如何将以下字符串转换为datetime对象?
"Jun 1 2005 1:33PM"
如何将以下字符串转换为datetime对象?
"Jun 1 2005 1:33PM"
当前回答
In [34]: import datetime
In [35]: _now = datetime.datetime.now()
In [36]: _now
Out[36]: datetime.datetime(2016, 1, 19, 9, 47, 0, 432000)
In [37]: print _now
2016-01-19 09:47:00.432000
In [38]: _parsed = datetime.datetime.strptime(str(_now),"%Y-%m-%d %H:%M:%S.%f")
In [39]: _parsed
Out[39]: datetime.datetime(2016, 1, 19, 9, 47, 0, 432000)
In [40]: assert _now == _parsed
其他回答
我已经完成了一个项目,可以转换一些非常整洁的表达式。查看时间字符串。
以下是一些示例:
pip install timestring
>>> import timestring
>>> timestring.Date('monday, aug 15th 2015 at 8:40 pm')
<timestring.Date 2015-08-15 20:40:00 4491909392>
>>> timestring.Date('monday, aug 15th 2015 at 8:40 pm').date
datetime.datetime(2015, 8, 15, 20, 40)
>>> timestring.Range('next week')
<timestring.Range From 03/10/14 00:00:00 to 03/03/14 00:00:00 4496004880>
>>> (timestring.Range('next week').start.date, timestring.Range('next week').end.date)
(datetime.datetime(2014, 3, 10, 0, 0), datetime.datetime(2014, 3, 14, 0, 0))
Use:
emp = pd.read_csv("C:\\py\\programs\\pandas_2\\pandas\\employees.csv")
emp.info()
它显示“开始日期时间”列和“上次登录时间”都是数据帧中的“对象=字符串”:
<class 'pandas.core.frame.DataFrame'>
RangeIndex: 1000 entries, 0 to 999
Data columns (total 8 columns):
First Name 933 non-null object
Gender 855 non-null object
Start Date 1000 non-null object
Last Login Time 1000 non-null object
Salary 1000 non-null int64
Bonus % 1000 non-null float64
Senior Management 933 non-null object
Team 957 non-null object
dtypes: float64(1), int64(1), object(6)
memory usage: 62.6+ KB
通过使用read_csv中的parse_dates选项,可以将字符串datetime转换为panda datetime格式。
emp = pd.read_csv("C:\\py\\programs\\pandas_2\\pandas\\employees.csv", parse_dates=["Start Date", "Last Login Time"])
emp.info()
输出:
<class 'pandas.core.frame.DataFrame'>
RangeIndex: 1000 entries, 0 to 999
Data columns (total 8 columns):
First Name 933 non-null object
Gender 855 non-null object
Start Date 1000 non-null datetime64[ns]
Last Login Time 1000 non-null datetime64[ns]
Salary 1000 non-null int64
Bonus % 1000 non-null float64
Senior Management 933 non-null object
Team 957 non-null object
dtypes: datetime64[ns](2), float64(1), int64(1), object(4)
memory usage: 62.6+ KB
这里没有提到但很有用的一点:在当天添加后缀。我解耦了后缀逻辑,这样你就可以将它用于任何你喜欢的数字,而不仅仅是日期。
import time
def num_suffix(n):
'''
Returns the suffix for any given int
'''
suf = ('th','st', 'nd', 'rd')
n = abs(n) # wise guy
tens = int(str(n)[-2:])
units = n % 10
if tens > 10 and tens < 20:
return suf[0] # teens with 'th'
elif units <= 3:
return suf[units]
else:
return suf[0] # 'th'
def day_suffix(t):
'''
Returns the suffix of the given struct_time day
'''
return num_suffix(t.tm_mday)
# Examples
print num_suffix(123)
print num_suffix(3431)
print num_suffix(1234)
print ''
print day_suffix(time.strptime("1 Dec 00", "%d %b %y"))
print day_suffix(time.strptime("2 Nov 01", "%d %b %y"))
print day_suffix(time.strptime("3 Oct 02", "%d %b %y"))
print day_suffix(time.strptime("4 Sep 03", "%d %b %y"))
print day_suffix(time.strptime("13 Nov 90", "%d %b %y"))
print day_suffix(time.strptime("14 Oct 10", "%d %b %y"))
我个人喜欢使用解析器模块的解决方案,这是这个问题的第二个答案,非常漂亮,因为您不必构造任何字符串文字就能使其工作。但是,一个缺点是它比strptime的公认答案慢了90%。
from dateutil import parser
from datetime import datetime
import timeit
def dt():
dt = parser.parse("Jun 1 2005 1:33PM")
def strptime():
datetime_object = datetime.strptime('Jun 1 2005 1:33PM', '%b %d %Y %I:%M%p')
print(timeit.timeit(stmt=dt, number=10**5))
print(timeit.timeit(stmt=strptime, number=10**5))
输出:
10.702968013429021.3627995655316933
只要你不反复做一百万次,我仍然认为解析器方法更方便,并且可以自动处理大多数时间格式。
In [34]: import datetime
In [35]: _now = datetime.datetime.now()
In [36]: _now
Out[36]: datetime.datetime(2016, 1, 19, 9, 47, 0, 432000)
In [37]: print _now
2016-01-19 09:47:00.432000
In [38]: _parsed = datetime.datetime.strptime(str(_now),"%Y-%m-%d %H:%M:%S.%f")
In [39]: _parsed
Out[39]: datetime.datetime(2016, 1, 19, 9, 47, 0, 432000)
In [40]: assert _now == _parsed