phrases = filter_word(word, phrases)
    for phrase, clicks in phrases:
        for word in phrase.split(" "):
            if word in visited:
                continue
            visited.append(word)
            root.unprocessed_children.append(Node(word, clicks))


def identify_root(phrases, depth, lower_score, ignore=()):
    words = count(phrases, ignore=ignore).most_common()
    print words[0:10]
    trees = []
    for root in words:
        trees.append(build_tree(root, phrases, depth, lower_score))
        return
    return trees

But I am lost in build_tree as to actually go down and create more children upto depth.

Answer 1

If I understood your meaning, this recursive code should do it.

def build_node(word, clicks, phrases, depth, ignore):        
    node = Node(word, clicks)
    node.children = build_children(list(filter_word(word, phrases)), 
        depth-1, 
        ignore + (word,))
    return node

def build_children(phrases, depth, ignore):
    if depth > 0:
        words = count(phrases, ignore=ignore).most_common()
        return [build_node(word, clicks, phrases, depth, ignore) 
                for word, clicks in words]
    else:
        return []

def identify_root(phrases, depth, ignore=()):
    return build_children(phrases, depth, ignore)