Python中单词列表中的共现矩阵

Question

我有一个名单列表,如:

names = ['A', 'B', 'C', 'D']

和文件清单,在每个文件中提到了一些这些名称.

document =[['A', 'B'], ['C', 'B', 'K'],['A', 'B', 'C', 'D', 'Z']]

我想得到一个输出作为共现矩阵,如:

  A  B  C  D
A 0  2  1  1
B 2  0  2  1
C 1  2  0  1
D 1  1  1  0

在R中有一个针对这个问题的解决方案(创建共生矩阵),但我无法在Python中实现.我想在熊猫中做到这一点,但还没有进展!

Answer 1

另一种选择是使用构造 csr_matrix((data, (row_ind, col_ind)), [shape=(M, N)])从scipy.sparse.csr_matrix那里data，row_ind并col_ind满足关系a[row_ind[k], col_ind[k]] = data[k]。

诀窍是生成row_ind并col_ind遍历文档并创建元组列表（doc_id，word_id）。data将仅仅是相同长度向量的向量。

将docs-words矩阵与其转置相乘会得到共现矩阵。

此外，这在运行时间和内存使用方面都是高效的，因此它还应该处理大型的程序集。

import numpy as np
import itertools
from scipy.sparse import csr_matrix


def create_co_occurences_matrix(allowed_words, documents):
    print(f"allowed_words:\n{allowed_words}")
    print(f"documents:\n{documents}")
    word_to_id = dict(zip(allowed_words, range(len(allowed_words))))
    documents_as_ids = [np.sort([word_to_id[w] for w in doc if w in word_to_id]).astype('uint32') for doc in documents]
    row_ind, col_ind = zip(*itertools.chain(*[[(i, w) for w in doc] for i, doc in enumerate(documents_as_ids)]))
    data = np.ones(len(row_ind), dtype='uint32')  # use unsigned int for better memory utilization
    max_word_id = max(itertools.chain(*documents_as_ids)) + 1
    docs_words_matrix = csr_matrix((data, (row_ind, col_ind)), shape=(len(documents_as_ids), max_word_id))  # efficient arithmetic operations with CSR * CSR
    words_cooc_matrix = docs_words_matrix.T * docs_words_matrix  # multiplying docs_words_matrix with its transpose matrix would generate the co-occurences matrix
    words_cooc_matrix.setdiag(0)
    print(f"words_cooc_matrix:\n{words_cooc_matrix.todense()}")
    return words_cooc_matrix, word_to_id 

运行示例：

allowed_words = ['A', 'B', 'C', 'D']
documents = [['A', 'B'], ['C', 'B', 'K'],['A', 'B', 'C', 'D', 'Z']]
words_cooc_matrix, word_to_id = create_co_occurences_matrix(allowed_words, documents)

输出：

allowed_words:
['A', 'B', 'C', 'D']

documents:
[['A', 'B'], ['C', 'B', 'K'], ['A', 'B', 'C', 'D', 'Z']]

words_cooc_matrix:
[[0 2 1 1]
 [2 0 2 1]
 [1 2 0 1]
 [1 1 1 0]]

Answer 2

from collections import OrderedDict

document = [['A', 'B'], ['C', 'B'], ['A', 'B', 'C', 'D']]
names = ['A', 'B', 'C', 'D']

occurrences = OrderedDict((name, OrderedDict((name, 0) for name in names)) for name in names)

# Find the co-occurrences:
for l in document:
    for i in range(len(l)):
        for item in l[:i] + l[i + 1:]:
            occurrences[l[i]][item] += 1

# Print the matrix:
print(' ', ' '.join(occurrences.keys()))
for name, values in occurrences.items():
    print(name, ' '.join(str(i) for i in values.values()))

输出;

  A B C D
A 0 2 1 1 
B 2 0 2 1 
C 1 2 0 1 
D 1 1 1 0 

Answer 3

您也可以使用矩阵技巧来找到共生矩阵。希望当你有更大的词汇量时，这能很好地工作。

import scipy.sparse as sp
voc2id = dict(zip(names, range(len(names))))
rows, cols, vals = [], [], []
for r, d in enumerate(document):
    for e in d:
        if voc2id.get(e) is not None:
            rows.append(r)
            cols.append(voc2id[e])
            vals.append(1)
X = sp.csr_matrix((vals, (rows, cols)))

现在，你可以通过简单的乘法发现共现矩阵X.T与X

Xc = (X.T * X) # coocurrence matrix
Xc.setdiag(0)
print(Xc.toarray())

Answer 4

我们可以使用 极大地简化这个过程NetworkX。以下names是我们要考虑的节点，其中的列表包含document要连接的节点。

我们可以连接每个子列表中长度为 2 的节点combinations，并创建一个MultiGraph来考虑共现：

import networkx as nx
from itertools import combinations

G = nx.from_edgelist((c for n_nodes in document for c in combinations(n_nodes, r=2)),
                     create_using=nx.MultiGraph)
nx.to_pandas_adjacency(G, nodelist=names, dtype='int')

   A  B  C  D
A  0  2  1  1
B  2  0  2  1
C  1  2  0  1
D  1  1  1  0

Answer 5

显然，可以针对您的目的进行扩展，但是它会执行以下常规操作：

import math

for a in 'ABCD':
    for b in 'ABCD':
        count = 0

        for x in document:
            if a != b:
                if a in x and b in x:
                    count += 1

            else:
                n = x.count(a)
                if n >= 2:
                    count += math.factorial(n)/math.factorial(n - 2)/2

        print '{} x {} = {}'.format(a, b, count)

Answer 6

这是使用的另一个解决方案itertools和模块中的Counter类collections.

import numpy
import itertools
from collections import Counter

document =[['A', 'B'], ['C', 'B'],['A', 'B', 'C', 'D']]

# Get all of the unique entries you have
varnames = tuple(sorted(set(itertools.chain(*document))))

# Get a list of all of the combinations you have
expanded = [tuple(itertools.combinations(d, 2)) for d in document]
expanded = itertools.chain(*expanded)

# Sort the combinations so that A,B and B,A are treated the same
expanded = [tuple(sorted(d)) for d in expanded]

# count the combinations
c = Counter(expanded)


# Create the table
table = numpy.zeros((len(varnames),len(varnames)), dtype=int)

for i, v1 in enumerate(varnames):
    for j, v2 in enumerate(varnames[i:]):        
        j = j + i 
        table[i, j] = c[v1, v2]
        table[j, i] = c[v1, v2]

# Display the output
for row in table:
    print(row)

输出(可以很容易变成DataFrame)是:

[0 2 1 1]
[2 0 2 1]
[1 2 0 1]
[1 1 1 0]