在luis.ai获得1个单词的2个实体

Question

我正在使用

from datetime import datetime
from django.conf.urls import url
import django.contrib.auth.views

import app.forms
import app.views

# Uncomment the next lines to enable the admin:
# from django.conf.urls import include
# from django.contrib import admin
# admin.autodiscover()

urlpatterns = [
    # Examples:
    url(r'^$', app.views.home, name='home'),
    url(r'^contact$', app.views.contact, name='contact'),
    url(r'^about', app.views.about, name='about'),
    url(r'^Register', app.views.Register, name='Register'),
    url(r'^problemreports$', app.views.problemreports, name='problemreports'),
    url(r'^problemreports/NewItem', app.views.PRNewItem, name='PRNewItem'),
    url(r'^problemreports/MyProblemReports', app.views.myPRs, name='myPRs'),
    url(r'^problemreports/Closed', app.views.Closed, name='Closed'),
    url(r'^problemreports/SearchPRs', app.views.SearchPRs, name='SearchPRs'),
    url(r'^problemreports/(\d{1,10})', app.views.EditPR, name='EditPR'),
    url(r'^login/$',
        django.contrib.auth.views.login,
        {
            'template_name': 'app/login.html',
            'authentication_form': app.forms.BootstrapAuthenticationForm,
            'extra_context':
            {
                'title': 'Log in',
                'year': datetime.now().year,
            }
        },
        name='login'),
    url(r'^logout$',
        django.contrib.auth.views.logout,
        {
            'next_page': '/',
        },
        name='logout'),

    # Uncomment the admin/doc line below to enable admin documentation:
    # url(r'^admin/doc/', include('django.contrib.admindocs.urls')),

    # Uncomment the next line to enable the admin:
    # url(r'^admin/', include(admin.site.urls)),
]

，我有两个实体：

• 列表实体
• 一个简单的实体。

当我查询from datetime import datetime from django.conf.urls import url import django.contrib.auth.views import app.forms import app.views # Uncomment the next lines to enable the admin: # from django.conf.urls import include # from django.contrib import admin # admin.autodiscover() urlpatterns = [ # Examples: url(r'^$', app.views.home, name='home'), url(r'^contact$', app.views.contact, name='contact'), url(r'^about', app.views.about, name='about'), url(r'^Register', app.views.Register, name='Register'), url(r'^problemreports$', app.views.problemreports, name='problemreports'), url(r'^problemreports/NewItem', app.views.PRNewItem, name='PRNewItem'), url(r'^problemreports/MyProblemReports', app.views.myPRs, name='myPRs'), url(r'^problemreports/Closed', app.views.Closed, name='Closed'), url(r'^problemreports/SearchPRs', app.views.SearchPRs, name='SearchPRs'), url(r'^problemreports/(\d{1,10})', app.views.EditPR, name='EditPR'), url(r'^login/$', django.contrib.auth.views.login, { 'template_name': 'app/login.html', 'authentication_form': app.forms.BootstrapAuthenticationForm, 'extra_context': { 'title': 'Log in', 'year': datetime.now().year, } }, name='login'), url(r'^logout$', django.contrib.auth.views.logout, { 'next_page': '/', }, name='logout'), # Uncomment the admin/doc line below to enable admin documentation: # url(r'^admin/doc/', include('django.contrib.admindocs.urls')), # Uncomment the next line to enable the admin: # url(r'^admin/', include(admin.site.urls)), ] 在一个单词中匹配时，我得到两个实体。

我只想要简单的实体，但它也匹配列表实体，因为有一个内部字符串匹配列表实体的同义词列表中的单词。 有没有办法只获得匹配整个单词的实体，而不仅仅是其中的一部分。

Answer 1

没有办法做到这一点。但是，在您的代码中，您可以根据您的实体Type进行过滤，并只关注价值为＃34; Pets＆＃34;的那个。