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所以我有一组测试,我想测试解决方案的多个版本。目前我有双参数化
import pytest
import product_not_at_index
functions_to_test = [
product_not_at_index.product_not_at_index_n_squared,
product_not_at_index.product_not_at_index,
]
def run_test(function_input, expected_result, test_func):
actual_result = test_func(function_input)
assert actual_result == expected_result
@pytest.mark.parametrize("test_func", functions_to_test)
def test_empty_list(test_func):
input_data = []
expected_result = []
run_test(input_data, expected_result, test_func)
@pytest.mark.parametrize("test_func", functions_to_test)
def test_single_item(test_func):
input_data = [1]
expected_result = [1]
run_test(input_data, expected_result, test_func)
@pytest.mark.parametrize("test_func", functions_to_test)
def test_one_times_one(test_func):
input_data = [1, 1]
expected_result = [1, 1]
run_test(input_data, expected_result, test_func)
@pytest.mark.parametrize("test_func", functions_to_test)
def test_normal_use_case(test_func):
input_data = [1, 7, 3, 4]
expected_result = [84, 12, 28, 21]
run_test(input_data, expected_result, test_func)
而且这个工程很好。但看着我的解决方案,我发现我的所有测试都具有相同的基本代码集。我怎样才能参数化一个函数两次,以便我可以只有一个测试函数,并停止重复自己?
我认为我可以做类似
import pytest
import product_not_at_index
functions_to_test = [product_not_at_index.product_not_at_index_n_squared]
test_data = [
[], [],
[1], [1],
[1, 1], [1, 1],
[1, 7, 3, 4], [84, 12, 28, 21],
]
@pytest.mark.parametrize("function_input,expected_result", test_data)
@pytest.mark.parametrize("test_func", functions_to_test)
def test_run(function_input, expected_result, test_func):
actual_result = test_func(function_input)
assert actual_result == expected_result
但只是返回该错误
E assert 0 == 2
E + where 0 = len([])
E + and 2 = len(['function_input', 'expected_result'])
[是,堆栈交换明确鼓励](https://stackoverflow.com/help/self-answer) – AlexLordThorsen