Question

我有一个严格固定的测试功能，它会因某些灯具输入而失败（应该如此）。我该如何表明这一点？这就是我现在所做的，也许还有更好的方法。我对py.test很陌生，所以我很感激任何提示。

下一部分是所有输入装置。仅供参考，example_datapackage_path在conf.test

中定义

@pytest.fixture(params=[None, 'pooled_col', 'phenotype_col'])
def metadata_key(self, request):
    return request.param

@pytest.fixture(params=[None, 'feature_rename_col'])
def expression_key(self, request):
    return request.param

@pytest.fixture(params=[None, 'feature_rename_col'])
def splicing_key(self, request):
    return request.param

@pytest.fixture
def datapackage(self, example_datapackage_path, metadata_key,
                expression_key, splicing_key):
    with open(example_datapackage_path) as f:
        datapackage = json.load(f)
    datatype_to_key = {'metadata': metadata_key,
                       'expression': expression_key,
                       'splicing': splicing_key}
    for datatype, key in datatype_to_key.iteritems():
        if key is not None:
            resource = name_to_resource(datapackage, datatype)
            if key in resource:
                resource.pop(key)
    return datapackage

@pytest.fixture
def datapackage_dir(self, example_datapackage_path):
    return os.path.dirname(example_datapackage_path)

这是测试本身。

def test_from_datapackage(self, datapackage, datapackage_dir):
    import flotilla
    from flotilla.external import get_resource_from_name

    study = flotilla.Study.from_datapackage(datapackage, datapackage_dir,
                                            load_species_data=False)

    metadata_resource = get_resource_from_name(datapackage, 'metadata')
    expression_resource = get_resource_from_name(datapackage,
                                                 'expression')
    splicing_resource = get_resource_from_name(datapackage, 'splicing')

    phenotype_col = 'phenotype' if 'phenotype_col' \
        not in metadata_resource else metadata_resource['phenotype_col']
    pooled_col = None if 'pooled_col' not in metadata_resource else \
        metadata_resource['pooled_col']
    expression_feature_rename_col = 'gene_name' if \
        'feature_rename_col' not in expression_resource \
        else expression_resource['feature_rename_col']
    splicing_feature_rename_col = 'gene_name' if \
        'feature_rename_col' not in splicing_resource \
        else splicing_resource['feature_rename_col']

    assert study.metadata.phenotype_col == phenotype_col
    assert study.metadata.pooled_col == pooled_col
    assert study.expression.feature_rename_col \
           == expression_feature_rename_col
    assert study.splicing.feature_rename_col == splicing_feature_rename_col

我想要做的是metadata_key，比如当参数为pooled_col或phenotype_col时，它会失败。我查看了pytest: Skip and xfail: dealing with tests that can not succeed，但它只讨论skip和xfail进行参数化测试，但没有讨论固定装置。

Answer 1

在您的datapackage或expression_key灯具中，您可以按here所述使用pytest.xfail和pytest.skip。例如：

@pytest.fixture
def datapackage(self, example_datapackage_path, metadata_key,
                expression_key, splicing_key):
    if metadata_key == 'pooled_col':
        pytest.skip('metadata key is "pooled_col"')
    ...

您还可以在灯具参数中使用pytest.mark.xfail，如下例所示：

@pytest.fixture(params=['a', pytest.mark.xfail('b'), 'c'])
def fx1(request):
    return request.param


def test_spam(fx1):
    assert fx1

如果您希望跳过这些测试，这似乎有效：

@pytest.fixture(
    params=['a', pytest.mark.skipif(True, reason='reason')('b'), 'c'])
def fx1(request):
    return request.param


def test_spam(fx1):
    assert fx1

在pytest中，如何跳过或者忽略某些灯具？

1 个答案: