我有 numpy 字符串数组(附注:为什么字符串表示为对象?!)
t = array(['21/02/2014 08:40:00 AM', '11/02/2014 10:50:00 PM',
'07/04/2014 05:50:00 PM', '17/02/2014 10:20:00 PM',
'07/03/2014 06:10:00 AM', '02/03/2014 12:25:00 PM',
'05/02/2014 03:20:00 AM', '31/01/2014 12:30:00 AM',
'28/02/2014 01:25:00 PM'], dtype=object)
我想将它转换为具有日分辨率的 numpy.datetime64,但是我找到的唯一解决方案是:
t = [datetime.strptime(tt,"%d/%m/%Y %H:%M:%S %p") for tt in t]
t = np.array(t,dtype='datetime64[us]').astype('datetime64[D]')
还有比这更丑陋的吗?为什么我需要遍历 native Python 列表? 必须有另一种方式......
顺便说一句,我找不到在 numpy/pandas 中绘制日期直方图的方法
最佳答案
日期格式是问题所在,01/01/2015
不明确,如果它在 ISO 8601 中,您可以使用 numpy 直接解析它,在您的情况下,因为您只需要日期然后拆分并且重新排列数据会明显更快:
t = np.array([datetime.strptime(d.split(None)[0], "%d/%m/%Y")
for d in t],dtype='datetime64[us]').astype('datetime64[D]')
一些时序,先解析后重排:
In [36]: %%timeit
from datetime import datetime
t = np.array(['21/02/2014 08:40:00', '11/02/2014 10:50:00 PM',
'07/04/2014 05:50:00 PM', '17/02/2014 10:20:00 PM',
'07/03/2014 06:10:00 AM', '02/03/2014 12:25:00 PM',
'05/02/2014 03:20:00 AM', '31/01/2014 12:30:00 AM',
'28/02/2014 01:25:00 PM']*10000)
t1 = np.array([np.datetime64("{}-{}-{}".format(c[:4], b, a)) for a, b, c in (s.split("/", 2) for s in t)])
....:
10 loops, best of 3: 125 ms per loop
您的代码:
In [37]: %%timeit
from datetime import datetime
t = np.array(['21/02/2014 08:40:00 AM', '11/02/2014 10:50:00 PM',
'07/04/2014 05:50:00 PM', '17/02/2014 10:20:00 PM',
'07/03/2014 06:10:00 AM', '02/03/2014 12:25:00 PM',
'05/02/2014 03:20:00 AM', '31/01/2014 12:30:00 AM',
'28/02/2014 01:25:00 PM']*10000)
t = [datetime.strptime(tt,"%d/%m/%Y %H:%M:%S %p") for tt in t]
t = np.array(t,dtype='datetime64[us]').astype('datetime64[D]')
....:
1 loops, best of 3: 1.56 s per loop
两者给出相同结果的显着差异:
In [48]: t = np.array(['21/02/2014 08:40:00 AM', '11/02/2014 10:50:00 PM',
'07/04/2014 05:50:00 PM', '17/02/2014 10:20:00 PM',
'07/03/2014 06:10:00 AM', '02/03/2014 12:25:00 PM',
'05/02/2014 03:20:00 AM', '31/01/2014 12:30:00 AM',
'28/02/2014 01:25:00 PM'] * 10000)
In [49]: t1 = [datetime.strptime(tt,"%d/%m/%Y %H:%M:%S %p") for tt in t]
t1 = np.array(t1,dtype='datetime64[us]').astype('datetime64[D]')
....:
In [50]: t2 = np.array([np.datetime64("{}-{}-{}".format(c[:4], b, a)) for a, b, c in (s.split("/", 2) for s in t)])
In [51]: (t1 == t2).all()
Out[51]: True
关于python - 在numpy中将文本转换为datetime64,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/25471299/