使用参数命名URL时的Django - NoReVeMeCATCH错误

我试图用一个带有参数的URL来构建这个链接:

http://127.0.0.1:8000/wakemeup/admin/list/colegio

参数是后面的字符串 .../list/ (在这种情况下,“Celejo”)。我有一个 print() 中的语句 admin_list 视图,它显示参数得到正确提取。但是,当我尝试构建url时:

<a href="{% url 'wakemeup:admin_list' list_type=colegio %}">Colegios</a>

我得到这个错误:

NoReverseMatch at /wakemeup/admin/list/colegio
Reverse for 'admin_list' with keyword arguments '{'list_type': ''}' not found. 1 pattern(s) tried: ['wakemeup/admin/list/(?P<list_type>\\w+)$']
Request Method: GET
Request URL:    http://127.0.0.1:8000/wakemeup/admin/list/colegio
Django Version: 2.0.1
Exception Type: NoReverseMatch
Exception Value:    
Reverse for 'admin_list' with keyword arguments '{'list_type': ''}' not found. 1 pattern(s) tried: ['wakemeup/admin/list/(?P<list_type>\\w+)$']

VIEW

def admin_list(request, list_type):
    print(list_type)
    return index(request)

URLS.Py

url(r'^admin/list/(?P<list_type>\w+)$', views.admin_list, name="admin_list"),

我还尝试使用一个未命名的参数,但这不起作用。