How to send additional attribute to assigned method?

Question!

I found in internet such solution as:

def is_owner(self):
    if self.request.user.profile_url == self.kwargs['profile_url']:
        return True
    else:
        raise PermissionDenied

class CompanyProfileUpdateView(LoginRequiredMixin, UserPassesTestMixin, UpdateView):
    model = CompanyProfile
    template_name = 'profiles/create.html'
    fields = ['user', 'name']

    test_func = is_owner

Could anyone tell me, how to send additional value to the method? I want to have something like:

def is_owner(self, profile_type):
    if profile_type == 'user':
        if self.request.user.profile_url == self.kwargs['profile_url']:
            return True
        else:
            raise PermissionDenied

    else:
        # some code

test_func = is_owner('user') obviously is not working because there is no self

p.s. test_func is a method of UserPassesTestMixin class



Answers

You can't without making changes to UserPassesTestMixin. An easier solution may be to supply a kwarg to the view in the urls.py file or create a new subclass of the View with a different profile_type property on the class.

For example:

class CompanyProfileUpdateView(LoginRequiredMixin, UserPassesTestMixin, UpdateView):
    model = CompanyProfile
    template_name = 'profiles/create.html'
    fields = ['user', 'name']

    def test_func(self):
        if self.kwargs['profile_type'] == 'user':
            if self.request.user.profile_url == self.kwargs['profile_url']:
                return True
            else:
                raise PermissionDenied
        else:
            # something
            pass

urlpatterns += [
    url('^$', views.CompanyProfileUpdateView.as_view(), name='update_user', kwargs={'profile_type': 'user'}),
    url('^$', views.CompanyProfileUpdateView.as_view(), name='update_other', kwargs={'profile_type': 'other type'})
]

Second option:

class CompanyProfileUpdateView(LoginRequiredMixin, UserPassesTestMixin, UpdateView):
    model = CompanyProfile
    template_name = 'profiles/create.html'
    fields = ['user', 'name']

    profile_type = 'user'
    def test_func(self):
        if self.profile_type == 'user':
            if self.request.user.profile_url == self.kwargs['profile_url']:
                return True
            else:
                raise PermissionDenied
        else:
            # something
            pass

class OtherProfileUpdateView(CompanyProfileUpdateView):
    profile_type = 'other type'


There is a standard trick for computing the Fibonacci numbers that can easily be adapted to your problem. The naive definition for Fibonacci numbers is:

fibFunction :: Int -> Integer
fibFunction 0 = 1
fibFunction 1 = 1
fibFunction n = fibFunction (n-2) + fibFunction (n-1)

However, this is very costly: since all the leaves of the recursion are 1, if fib x = y, then we must perform y recursive calls! Since the Fibonacci numbers grow exponentially, this is a bad state of affairs to be in. But with dynamic programming, we can share the computations needed in the two recursive calls. The pleasing one-liner for this looks like this:

fibList :: [Integer]
fibList = 1 : 1 : zipWith (+) fibList (tail fibList)

This may look a bit puzzling at first; here the fibList argument to zipWith serves as the recursion on two indices ago, while the tail fibList argument serves as the recursion on one index ago, which gives us both the fib (n-2) and fib (n-1) values. The two 1s at the beginning are of course the base cases. There are other good questions here on SO that explain this technique in further detail, and you should study this code and those answers until you feel you understand how it works and why it is very fast.

If necessary, one can recover the Int -> Integer type signature from this using (!!).

Let's try to apply this technique to your function. As with computing Fibonacci numbers, you need the previous and second-to-last values; and additionally need the current index. That can be done by including [2..] in the call to zipWith. Here's how it would look:

waves :: [Integer]
waves = 1 : 1 : zipWith3 thisWave [2..] waves (tail waves) where
    thisWave n back2 back1 = ((3 * n - 3) * back2 + (2 * n + 1) * back1) `div` (n + 2)

As before, one can recover the function version with (!!) or genericIndex (if one really needs Integer indices). We can confirm that it computes the same function (but faster, and using less memory) in ghci:

> :set +s
> map wave [0..30]
[1,1,2,4,9,21,51,127,323,835,2188,5798,15511,41835,113634,310572,853467,2356779,6536382,18199284,50852019,142547559,400763223,1129760415,3192727797,9043402501,25669818476,73007772802,208023278209,593742784829,1697385471211]
(6.00 secs, 3,334,097,776 bytes)
> take 31 waves
[1,1,2,4,9,21,51,127,323,835,2188,5798,15511,41835,113634,310572,853467,2356779,6536382,18199284,50852019,142547559,400763223,1129760415,3192727797,9043402501,25669818476,73007772802,208023278209,593742784829,1697385471211]
(0.00 secs, 300,696 bytes)


With n=30, you need to compute wave 29 and wave 28, which, in turn, needs to compute wave 28, wave 27 twice and wave 26 and so forth, this quickly goes in the billions.

You can employ the same trick that is used in computation of the fibonacci numbers:

wave 0 = 1
wave 1 = 1
wave n = helper 1 1 2
    where
       helper x y k | k <n      = helper y z (k+1)
                    | otherwise = z
                    where z = ((3*k-3) * x + (2*k+1) * y) `div` (k+2)

This runs in linear time, and the helper has, for every k the values for wave (k-2) and wave (k-1) ready.

By : Ingo


This video can help you solving your question :)
By: admin