1. 问题背景
在编写测试用例来测试 FilterForm 时，遇到了以下问题：
class MyTestCreateFilter(TestCase):
def test_createfilter(self):
self.client = Client()
self.user = User.objects.create_superuser(username='bizi111',email='test5@example.com',password='bizi111')
self.user = authenticate(username='bizi111', password='bizi111')
#print self.user
self.factory = RequestFactory()
request = self.factory.get('/filter/new')
request.user = self.user
response = create_or_edit_filter(request)
self.assertEqual(response.status_code, 200)
filterform = FilterForm()
#print filterform.is_valid()
# http://jshk.com.cn/mb/reg.asp?kefu=zhangyajie
form_data = {'keyword': 'it','industry':'it','zip_code':'50005','distance':30}
filteform = FilterForm(form_data)
self.assertEqual(filteform.is_valid(), True)
其中 FilterForm 有一个 clean 方法：
def clean(self):
#print self.instance.user_profile
user_profile = self.instance.user_profile
keyword = self.cleaned_data.get("keyword")
if Filter.objects.filter(user_profile=user_profile, keyword=keyword).exclude(id=self.instance.id).count() > 0:
msg = u"A filter with that keyword already exists!"
self._errors["keyword"] = self.error_class([msg])
return self.cleaned_data
在运行测试时，出现错误：
File "/home/suma/workspace2/bizintro/bizintro/forms.py", line 80, in
clean
user_profile = self.instance.user_profile File "/usr/local/lib/python2.7/dist-packages/django/db/models/fields/related.py",
line 343, in get
raise self.field.rel.to.DoesNotExist DoesNotExist
2. 解决方案
根据错误信息，可以发现问题是 FilterForm 是一个绑定表单，需要有一个模型实例作为上下文。在测试用例中，没有为 FilterForm 设置模型实例。
为了解决这个问题，可以在测试用例中添加以下代码：
filterform = FilterForm()
#print filterform.is_valid()
form_data = {'keyword': 'it','industry':'it','zip_code':'50005','distance':30}
filterform = FilterForm(form_data)
filterform.instance = # Some object here...
self.assertEqual(filterform.is_valid(), True)
其中，需要将 # Some object here... 替换为一个有效的模型实例。
代码例子：
class MyTestCreateFilter(TestCase):
def test_createfilter(self):
self.client = Client()
self.user = User.objects.create_superuser(username='bizi111',email='test5@example.com',password='bizi111')
self.user = authenticate(username='bizi111', password='bizi111')
#print self.user
self.factory = RequestFactory()
request = self.factory.get('/filter/new')
request.user = self.user
response = create_or_edit_filter(request)
self.assertEqual(response.status_code, 200)
filterform = FilterForm()
#print filterform.is_valid()
form_data = {'keyword': 'it','industry':'it','zip_code':'50005','distance':30}
filterform = FilterForm(form_data)
filterform.instance = Filter() # 创建一个 Filter 模型实例
self.assertEqual(filterform.is_valid(), True)
按照上述代码，就可以成功运行测试用例，并不会出现 DoesNotExist 的错误。
标题: 线段交点检测及其解决方案
-
问题背景
给定两个列表，我们希望找出它们在相同索引处相交的点。例如，如果我们提供两个列表 [9, 8, 7, 6, 5] 和 [3, 4, 5, 6, 7]，目标是找到它们在索引 3 处相交的点。常见的解决方案涉及遍历并比较两个列表中的每个元素，但我们希望探索更具数学性、高效的方法。
-
解决方案
- 集合交集法：一种常用方法是使用集合的交集运算。我们可以将每个列表的坐标视为一个集合，计算它们的交集。例如，我们计算 (9, 0), (8, 1), (7, 2), (6, 3), (5, 4) 和 (3, 0), (4, 1), (5, 2), (6, 3), (7, 4) 的交集，发现 (6, 3) 和 (7, 4) 同时出现在两个列表中。因此，我们找到这两个列表在索引 3 和 4 处相交。
- 线性方程法：另一种方法是将列表中的元素视为线段，使用线性方程求解线段相交点。我们可以构造一个线性方程组，其中每个方程代表列表中的一条线段。求解该方程组，可以得到两个线段的交点。例如，我们构造方程组 y = 9 - x、y = 3 + x。求解得 x = 6，y = 3。因此，这两个列表在点 (6, 3) 处相交。
代码例子：
# 集合交集法
def find_intersection_by_set(list1, list2):
"""
Find the intersection points of two lists using set intersection.
Args:
list1: The first list of coordinates.
list2: The second list of coordinates.
Returns:
A list of tuples representing the intersection points.
"""
intersecting_points = set(enumerate(list1)).intersection(set(enumerate(list2)))
return [(index, value) for index, value in intersecting_points]
# 线性方程法
def find_intersection_by_linear_equation(list1, list2):
"""
Find the intersection points of two lists using linear equation.
Args:
list1: The first list of coordinates.
list2: The second list of coordinates.
Returns:
A list of tuples representing the intersection points.
"""
intersecting_points = []
for i, (A0, B0, A1, B1) in enumerate(zip(list1, list2, list1[1:], list2[1:])):
# Check integer intersections
if A0 == B0:
intersecting_points.append((i, A0))
if A1 == B1:
intersecting_points.append((i + 1, A1))
# Check for intersection between points
if (A0 > B0 and A1 < B1) or (A0 < B0 and A1 > B1):
intersection_index = solve_linear_equation(A0, A1, B0, B1)
intersecting_points.append((i + intersection_index, (A1 - A0) * intersection_index + A0))
return intersecting_points
def solve_linear_equation(A0, A1, B0, B1):
"""
Solve the linear equation (A0, A1) and (B0, B1) and return the intersection index.
Args:
A0: The y-coordinate of the first point of the first line.
A1: The y-coordinate of the second point of the first line.
B0: The y-coordinate of the first point of the second line.
B1: The y-coordinate of the second point of the second line.
Returns:
The intersection index.
"""
return (B0 - A0) / (A1 - A0)
最后，根据问题的情况，我们可以使用任一方法来找到列表 [9, 8, 7, 6, 5] 和 [3, 4, 5, 6, 7] 在索引 3 处的交点。
