上一个问题：Make a new column based from CASE and GROUP BY result
从上一个问题我得到了如下结果
+------------------------+------+
| id_laporan_rekomendasi | test |
+------------------------+------+
| 1 | 1 |
| 2 | 1 |
| 3 | 1 |
| 4 | 0 |
| 5 | 1 |
| 6 | 0 |
+------------------------+------+
查询如下（感谢Gordon Linoff's answer）：
SELECT t1.id_laporan_rekomendasi,
MAX( t1.status = 3 ) as test
FROM test t1
GROUP BY t1.id_laporan_rekomendasi;
现在，我还有其他问题。在列id_laporan_rekomendasi的左侧，有一个名为id_laporan的列，因此该表将如下所示。
+------------+------------------------+------+
| id_laporan | id_laporan_rekomendasi | test |
+------------+------------------------+------+
| 3 | 2 | 0 |
| 3 | 6 | 1 |
| 8 | 3 | 1 |
| 8 | 4 | 1 |
| 7 | 1 | 1 |
| 7 | 5 | 1 |
+------------+------------------------+------+
我需要再次分组id_laporan并添加一个名为test2的新列，就像来自列test的列至少具有值0一样，结果将是{ {1}}，如果除0之外没有其他值（例如0），则会在1列中为我们提供值1。
我尝试过很多可能的问题：比如
test2但它给了我结果
SELECT t2.id_laporan, MAX( t2.test = 1 )
FROM (SELECT t1.id_laporan, t1.id_laporan_rekomendasi, MAX( t1.status = 3 ) as test
FROM test t1
GROUP BY t1.id_laporan_rekomendasi) t2
GROUP BY t2.id_laporan
我将最大值更改为+------------+--------------------+
| id_laporan | MAX( t2.test = 1 ) |
+------------+--------------------+
| 3 | 1 |
| 7 | 1 |
| 8 | 1 |
+------------+--------------------+
，它给我的结果如下：
MAX (t2.test = 0 )我期待结果如下：
+------------+--------------------+
| id_laporan | MAX( t2.test = 0 ) |
+------------+--------------------+
| 3 | 1 |
| 7 | 0 |
| 8 | 0 |
+------------+--------------------+
答案 0 :(得分：0)
我终于得到了解决方案，
我只是将assert array_nan_close(
np.array([1.3, np.nan, 3.4, np.nan]), np.array([1.3000001, 2, 3.4, 4])
)
assert not array_nan_close(
np.array([1.1, 4.0, 3.5, np.nan]), np.array([1, 2, 3, 4])
)
函数更改为MAX函数，所以它就像
MIN
完整查询
MIN( t2.test = 1 )使用MIN功能，当列SELECT t2.id_laporan, MIN( t2.test = 1 )
FROM (SELECT t1.id_laporan, t1.id_laporan_rekomendasi, MAX( t1.status = 3 ) as test
FROM test t1
GROUP BY t1.id_laporan_rekomendasi) t2
GROUP BY t2.id_laporan
的值低于0时，它将返回test。