在 Python Pandas DataFrame 中将 timedelta64[ns] 列转换为秒

Convert timedelta64[ns] column to seconds in Python Pandas DataFrame

提问by Nyxynyx

A pandas DataFrame column durationcontains timedelta64[ns]as shown. How can you convert them to seconds?

Pandas DataFrame 列duration包含timedelta64[ns]如图所示。如何将它们转换为秒？

0   00:20:32
1   00:23:10
2   00:24:55
3   00:13:17
4   00:18:52
Name: duration, dtype: timedelta64[ns]

I tried the following

我尝试了以下

print df[:5]['duration'] / np.timedelta64(1, 's')

but got the error

但得到了错误

Traceback (most recent call last):
  File "test.py", line 16, in <module>
    print df[0:5]['duration'] / np.timedelta64(1, 's')
  File "C:\Python27\lib\site-packages\pandas\core\series.py", line 130, in wrapper
    "addition and subtraction, but the operator [%s] was passed" % name)
TypeError: can only operate on a timedeltas for addition and subtraction, but the operator [__div__] was passed

Also tried

也试过

print df[:5]['duration'].astype('timedelta64[s]')

but received the error

但收到错误

Traceback (most recent call last):
  File "test.py", line 17, in <module>
    print df[:5]['duration'].astype('timedelta64[s]')
  File "C:\Python27\lib\site-packages\pandas\core\series.py", line 934, in astype
    values = com._astype_nansafe(self.values, dtype)
  File "C:\Python27\lib\site-packages\pandas\core\common.py", line 1653, in _astype_nansafe
    raise TypeError("cannot astype a timedelta from [%s] to [%s]" % (arr.dtype,dtype))
TypeError: cannot astype a timedelta from [timedelta64[ns]] to [timedelta64[s]]

采纳答案by unutbu

This works properly in the current version of Pandas (version 0.14):

这在当前版本的 Pandas（0.14 版）中正常工作：

In [132]: df[:5]['duration'] / np.timedelta64(1, 's')
Out[132]: 
0    1232
1    1390
2    1495
3     797
4    1132
Name: duration, dtype: float64

Here is a workaround for older versions of Pandas/NumPy:

这是旧版本的 Pandas/NumPy 的解决方法：

In [131]: df[:5]['duration'].values.view('<i8')/10**9
Out[131]: array([1232, 1390, 1495,  797, 1132], dtype=int64)

timedelta64 and datetime64 data are stored internally as 8-byte ints (dtype '<i8'). So the above views the timedelta64s as 8-byte ints and then does integer division to convert nanoseconds to seconds.

timedelta64 和 datetime64 数据在内部存储为 8 字节整数 (dtype '<i8')。因此，上面将 timedelta64s 视为 8 字节整数，然后进行整数除法以将纳秒转换为秒。

Note that you need NumPy version 1.7 or newerto work with datetime64/timedelta64s.

请注意，您需要 NumPy 1.7 或更高版本才能使用 datetime64/timedelta64s。

回答by Gunay Anach

Just realized it's an old thread, anyway leaving it here if wanderers like me clicks only on top 5 results on the search engine and ends up here.

刚刚意识到这是一个旧线程，无论如何，如果像我这样的流浪者只点击搜索引擎上的前 5 个结果并最终到达这里，就将它留在这里。

Make sure that your types are correct.

确保您的类型正确。

• If you want to convert datetimeto seconds, just sum up seconds for each hour, minute and seconds of the datetime object if its for duration within one date.

  • • hours - hours x 3600 = seconds
  • • minutes - minutes x 60 = seconds
  • • seconds - seconds

• 如果要将datetime转换为seconds，只需将 datetime 对象的每一小时、分钟和秒的秒数相加（如果它的持续时间在一个日期内）。

  • • 小时 - 小时 x 3600 = 秒
  • • 分钟 - 分钟 x 60 = 秒
  • • 秒 - 秒

linear_df['duration'].dt.hour*3600 + linear_df['duration'].dt.minute*60 + linear_df['duration'].dt.second

linear_df['duration'].dt.hour*3600 + linear_df['duration'].dt.minute*60 + linear_df['duration'].dt.second

• If you want to convert timedeltato secondsuse the one bellow.

• 如果要将timedelta转换为秒，请使用下面的一个。

linear_df[:5]['duration'].astype('timedelta64[s]')

linear_df[:5]['duration'].astype('timedelta64[s]')

I got it to work like this:

我让它像这样工作：

start_dt and end_dt columns are in this format:

start_dt 和 end_dt 列采用以下格式：

import datetime

linear_df[:5]['start_dt']

0   1970-02-22 21:32:48.000
1   2016-12-30 17:47:33.216
2   2016-12-31 09:33:27.931
3   2016-12-31 09:52:53.486
4   2016-12-31 10:29:44.611
Name: start_dt, dtype: datetime64[ns]

Had my duration in timedelta64[ns] format, which was subtraction of startand enddatetime values.

我的持续时间为 timedelta64[ns] 格式，这是减去开始和结束日期时间值。

linear_df['duration'] = linear_df['end_dt'] - linear_df['start_dt']

Resulted duration column look like this

结果持续时间列如下所示

linear_df[:5]['duration']

0          0 days 00:00:14
1   2 days 17:44:50.558000
2   0 days 15:37:28.418000
3   0 days 18:45:45.727000
4   0 days 19:21:27.159000
Name: duration, dtype: timedelta64[ns]

Using pandas I had my duration seconds between two dates in float. Easier to compare or filter your duration afterwards.

使用熊猫，我的两个日期之间的持续时间秒数处于浮动状态。之后更容易比较或过滤您的持续时间。

linear_df[:5]['duration'].astype('timedelta64[s]')

0        14.0
1    236690.0
2     56248.0
3     67545.0
4     69687.0
Name: duration, dtype: float64

In my case if I want to get all duration which is more than 1 second.

就我而言，如果我想获得超过 1 秒的所有持续时间。

Hope it helps.

希望能帮助到你。

回答by Pardhu

We can simply use the pandas apply()function

我们可以简单地使用 pandas apply()函数

def get_seconds(time_delta):
    return time_delta.seconds

def get_microseconds(time_delta):
    return time_delta.micro_seconds

time_delta_series = df['duration']

converted_series = time_delta_series.apply(get_seconds)
print(converted_series)

回答by wwii

Use the Series dt accessorto get access to the methods and attributes of a datetime (timedelta) series.

使用Series dt 访问器来访问日期时间 (timedelta) 系列的方法和属性。

>>> s
0   -1 days +23:45:14.304000
1   -1 days +23:46:57.132000
2   -1 days +23:49:25.913000
3   -1 days +23:59:48.913000
4            00:00:00.820000
dtype: timedelta64[ns]
>>>
>>> s.dt.total_seconds()
0   -885.696
1   -782.868
2   -634.087
3    -11.087
4      0.820
dtype: float64

---

---

There are other Pandas Series Accessorsfor String, Categorical, and Sparse data types.

还有其他用于字符串、分类和稀疏数据类型的Pandas系列访问器。

回答by AntoineP

Use the 'total_seconds()' function :

使用“total_seconds()”函数：

df['durationSeconds'] = df['duration'].dt.total_seconds()