python - 如何在 Python 中访问由嵌套 defaultdict 创建的矩阵的一行？

Question

我正在 Python 中创建一个单词共现矩阵，并使用嵌套的 defaultdicts 来创建该矩阵。我已经成功创建了矩阵并存储了字数，但是现在在尝试从嵌套的默认字典中获取向量(矩阵行)时遇到了麻烦。

这是我用来初始化矩阵的代码行:

matrix = collections.defaultdict(lambda: collections.defaultdict(int))

以下是我用来将字数统计放入矩阵中的行:

matrix[target_word_id][collocated_word_id] += 1

matrix[collocated_word_id][target_word_id] += 1

这就是我尝试访问矩阵中与给定单词 id 对应的行的方式:

vector1 = matrix[word1_id]

当我打印 vector1 来测试我的工作时，这就是我得到的输出:

defaultdict(<class 'int'>, {})

该类的完整代码在这里。我从一个单独的主类调用这些函数:

class Create_vector():

    def build_vocab(self, corpus):
        vocab = collections.defaultdict(int)
        i = 1

        for line in corpus:
            token = line.strip()
            if token not in vocab:
                vocab[token] = i
                i += 1

        return vocab



    def build_cooccurrence(self, corpus, vocab, window):

        matrix = collections.defaultdict(lambda: collections.defaultdict(int))

        for x, line in enumerate(corpus):

            if x % 100000 == 0:
                print('Building cooccurrence matrix: on line %i', x)
            tokens = line.strip()
            token_ids = [vocab[token] for token in tokens]

            for i, target_word_id in enumerate(token_ids):

                collocated_word_ids = token_ids[min(0, target_word_id - window): target_word_id]

                for j, collocated_word_id in enumerate(collocated_word_ids):

                    matrix[target_word_id][collocated_word_id] += 1

                    matrix[collocated_word_id][target_word_id] += 1

        return matrix


    def get_vector(self, matrix, vocab, weight, word1, word2):

        if weight == 'FREQ':

            if word1 in vocab:
                word1_id = vocab[word1]
                vector1 = matrix[word1_id]
                pprint.pprint(vector1)

主类在这里:

import nltk
    import sys
    from nltk.corpus import stopwords
    import create_vector
    import pprint
    import string


def main():
    brown_words = list(nltk.corpus.brown.words())
    window = int(sys.argv[1])
    weight = sys.argv[2]
    brown_words_lower = [word.lower() for word in brown_words]
    brown_words_only = [w for w in brown_words_lower if w not in string.punctuation]
    stops = set(stopwords.words('english'))
    brown_words_filtered = [w for w in brown_words_only if w not in stops]

    vector = create_vector.Create_vector()

    vocab = vector.build_vocab(brown_words_filtered)
    cooccurrence = vector.build_cooccurrence(brown_words_filtered, vocab, window)

    for line in text:
        words = line.split(',')
        word1 = words[0]
        word2 = words[1]
        vector1, vector2 = vector.get_vector(cooccurrence, vocab, weight, word1, word2)

运行它的命令是:python3.4 main.py 2 频率

Answer 1

无法重现:

>>> import collections
>>> matrix = collections.defaultdict(lambda: collections.defaultdict(int))
>>> matrix[2][3] += 1
>>> matrix[3][2] += 1
>>> vector1 = matrix[2]
>>> vector1
defaultdict(<type 'int'>, {3: 1})

您确定 word1_id 中的值等于已插入矩阵中的值吗？您能否发布完整的代码，而不仅仅是其中的片段？