一尘不染

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

python

一个pandas DataFrame列duration包含timedelta64[ns]如下所示。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]

我尝试了以下

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

但是得到了错误

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

也尝试过

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

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]]

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2020-12-20

共1个答案

一尘不染

在当前版本的Pandas(版本0.14)中,这可以正常工作:

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

这是较旧版本的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和datetime64数据在内部存储为8字节整数(dtype
'<i8')。因此,以上将timedelta64s视为8字节整数,然后进行整数除法将纳秒转换为秒。

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

2020-12-20