在Python中有条件地填充新列

Question

我有一个表，如下所示：

+--------+--------+-----------+------------+
| Binary | Status | Inception | New Status |
+--------+--------+-----------+------------+
|      1 | Raising|      2016 |            |
|      1 | Raising|      2017 |            |
|      0 | Failed |      2018 |            |
|      1 | Closed |      2021 |            |
|      1 | Raising|      2020 |            |

我一直在按如下方式填充New Status列：

q = df.query('Binary == 1 and Inception <= 2022 - 3 and Status != "Closed" and Status != "Failed"')
my_query_index = q.index
my_query_index
df.iloc[my_query_index, 3] = "Closed"
df

然而，我想用一种更动态的方式来做这件事，在第一种情况下使用<，在第二种情况下则使用<=。在这种情况下，如果今天的日期恰好在今年上半年，我将使用查询

q = df.query('Binary == 1 and Inception < 2022 - 3 and Status != "Closed" and Status != "Failed"')
my_query_index = q.index
my_query_index
df.iloc[my_query_index, 3] = "Closed"
df

如果今天的日期恰好在下半年，请使用

q = df.query('Binary == 1 and Inception <= 2022 - 3 and Status != "Closed" and Status != "Failed"')
my_query_index = q.index
my_query_index
df.iloc[my_query_index, 3] = "Closed"
df

考虑到today = datetime.today().strftime('%Y-%m-%d')，我想用today替换Inception(在代码中(。我不知道如何将这两种情况合并到代码中，也不确定如何将其合并到查询中。

Answer 1

您可以使用str格式进行查询

today = datetime.today().strftime('%Y-%m-%d')
half_year = '2022-06-01'
if today < half_year:
query = 'Binary == 1 and Inception < 2022 - 3 and Status != "Closed" and Status != "Failed"'
else:
query = 'Binary == 1 and Inception <= 2022 - 3 and Status != "Closed" and Status != "Failed"'

然后你的常规代码

q = df.query(query)
my_query_index = q.index
my_query_index
df.iloc[my_query_index, 3] = "Closed"
df