为什么日期格式会随着 np.where 改变以及如何阻止它发生

Question

我有来自数据帧的以下数据子集。

{'NID': {131598: '215026851',
  131599: '215026851',
  131600: '215026851',
  131601: '215026851',
  131602: '215026851',
  131603: '215026851',
  131604: '215026851',
  131605: '215026851',
  131606: '215026851'},
 'AbCode': {131598: 0,
  131599: 0,
  131600: 0,
  131601: 0,
  131602: 0,
  131603: 1,
  131604: 0,
  131605: 0,
  131606: 0},
 'ABdat': {131598: Timestamp('2018-01-24 00:00:00'),
  131599: Timestamp('2019-01-25 00:00:00'),
  131600: NaT,
  131601: Timestamp('2019-11-08 00:00:00'),
  131602: Timestamp('2020-01-24 00:00:00'),
  131603: Timestamp('2020-02-15 00:00:00'),
  131604: Timestamp('2020-10-16 00:00:00'),
  131605: Timestamp('2020-10-26 00:00:00'),
  131606: NaT}}

格式化数据后，如下所示，

          NID     AbCode  ABdat
131598  215026851   0   2018-01-24
131599  215026851   0   2019-01-25
131600  215026851   0   NaT
131601  215026851   0   2019-11-08
131602  215026851   0   2020-01-24
131603  215026851   1   2020-02-15
131604  215026851   0   2020-10-16
131605  215026851   0   2020-10-26
131606  215026851   0   NaT 

我想将 AbCode = 0 的 ABdat 替换为 Missing (NaT)，并将 AbCode = 1 的 ABdat 替换为 ABdat-7days

我在下面编写了以下 np.where 代码来执行此操作。

breed_info['ABdat'] = np.where(breed_info.AbCode == 1, breed_info['ABdat'] - pd.DateOffset(days=7), breed_info['ABdat'].isnull)

输出如下所示，

          NID   AbCode       ABdat
131598  215026851   0   <bound method Series.isnull of 49017 ...
131599  215026851   0   <bound method Series.isnull of 49017 ...
131600  215026851   0   <bound method Series.isnull of 49017 ...
131601  215026851   0   <bound method Series.isnull of 49017 ...
131602  215026851   0   <bound method Series.isnull of 49017 ...
131603  215026851   1   1581120000000000000
131604  215026851   0   <bound method Series.isnull of 49017 ...
131605  215026851   0   <bound method Series.isnull of 49017 ...
131606  215026851   0   <bound method Series.isnull of 49017 ...

您能否告知为什么日期格式发生变化以及如何避免这种情况发生？

谢谢

Answer 1

最简单的是使用一些 pandas 解决方案和 pandas 方法，例如 Series.where：

breed_info['ABdat'] = (breed_info['ABdat'] - pd.DateOffset(days=7))
                                                        .where(breed_info.AbCode == 1)

使用 np.where 带有帮助器 Series 的黑客解决方案：

breed_info['ABdat'] = np.where(breed_info.AbCode == 1,
                               breed_info['ABdat'] - pd.DateOffset(days=7),
                                pd.Series(pd.NaT, index=breed_info.index))
print (breed_info)
              NID  AbCode      ABdat
131598  215026851       0        NaT
131599  215026851       0        NaT
131600  215026851       0        NaT
131601  215026851       0        NaT
131602  215026851       0        NaT
131603  215026851       1 2020-02-08
131604  215026851       0        NaT
131605  215026851       0        NaT
131606  215026851       0        NaT

因为如果传递 pd.NAT它返回下划线numpy数组（以纳秒为单位）：

breed_info['ABdat'] = np.where(breed_info.AbCode == 1, 
                               breed_info['ABdat'] - pd.DateOffset(days=7),
                               pd.NaT)
print (breed_info)
              NID  AbCode                ABdat
131598  215026851       0                  NaT
131599  215026851       0                  NaT
131600  215026851       0                  NaT
131601  215026851       0                  NaT
131602  215026851       0                  NaT
131603  215026851       1  1581120000000000000
131604  215026851       0                  NaT
131605  215026851       0                  NaT
131606  215026851       0                  NaT

---

我认为原因是错误。