Pandas - 如何将 RangeIndex 转换为 DateTimeIndex
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Pandas - how to convert RangeIndex into DateTimeIndex
提问by rioZg
I have the following dataframe. It is OHLC one-minute data. Obviously I need the T column to become and index in order to use time-series functionallity
我有以下数据框。它是 OHLC 一分钟数据。显然我需要 T 列成为和索引才能使用时间序列功能
C H L O T V
CHLOTV
13712 6873.0 6873.0 6873.0 6873.0 2018-01-13T17:17:00 799.448421
13713 6878.0 6878.0 6875.0 6875.0 2018-01-13T17:18:00 1707.578666
13714 6880.0 6880.0 6825.0 6825.0 2018-01-13T17:21:00 481.245707
13715 6876.0 6876.0 6876.0 6876.0 2018-01-13T17:22:00 839.177283
13716 6870.0 6878.0 6830.0 6878.0 2018-01-13T17:23:00 4336.830277
I used:
我用了：
df['T'] = pd.to_datetime(df['T'])
So far so good! The T column is now recognised as a date
到现在为止还挺好！T 列现在被识别为日期
Check:
查看：
<class 'pandas.core.frame.DataFrame'>
DatetimeIndex: 13717 entries, 1970-01-01 00:00:00 to 1970-01-01 00:00:00.000013716
Data columns (total 7 columns):
BV 13717 non-null float64
C 13717 non-null float64
H 13717 non-null float64
L 13717 non-null float64
O 13717 non-null float64
T 13717 non-null datetime64[ns]
V 13717 non-null float64
dtypes: datetime64[ns](1), float64(6)
memory usage: 857.3 KB
And now comes the fun and unexplainable part:
现在是有趣且无法解释的部分：
df.set_index(df['T'])
C H L O T V
T
2018-01-03 17:27:00 5710.0 5710.0 5663.0 5667.0 2018-01-03 17:27:00 3863.030204
2018-01-03 17:28:00 5704.0 5710.0 5663.0 5710.0 2018-01-03 17:28:00 1208.627542
2018-01-03 17:29:00 5699.0 5699.0 5663.0 5663.0 2018-01-03 17:29:00 1755.123688
Still looks good, but when I check the type of index I get:
看起来仍然不错，但是当我检查索引类型时，我得到：
RangeIndex(start=0, stop=13717, step=1)
And now if I try:
现在，如果我尝试：
df.index = pd.to_datetime(df.index)
I end up with:
我最终得到：
DatetimeIndex([ '1970-01-01 00:00:00',
'1970-01-01 00:00:00.000000001',
'1970-01-01 00:00:00.000000002',
'1970-01-01 00:00:00.000000003',
'1970-01-01 00:00:00.000000004' and so on...
which is evidently wrong.
这显然是错误的。
The questions are: 1. Why don't I get the normal DateTimeIndex if I am converting a date to index?
问题是： 1. 如果我将日期转换为索引，为什么我得不到正常的 DateTimeIndex？
- How can I convert that RangeIndex to DateTimeIndex with correct timestamps?
- 如何使用正确的时间戳将该 RangeIndex 转换为 DateTimeIndex？
Thanks!
谢谢！
回答by jezrael
If input data are csvthe simpliest is use parameters parse_datesand index_colin read_csv:
如果输入数据是csv最简单的使用参数parse_dates和index_colin read_csv：
df = pd.read_csv(file, parse_dates=['T'], index_col=['T'])
If not, then use your solution, don't forget assign back output of set_indexand if need drop column Talso after DatetimeIndexuse Tinstead df['T']:
如果没有，则使用您的解决方案，不要忘记分配后出set_index，如果需要删除列T后也DatetimeIndex使用T，而不是df['T']：
df['T'] = pd.to_datetime('T')
df = df.set_index('T')
#alternative solution
#df.set_index('T', inplace=True)
Why don't I get the normal DateTimeIndex if I am converting a date to index?
如果我将日期转换为索引，为什么我得不到正常的 DateTimeIndex？
Because your index is default (0,1,2..), so df.index = pd.to_datetime(df.index)parse integerss like nsand get weird datetimes.
因为您的索引是默认的 ( 0,1,2..)，所以df.index = pd.to_datetime(df.index)解析integerssns并获得奇怪的日期时间。
