Python 熊猫：使用DateTimeIndex列表从DataFrame中提取多行

我有一个熊猫数据帧，索引如下，频率为秒：

DatetimeIndex(['2015-12-28 05:20:05', '2015-12-28 05:20:06',
               '2015-12-28 05:20:07', '2015-12-28 05:20:08',
               ...
               '2015-12-28 21:19:55', '2015-12-28 21:19:56',
               '2015-12-28 21:19:57', '2015-12-28 21:19:58']

我希望在给定日期时间字符串列表的情况下一次提取多行。我试过：

df.loc[['2015-12-28 08:32:39', '2015-12-28 08:32:48']]

但我得到了以下错误：

KeyError: "None of [['2015-12-28 08:32:39', '2015-12-28 08:32:48']] are in the [index]"

---

您可以通过或将

列表

转换为

日期时间

：

或：

样本：

idx = pd.DatetimeIndex(['2015-12-28 05:20:05', '2015-12-28 05:20:06',
               '2015-12-28 05:20:07', '2015-12-28 05:20:08',
               '2015-12-28 21:19:55', '2015-12-28 21:19:56',
               '2015-12-28 21:19:57', '2015-12-28 21:19:58'])
df = pd.DataFrame({'s': range(8)}, index=idx)  
print (df)
                     s
2015-12-28 05:20:05  0
2015-12-28 05:20:06  1
2015-12-28 05:20:07  2
2015-12-28 05:20:08  3
2015-12-28 21:19:55  4
2015-12-28 21:19:56  5
2015-12-28 21:19:57  6
2015-12-28 21:19:58  7

d = ['2015-12-28 05:20:05', '2015-12-28 21:19:58']
print (df.loc[pd.to_datetime(d)])
                     s
2015-12-28 05:20:05  0
2015-12-28 21:19:58  7

print (df.loc[pd.to_datetime(d)])
                     s
2015-12-28 05:20:05  0
2015-12-28 21:19:58  7

print (df.loc[pd.to_datetime(d)])

idx = pd.DatetimeIndex(['2015-12-28 05:20:05', '2015-12-28 05:20:06',
               '2015-12-28 05:20:07', '2015-12-28 05:20:08',
               '2015-12-28 21:19:55', '2015-12-28 21:19:56',
               '2015-12-28 21:19:57', '2015-12-28 21:19:58'])
df = pd.DataFrame({'s': range(8)}, index=idx)  
print (df)
                     s
2015-12-28 05:20:05  0
2015-12-28 05:20:06  1
2015-12-28 05:20:07  2
2015-12-28 05:20:08  3
2015-12-28 21:19:55  4
2015-12-28 21:19:56  5
2015-12-28 21:19:57  6
2015-12-28 21:19:58  7

d = ['2015-12-28 05:20:05', '2015-12-28 21:19:58']
print (df.loc[pd.to_datetime(d)])
                     s
2015-12-28 05:20:05  0
2015-12-28 21:19:58  7

print (df.loc[pd.to_datetime(d)])
                     s
2015-12-28 05:20:05  0
2015-12-28 21:19:58  7