Python – PyCharm doesn’t detect interpreter

There isn't a hugely elegant way to handle import cycles in general, I'm afraid. Your choices are to either redesign your code to remove the cyclic dependency, or if it isn't feasible, do something like this:

# some_file.py

from typing import TYPE_CHECKING
if TYPE_CHECKING:
    from main import Main

class MyObject(object):
    def func2(self, some_param: 'Main'):
        ...

The TYPE_CHECKING constant is always False at runtime, so the import won't be evaluated, but mypy (and other type-checking tools) will evaluate the contents of that block.

We also need to make the Main type annotation into a string, effectively forward declaring it since the Main symbol isn't available at runtime.

If you are using Python 3.7+, we can at least skip having to provide an explicit string annotation by taking advantage of PEP 563:

# some_file.py

from __future__ import annotations
from typing import TYPE_CHECKING
if TYPE_CHECKING:
    from main import Main

class MyObject(object):
    # Hooray, cleaner annotations!
    def func2(self, some_param: Main):
        ...

The from __future__ import annotations import will make all type hints be strings and skip evaluating them. This can help make our code here mildly more ergonomic.

All that said, using mixins with mypy will likely require a bit more structure then you currently have. Mypy recommends an approach that's basically what deceze is describing -- to create an ABC that both your Main and MyMixin classes inherit. I wouldn't be surprised if you ended up needing to do something similar in order to make Pycharm's checker happy.

Yes, it is possible and you should also use a doubly block circulant matrix (which is a special case of Toeplitz matrix). I will give you an example with a small size of kernel and the input, but it is possible to construct Toeplitz matrix for any kernel. So you have a 2d input x and 2d kernel k and you want to calculate the convolution x * k. Also let's assume that k is already flipped. Let's also assume that x is of size n×n and k is m×m.

So you unroll k into a sparse matrix of size (n-m+1)^2 × n^2, and unroll x into a long vector n^2 × 1. You compute a multiplication of this sparse matrix with a vector and convert the resulting vector (which will have a size (n-m+1)^2 × 1) into a n-m+1 square matrix.

I am pretty sure this is hard to understand just from reading. So here is an example for 2×2 kernel and 3×3 input.

*

Here is a constructed matrix with a vector:

which is equal to .

And this is the same result you would have got by doing a sliding window of k over x.