Python NLTK的命令会吐出已识别单词的音素。例如'see'-> [u'S',u'IY1'],但是对于无法识别的单词会给出错误。例如'seasee'->错误。
import nltk
arpabet = nltk.corpus.cmudict.dict()
for word in ('s', 'see', 'sea', 'compute', 'comput', 'seesea'):
try:
print arpabet[word][0]
except Exception as e:
print e
#Output
[u'EH1', u'S']
[u'S', u'IY1']
[u'S', u'IY1']
[u'K', u'AH0', u'M', u'P', u'Y', u'UW1', u'T']
'comput'
'seesea'
是否有没有那个限制但能够找到/猜测任何真实或虚构单词的音素的模块?
如果没有,我有什么办法可以对其编程?我正在考虑做循环以测试单词的递增部分。例如,在“ seasee”中,第一个循环使用“ s”,下一个循环使用“ se”,第三个循环使用“ sea” ...等等,并运行命令。尽管问题是我不知道该如何发信号,但这是需要考虑的正确音素。例如,“ seasee”中的“ s”和“ sea”都将输出一些有效音素。
工作进程:
import nltk
arpabet = nltk.corpus.cmudict.dict()
for word in ('s', 'see', 'sea', 'compute', 'comput', 'seesea', 'darfasasawwa'):
try:
phone = arpabet[word][0]
except:
try:
counter = 0
for i in word:
substring = word[0:1+counter]
counter += 1
try:
print substring, arpabet[substring][0]
except Exception as e:
print e
except Exception as e:
print e
#Output
c [u'S', u'IY1']
co [u'K', u'OW1']
com [u'K', u'AA1', u'M']
comp [u'K', u'AA1', u'M', u'P']
compu [u'K', u'AA1', u'M', u'P', u'Y', u'UW0']
comput 'comput'
s [u'EH1', u'S']
se [u'S', u'AW2', u'TH', u'IY1', u'S', u'T']
see [u'S', u'IY1']
sees [u'S', u'IY1', u'Z']
seese [u'S', u'IY1', u'Z']
seesea 'seesea'
d [u'D', u'IY1']
da [u'D', u'AA1']
dar [u'D', u'AA1', u'R']
darf 'darf'
darfa 'darfa'
darfas 'darfas'
darfasa 'darfasa'
darfasas 'darfasas'
darfasasa 'darfasasa'
darfasasaw 'darfasasaw'
darfasasaww 'darfasasaww'
darfasasawwa 'darfasasawwa'
最佳答案
我遇到了同样的问题,并通过递归方式对未知对象进行分区来解决它(请参见wordbreak
)
import nltk
from functools import lru_cache
from itertools import product as iterprod
try:
arpabet = nltk.corpus.cmudict.dict()
except LookupError:
nltk.download('cmudict')
arpabet = nltk.corpus.cmudict.dict()
@lru_cache()
def wordbreak(s):
s = s.lower()
if s in arpabet:
return arpabet[s]
middle = len(s)/2
partition = sorted(list(range(len(s))), key=lambda x: (x-middle)**2-x)
for i in partition:
pre, suf = (s[:i], s[i:])
if pre in arpabet and wordbreak(suf) is not None:
return [x+y for x,y in iterprod(arpabet[pre], wordbreak(suf))]
return None