[英]Convert pandas DateTimeIndex to Unix Time?
What is the idiomatic way of converting a pandas DateTimeIndex to (an iterable of) Unix Time?将 pandas DateTimeIndex 转换为(可迭代的)Unix 时间的惯用方法是什么? This is probably not the way to go:
这可能不是要走的路:
[time.mktime(t.timetuple()) for t in my_data_frame.index.to_pydatetime()]
As DatetimeIndex
is ndarray
under the hood, you can do the conversion without a comprehension (much faster). 由于
DatetimeIndex
是ndarray
,你可以在没有理解的情况下进行转换(更快)。
In [1]: import numpy as np
In [2]: import pandas as pd
In [3]: from datetime import datetime
In [4]: dates = [datetime(2012, 5, 1), datetime(2012, 5, 2), datetime(2012, 5, 3)]
...: index = pd.DatetimeIndex(dates)
...:
In [5]: index.astype(np.int64)
Out[5]: array([1335830400000000000, 1335916800000000000, 1336003200000000000],
dtype=int64)
In [6]: index.astype(np.int64) // 10**9
Out[6]: array([1335830400, 1335916800, 1336003200], dtype=int64)
%timeit [t.value // 10 ** 9 for t in index]
10000 loops, best of 3: 119 us per loop
%timeit index.astype(np.int64) // 10**9
100000 loops, best of 3: 18.4 us per loop
Note: Timestamp is just unix time with nanoseconds (so divide it by 10**9): 注意:时间戳只是unix时间,以纳秒为单位(因此除以10 ** 9):
[t.value // 10 ** 9 for t in tsframe.index]
For example: 例如:
In [1]: t = pd.Timestamp('2000-02-11 00:00:00')
In [2]: t
Out[2]: <Timestamp: 2000-02-11 00:00:00>
In [3]: t.value
Out[3]: 950227200000000000L
In [4]: time.mktime(t.timetuple())
Out[4]: 950227200.0
As @root points out it's faster to extract the array of values directly: 正如@root指出的那样,直接提取值数组更快:
tsframe.index.astype(np.int64) // 10 ** 9
A summary of other answers: 其他答案摘要:
df['<time_col>'].astype(np.int64) // 10**9
If you want to keep the milliseconds divide by 10**6
instead 如果你想保持毫秒除以
10**6
Complementing the other answers: //10**9
will do a flooring divide, which gives full past seconds rather than the nearest value in seconds. 补充其他答案:
//10**9
将执行地板划分,它会提供完整的过去秒数,而不是最接近的秒数值。 A simple way to get more reasonable rounding, if that is desired, is to add 5*10**8 - 1
before doing the flooring divide. 如果需要,一种简单的方法可以获得更合理的舍入,即在进行地板划分之前增加
5*10**8 - 1
。
To address the case of NaT, which above solutions will convert to large negative ints, in pandas>=0.24 a possible solution would be: 为了解决NaT的情况,上面的解决方案将转换为大的负面整数,在pandas> = 0.24中,可能的解决方案是:
def datetime_to_epoch(ser):
"""Don't convert NaT to large negative values."""
if ser.hasnans:
res = ser.dropna().astype('int64').astype('Int64').reindex(index=ser.index)
else:
res = ser.astype('int64')
return res // 10**9
In the case of missing values this will return the nullable int type 'Int64' (ExtensionType pd.Int64Dtype): 在缺少值的情况下,这将返回nullable int类型'Int64'(ExtensionType pd.Int64Dtype):
In [5]: dt = pd.to_datetime(pd.Series(["2019-08-21", "2018-07-28", np.nan]))
In [6]: datetime_to_epoch(dt)
Out[6]:
0 1566345600
1 1532736000
2 NaN
dtype: Int64
Otherwise a regular int64: 否则一个常规的int64:
In [7]: datetime_to_epoch(dt[:2])
Out[7]:
0 1566345600
1 1532736000
dtype: int64
If you have tried this on the datetime column of your dataframe:如果您在数据框的日期时间列上尝试过此操作:
dframe['datetime'].astype(np.int64) // 10**9
& that you are struggling with the following error: TypeError: int() argument must be a string, a bytes-like object or a number, not 'Timestamp'
you can just use these two lines : &您正在努力解决以下错误:
TypeError: int() argument must be a string, a bytes-like object or a number, not 'Timestamp'
您可以使用这两行:
dframe.index = pd.DatetimeIndex(dframe['datetime'])
dframe['datetime']= dframe.index.astype(np.int64)// 10**9
The code from the other answers其他答案的代码
dframe['datetime'].astype(np.int64) // 10**9
prints the following warning as of the time of my post:截至我发帖时打印以下警告:
FutureWarning: casting datetime64[ns] values to int64 with .astype(...) is deprecated and will raise in a future version.
FutureWarning:不推荐使用 .astype(...) 将 datetime64[ns] 值转换为 int64,并将在未来版本中提出。 Use .view(...) instead.
使用 .view(...) 代替。
So use the following instead:因此,请改用以下内容:
dframe['datetime'].view(np.int64) // 10 ** 9
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.