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答案

【问题标题】：python comparing two lists, treating each failure as an independent failure instead of a single failurepython比较两个列表，将每个故障视为独立故障而不是单个故障
【发布时间】：2014-12-15 23:06:16
【问题描述】：

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？我不想收到sublist 中每个元素的失败通知，即list_to_check_against 中的每个元素，并且只抑制sublist 元素的列表输出，因为列表很大，我在标准输出中看不到任何内容。

样本输出..

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

【问题讨论】：

标签： python list testing tdd unit-testing

【解决方案1】：

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？

【讨论】：

【解决方案2】：

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什么，可以打印差异

print set(list_to_check_against) - set(sublist)

【讨论】：