python比较两个列表，将每个故障视为独立故障而不是单个故障

Question

在单元测试中，我正在比较两个列表

class MyTest(unittest.TestCase):
    def setUp(self):
        self.list_to_check_against = ['hu','ge','li','st']

    def test_with_sub_list(self):
        #raise failure for each elenment in sublist not in list_to_check_against
        sublist = ['hu','go','le']
        #???

我该怎么做？ 我不会收到关于list_to_check_against ont sublist中每个元素失败的通知，并且因为列表很大并且我无法在stdout中看到任何内容，所以仅列出sublist列表才抑制列表的输出。

样品输出

2 out 3 failures.. 
'go' , 'le' ...

Answer 1

class MyTest(unittest.TestCase):
    def setUp(self):
        self.list_to_check_against = ['hu','ge','li','st']

    def test_with_sub_list(self):
        #raise failure for each elenment in sublist not in list_to_check_against
        sublist = ['hu','go','le']
        self.assertSequenceEqual(self.list_to_check_against, sublist)

如果您只想知道缺少的内容，可以打印出差异

print set(list_to_check_against) - set(sublist)

Answer 2

def test_with_sub_list(self):
        sublist = ['hu','go','le']
        missing_elems = set(sublist)-set(self.list_to_check_against)
        assert not missing_elems, 
             "Error: %s not in list_to_check_against"%missing_elems

也许？