The default values ​​for the Luigi and mocks parameter

Question

I am trying to mock what supplies the default value for the luigi parameter.

A dumb example showing what I'm trying to do:

Task:

import luigi
from bar import Bar

bar = Bar()

class Baz(luigi.Task):

    qux = luigi.Parameter(default=bar.bar())

    def baz(self):
        return self.qux;

    def foo(self):
        return bar.bar()

Unit Test Code:

import unittest
from mock import Mock, patch
from sut.baz import Baz

class TestMocking(unittest.TestCase):

    def test_baz_bar(self):
        self.assertEquals("bar", Baz().baz())

    @patch('sut.baz.bar')
    def test_patched_baz(self, mock_bar):
        mock_bar.bar = Mock(return_value="foo")
        self.assertEquals("foo", (Baz().baz()))

    @patch('sut.baz.bar')
    def test_patched_foo(self, mock_bar):
        mock_bar.bar = Mock(return_value="foo")
        self.assertEquals("foo", (Baz().foo()))

It seems that the logic of luigi.Parameter happens earlier than the patch.

This example test_patched_foopasses, but test_patched_bazfails. Thus, the patch occurs, but occurs after a call from a string luigi.Parameter(default=bar.bar()).

Is it possible to mock and correct something called in this way?

Answer 1

qux = luigi.Parameter(default=bar.bar()) __init__ Baz. __init__ , , __init__ , Baz. __init__ super:

class Baz(luigi.Task):

    def __init__(self, *args, **kwargs):
        super(Baz, self).__init__(*args, **kwargs)

        self.qux = luigi.Parameter(default=bar.bar())

    ...