我有一个名字列表，比如：
names = ['A', 'B', 'C', 'D']
以及一份文件清单，每份文件中都提到了其中一些名称。
document =[['A', 'B'], ['C', 'B', 'K'],['A', 'B', 'C', 'D', 'Z']]
我想获得一个输出作为共现矩阵，例如：
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中有一个解决方案(创建共现矩阵)，但我在Python中无法做到这一点。我正在考虑在熊猫中这样做，但还没有进展！
另一种选择是使用构造函数csr_matrix((data, (row_ind, col_ind)), [shape=(M, N)])来自scipy.sparse.csr_matrix，其中data、row_ind和col_ind满足 关系a[row_ind[k], col_ind[k]] = data[k].
诀窍是通过迭代文档并创建元组列表(doc_id、word_id)来生成row_ind和col_ind。data只是相同长度的向量。
将文档单词矩阵乘以其转置将得到共现矩阵。
此外，这在运行时和内存使用方面都是有效的，因此它还应该处理大型语料库。
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]]
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
显然，这可以根据您的目的进行扩展，但它会执行一般操作：
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)
您还可以使用矩阵技巧来查找共现矩阵。希望当你有更大的词汇量时，这很有效。
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，您可以通过简单的乘以X.TX来找到coocurrence 矩阵
Xc = (X.T * X) # coocurrence matrix
Xc.setdiag(0)
print(Xc.toarray())
这是使用itertools和collections模块中的Counter类的另一种解决方案。
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(可以很容易地转换为数据帧)为：
[0 2 1 1]
[2 0 2 1]
[1 2 0 1]
[1 1 1 0]
我们可以使用NetworkX大大简化这一点。这里names是我们需要考虑的节点，document中的列表包含要连接的节点。
我们可以连接每个子列表中的节点，长度为 2combinations，并创建一个MultiGraph来解释共现：
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
我遇到了同样的问题...所以我带着这个代码来了。此代码考虑上下文窗口，然后确定co_occurance矩阵。
希望这对您有所帮助...
def countOccurences(word,context_window):
"""
This function returns the count of context word.
"""
return context_window.count(word)
def co_occurance(feature_dict,corpus,window = 5):
"""
This function returns co_occurance matrix for the given window size. Default is 5.
"""
length = len(feature_dict)
co_matrix = np.zeros([length,length]) # n is the count of all words
corpus_len = len(corpus)
for focus_word in top_features:
for context_word in top_features[top_features.index(focus_word):]:
# print(feature_dict[context_word])
if focus_word == context_word:
co_matrix[feature_dict[focus_word],feature_dict[context_word]] = 0
else:
start_index = 0
count = 0
while(focus_word in corpus[start_index:]):
# get the index of focus word
start_index = corpus.index(focus_word,start_index)
fi,li = max(0,start_index - window) , min(corpus_len-1,start_index + window)
count += countOccurences(context_word,corpus[fi:li+1])
# updating start index
start_index += 1
# update [Aij]
co_matrix[feature_dict[focus_word],feature_dict[context_word]] = count
# update [Aji]
co_matrix[feature_dict[context_word],feature_dict[focus_word]] = count
return co_matrix
''对于 2 的窗口，data_corpus是由文本数据组成的系列，单词是由构建共现矩阵的单词组成的列表'''
"co_oc是共现矩阵">
co_oc=pd.DataFrame(index=words,columns=words)
for j in tqdm(data_corpus):
k=j.split()
for l in range(len(k)):
if l>=5 and l<(len(k)-6):
if k[l] in words:
for m in range(l-5,l+6):
if m==l:
continue
elif k[m] in words:
co_oc[k[l]][k[m]]+=1
elif l>=(len(k)-6):
if k[l] in words:
for m in range(l-5,len(k)):
if m==l:
continue
elif k[m] in words:
co_oc[k[l]][k[m]]+=1
else:
if k[l] in words:
for m in range(0,l+5):
if m==l:
continue
elif k[m] in words:
co_oc[k[l]][k[m]]+=1
print(co_oc.head())