你可以試試下面的代碼。在import collections, numpy

tokens=['He','is','not','lazy','intelligent','smart']

j=0

a=np.zeros((len(tokens),len(tokens)))

for pos,token in enumerate(tokens):

j+=pos+1

for token1 in tokens[pos+1:]:

count = 0

for sentence in [['He','is','not','lazy','He','is','intelligent','He','is','smart'] ]:

occurrences1 = [i for i,e in enumerate(sentence) if e == token1]

#print(token1,occurrences1)

occurrences2 = [i for i,e in enumerate(sentence) if e == token]

#print(token,occurrences2)

new1= np.repeat(occurrences1,len(occurrences2))

new2= np.asarray(occurrences2*len(occurrences1))

final_new= np.subtract(new1,new2)

final_abs_diff = np.absolute(final_new)

final_counts = collections.Counter(final_abs_diff)

count_1=final_counts[1]

count_2=final_counts[2]

count_0=final_counts[0]

count=count_1+count_2+count_0

a[pos][j]=count

#print(token,' ',pos,' ',token1,' ',j,' ',count)

j+=1

j=0

final_mat = a.T+a

print(final_mat)

輸出為：

^{pr2}$

總結

以上是生活随笔為你收集整理的python 共现矩阵_Python词组共现矩阵的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。