Question

我想比较每个SKU的“提供中”的平均收入与“提供中的平均收入”。

当我在sku上合并以下两个数据帧时，我为每个条目获得了多行，因为在第二个数据帧sku中不是唯一的。例如，sku = 1的每个实例将有两个条目，因为test_offer包含2个单独的sku 1报价。但是，在任何时候，一个SKU都只能有一个实时报价，这应验证条件：< / p>

 test_ga['day'] >= test_offer['start_day'] &  test_ga['day'] <= test_offer['end_day']

数据集1

test_ga = pd.DataFrame( {'day': {0: 1, 1: 2, 2: 3, 3: 4, 4: 5, 5: 6, 6: 7, 7: 8, 8: 1, 9: 2, 10: 3, 11: 4, 12: 5, 13: 6, 14: 7, 15: 8, 16: 1, 17: 2, 18: 3, 19: 4, 20: 5, 21: 6, 22: 7, 23: 8}, 
'sku': {0: 1, 1: 1, 2: 1, 3: 1, 4: 1, 5: 1, 6: 1, 7: 1, 8: 2, 9: 2, 10: 2, 11: 2, 12: 2, 13: 2, 14: 2, 15: 2, 16: 3, 17: 3, 18: 3, 19: 3, 20: 3, 21: 3, 22: 3, 23: 3}, 
'revenue': {0: 12, 1: 34, 2: 28, 3: 76, 4: 30, 5: 84, 6: 55, 7: 78, 8: 23, 9: 58, 10: 11, 11: 15, 12: 73, 13: 9, 14: 69, 15: 34, 16: 71, 17: 69, 18: 90, 19: 93, 20: 43, 21: 45, 22: 57, 23: 89}} )

数据集2

test_offer = pd.DataFrame( {'sku': {0: 1, 1: 1, 2: 2}, 
'offer_number': {0: 5, 1: 6, 2: 7}, 
'start_day': {0: 2, 1: 6, 2: 4}, 
'end_day': {0: 4, 1: 7, 2: 8}} )

预期产量

expected_output = pd.DataFrame( {'day': {0: 1, 1: 2, 2: 3, 3: 4, 4: 5, 5: 6, 6: 7, 7: 8, 8: 1, 9: 2, 10: 3, 11: 4, 12: 5, 13: 6, 14: 7, 15: 8}, 
'sku': {0: 1, 1: 1, 2: 1, 3: 1, 4: 1, 5: 1, 6: 1, 7: 1, 8: 2, 9: 2, 10: 2, 11: 2, 12: 2, 13: 2, 14: 2, 15: 2}, 
'offer': {0: float('nan'), 1: '5', 2: '5', 3: '5', 4: float('nan'), 5: '6', 6: '6', 7: float('nan'), 8: float('nan'), 9: float('nan'), 10: float('nan'), 11: '7', 12: '7', 13: '7', 14: '7', 15: '7'},
'start_day': {0: float('nan'), 1: '2', 2: '2', 3: '2', 4: float('nan'), 5: '6', 6: '6', 7: float('nan'), 8: float('nan'), 9: float('nan'), 10: float('nan'), 11: '4', 12: '4', 13: '4', 14: '4', 15: '4'}, 
'end_day': {0: float('nan'), 1: '4', 2: '4', 3: '4', 4: float('nan'), 5: '7', 6: '7', 7: float('nan'), 8: float('nan'), 9: float('nan'), 10: float('nan'), 11: '8', 12: '8', 13: '8', 14: '8', 15: '8'}, 
'revenue': {0: 12, 1: 34, 2: 28, 3: 76, 4: 30, 5: 84, 6: 55, 7: 78, 8: 23, 9: 58, 10: 11, 11: 15, 12: 73, 13: 9, 14: 69, 15: 34}} )

我确实确实找到了基于此SO answer的解决方案，但是花了我一段时间，问题还不清楚。

我认为即使找到解决方案，提出这个问题仍然有用。此外，可能还有更好的方法可以实现这些目标，而无需创建虚拟变量和对数据帧进行排序？

如果该问题重复，请告诉我，我将取消它。

Answer 1

一种可能的解决方案：

test_data = pd.merge(test_ga, test_offer, on = 'sku')
# I define if every row is in offer or not.
test_data['is_offer'] = np.where((test_data['day'] >= test_data['start_day']) & (test_data['day'] <= test_data['end_day']), 1, 0)
expected_output = test_data.sort_values(['sku','day','is_offer']).groupby(['day', 'sku']).tail(1)

，然后清理数据，为不在报价中的行添加Nan值。

expected_output['start_day'] = np.where(expected_output['is_offer'] == 0, np.NAN, expected_output['start_day'])
expected_output['end_day'] = np.where(expected_output['is_offer'] == 0, np.NAN, expected_output['end_day'])
expected_output['offer_number'] = np.where(expected_output['is_offer'] == 0, np.NAN, expected_output['offer_number'])
expected_output

合并熊猫数据框并根据条件删除重复的行

1 个答案: