在字符串列表上使用strptime().不能使用循环

Question

我有一个字符串列表(unicode).像这样:

>>> tstamp
[u'2017-08-08T08:51:20.465Z', u'2017-08-08T08:51:27.871Z', u'2017-08-08T08:51:33.399Z', u'2017-08-08T08:51:37.530Z', u'2017-08-08T08:51:47.248Z', u'2017-08-08T08:51:50.414Z', u'2017-08-08T08:51:54.707Z', u'2017-08-08T08:51:54.781Z']

我想将此字符串列表转换为日期时间对象列表.像这样:

>>> dtstamp
[datetime.datetime(2017, 8, 8, 8, 51, 20, 465000), datetime.datetime(2017, 8, 8, 8, 51, 27, 871000), datetime.datetime(2017, 8, 8, 8, 51, 33, 399000), datetime.datetime(2017, 8, 8, 8, 51, 37, 530000), datetime.datetime(2017, 8, 8, 8, 51, 47, 248000), datetime.datetime(2017, 8, 8, 8, 51, 50, 414000), datetime.datetime(2017, 8, 8, 8, 51, 54, 707000), datetime.datetime(2017, 8, 8, 8, 51, 54, 781000)]

我的解决方案非常粗糙,我希望在不使用任何循环的情况下进行此转换.转换速度至关重要.到目前为止,这是我的代码:

dtstamp = [0]*len(tstamp)
for i in range(0,len(tstamp)):
     dtstamp[i] = datetime.datetime.strptime(tstamp[i], '%Y-%m-%dT%H:%M:%S.%fZ')

它做我想做的事,但会很慢.我想过尝试这个,但不起作用:

dtstamp = datetime.datetime.strptime(tstamp, '%Y-%m-%dT%H:%M:%S.%fZ')

任何人都可以指出我正确的方向？

提前致谢!

Answer 1

只需pd.to_datetime在列表中使用就可以获得显着的加速.但是,即使您可以调整方法,我也不认为您每秒都会获得600,000次转换.

import pandas as pd
import datetime as dt

my_list = [u'2017-08-08T08:51:20.465Z', u'2017-08-08T08:51:27.871Z', u'2017-08-08T08:51:33.399Z', u'2017-08-08T08:51:37.530Z', u'2017-08-08T08:51:47.248Z', u'2017-08-08T08:51:50.414Z', u'2017-08-08T08:51:54.707Z', u'2017-08-08T08:51:54.781Z']
new_list = []
for x in xrange(100000):
    new_list.extend(my_list)

def basic_list_approach(the_list):
    return [dt.datetime.strptime(item, '%Y-%m-%dT%H:%M:%S.%fZ') for item in the_list]

def pandas_approach(the_list):
    converted = pd.to_datetime(the_list)
    return converted

%timeit basic_list_approach(new_list)
1 loop, best of 3: 12.6 s per loop

%timeit pandas_approach(new_list)
1 loop, best of 3: 1.45 s per loop