By using:
import os,os.path,time,shutil,datetime
print time.ctime(os.path.getmtime("/home/sulata/Documents/source/"))
print time.ctime(os.path.getmtime("/home/sulata/Documents/destination/"))
I'm getting the outputs:
Mon Apr 2 15:56:00 2018
Mon Apr 2 15:56:03 2018
I want to get the time without seconds.
The general method is:
import time
time.strftime(format)
example:
>>>time.strftime("%H:%M:%S")
20:08:40
In your case:
>>> time.strftime("%H:%M")
13:41
>>>time.strftime("%a %b %d %H:%M %Y")
'Mon Apr 02 13:27 2018'
if you want to remove the Zero, you could do something like this...
>>> time.strftime("%a %b "+str(int(time.strftime("%d"))) +" %H:%M %Y")
'Mon Apr 2 13:33 2018'
Use datetime to get the seconds.
Ex:
import os,os.path,time,shutil,datetime
print datetime.datetime.strptime(time.ctime(os.path.getmtime(r"/home/sulata/Documents/source/")), "%a %b %d %H:%M:%S %Y").second
JulioCamPlaz has already provided a decent enough solution, but if you want to maintain the date format, you can use regex for this:
import re
x = time.ctime(os.path.getmtime("/home/sulata/Documents/source/"))
print(re.sub(r':\d{1,2}\s', ' ', x))
What this does is that it removes the final two digits (seconds) of the time which are followed by a space, and replaces it with a space.
Although this may be an over-complicated method, it is short, and gives you the exact same date format without the seconds, and without altering it in any way.
getmtime returns a timestamp, so you can format it using strftime to whatever format you need:
>>> import datetime as dt
>>> import os
>>> mTime = os.path.getmtime("/tmp/xauth-1000-_0")
>>> dt.datetime.fromtimestamp(mTime).strftime("%Y-%m-%d %H:%M")
'2018-04-01 22:07'
Format identifiers can be found in the docs.
You can try this -
>>> import time, os
>>>
>>> x = time.gmtime(os.path.getmtime('/home/abhi/test.file'))
>>>
>>> x
time.struct_time(tm_year=2018, tm_mon=4, tm_mday=2, tm_hour=11, tm_min=19, tm_sec=43, tm_wday=0, tm_yday=92, tm_isdst=0)
>>> x.tm_sec
43
>>> x.tm_hour
11
>>>
Related
I have a date string with the format 'Mon Feb 15 2010'. I want to change the format to '15/02/2010'. How can I do this?
datetime module could help you with that:
datetime.datetime.strptime(date_string, format1).strftime(format2)
For the specific example you could do
>>> import datetime
>>> datetime.datetime.strptime('Mon Feb 15 2010', '%a %b %d %Y').strftime('%d/%m/%Y')
'15/02/2010'
>>>
You can install the dateutil library. Its parse function can figure out what format a string is in without having to specify the format like you do with datetime.strptime.
from dateutil.parser import parse
dt = parse('Mon Feb 15 2010')
print(dt)
# datetime.datetime(2010, 2, 15, 0, 0)
print(dt.strftime('%d/%m/%Y'))
# 15/02/2010
convert string to datetime object
from datetime import datetime
s = "2016-03-26T09:25:55.000Z"
f = "%Y-%m-%dT%H:%M:%S.%fZ"
out = datetime.strptime(s, f)
print(out)
output:
2016-03-26 09:25:55
>>> from_date="Mon Feb 15 2010"
>>> import time
>>> conv=time.strptime(from_date,"%a %b %d %Y")
>>> time.strftime("%d/%m/%Y",conv)
'15/02/2010'
As this question comes often, here is the simple explanation.
datetime or time module has two important functions.
strftime - creates a string representation of date or time from a datetime or time object.
strptime - creates a datetime or time object from a string.
In both cases, we need a formating string. It is the representation that tells how the date or time is formatted in your string.
Now lets assume we have a date object.
>>> from datetime import datetime
>>> d = datetime(2010, 2, 15)
>>> d
datetime.datetime(2010, 2, 15, 0, 0)
If we want to create a string from this date in the format 'Mon Feb 15 2010'
>>> s = d.strftime('%a %b %d %y')
>>> print s
Mon Feb 15 10
Lets assume we want to convert this s again to a datetime object.
>>> new_date = datetime.strptime(s, '%a %b %d %y')
>>> print new_date
2010-02-15 00:00:00
Refer This document all formatting directives regarding datetime.
#codeling and #user1767754 : The following two lines will work. I saw no one posted the complete solution for the example problem that was asked. Hopefully this is enough explanation.
import datetime
x = datetime.datetime.strptime("Mon Feb 15 2010", "%a %b %d %Y").strftime("%d/%m/%Y")
print(x)
Output:
15/02/2010
You may achieve this using pandas as well:
import pandas as pd
pd.to_datetime('Mon Feb 15 2010', format='%a %b %d %Y').strftime('%d/%m/%Y')
Output:
'15/02/2010'
You may apply pandas approach for different datatypes as:
import pandas as pd
import numpy as np
def reformat_date(date_string, old_format, new_format):
return pd.to_datetime(date_string, format=old_format, errors='ignore').strftime(new_format)
date_string = 'Mon Feb 15 2010'
date_list = ['Mon Feb 15 2010', 'Wed Feb 17 2010']
date_array = np.array(date_list)
date_series = pd.Series(date_list)
old_format = '%a %b %d %Y'
new_format = '%d/%m/%Y'
print(reformat_date(date_string, old_format, new_format))
print(reformat_date(date_list, old_format, new_format).values)
print(reformat_date(date_array, old_format, new_format).values)
print(date_series.apply(lambda x: reformat_date(x, old_format, new_format)).values)
Output:
15/02/2010
['15/02/2010' '17/02/2010']
['15/02/2010' '17/02/2010']
['15/02/2010' '17/02/2010']
Just for the sake of completion: when parsing a date using strptime() and the date contains the name of a day, month, etc, be aware that you have to account for the locale.
It's mentioned as a footnote in the docs as well.
As an example:
import locale
print(locale.getlocale())
>> ('nl_BE', 'ISO8859-1')
from datetime import datetime
datetime.strptime('6-Mar-2016', '%d-%b-%Y').strftime('%Y-%m-%d')
>> ValueError: time data '6-Mar-2016' does not match format '%d-%b-%Y'
locale.setlocale(locale.LC_ALL, 'en_US')
datetime.strptime('6-Mar-2016', '%d-%b-%Y').strftime('%Y-%m-%d')
>> '2016-03-06'
use datetime library
http://docs.python.org/library/datetime.html look up 9.1.7.
especiall strptime() strftime() Behavior¶
examples
http://pleac.sourceforge.net/pleac_python/datesandtimes.html
If you dont want to define the input date format then, Install dateparser (pip install dateparser) and,
from dateparser import parse
parse("Mon Feb 15 2010").strftime("%d/%m/%Y")
I have a date string with the format 'Mon Feb 15 2010'. I want to change the format to '15/02/2010'. How can I do this?
datetime module could help you with that:
datetime.datetime.strptime(date_string, format1).strftime(format2)
For the specific example you could do
>>> import datetime
>>> datetime.datetime.strptime('Mon Feb 15 2010', '%a %b %d %Y').strftime('%d/%m/%Y')
'15/02/2010'
>>>
You can install the dateutil library. Its parse function can figure out what format a string is in without having to specify the format like you do with datetime.strptime.
from dateutil.parser import parse
dt = parse('Mon Feb 15 2010')
print(dt)
# datetime.datetime(2010, 2, 15, 0, 0)
print(dt.strftime('%d/%m/%Y'))
# 15/02/2010
convert string to datetime object
from datetime import datetime
s = "2016-03-26T09:25:55.000Z"
f = "%Y-%m-%dT%H:%M:%S.%fZ"
out = datetime.strptime(s, f)
print(out)
output:
2016-03-26 09:25:55
>>> from_date="Mon Feb 15 2010"
>>> import time
>>> conv=time.strptime(from_date,"%a %b %d %Y")
>>> time.strftime("%d/%m/%Y",conv)
'15/02/2010'
As this question comes often, here is the simple explanation.
datetime or time module has two important functions.
strftime - creates a string representation of date or time from a datetime or time object.
strptime - creates a datetime or time object from a string.
In both cases, we need a formating string. It is the representation that tells how the date or time is formatted in your string.
Now lets assume we have a date object.
>>> from datetime import datetime
>>> d = datetime(2010, 2, 15)
>>> d
datetime.datetime(2010, 2, 15, 0, 0)
If we want to create a string from this date in the format 'Mon Feb 15 2010'
>>> s = d.strftime('%a %b %d %y')
>>> print s
Mon Feb 15 10
Lets assume we want to convert this s again to a datetime object.
>>> new_date = datetime.strptime(s, '%a %b %d %y')
>>> print new_date
2010-02-15 00:00:00
Refer This document all formatting directives regarding datetime.
#codeling and #user1767754 : The following two lines will work. I saw no one posted the complete solution for the example problem that was asked. Hopefully this is enough explanation.
import datetime
x = datetime.datetime.strptime("Mon Feb 15 2010", "%a %b %d %Y").strftime("%d/%m/%Y")
print(x)
Output:
15/02/2010
You may achieve this using pandas as well:
import pandas as pd
pd.to_datetime('Mon Feb 15 2010', format='%a %b %d %Y').strftime('%d/%m/%Y')
Output:
'15/02/2010'
You may apply pandas approach for different datatypes as:
import pandas as pd
import numpy as np
def reformat_date(date_string, old_format, new_format):
return pd.to_datetime(date_string, format=old_format, errors='ignore').strftime(new_format)
date_string = 'Mon Feb 15 2010'
date_list = ['Mon Feb 15 2010', 'Wed Feb 17 2010']
date_array = np.array(date_list)
date_series = pd.Series(date_list)
old_format = '%a %b %d %Y'
new_format = '%d/%m/%Y'
print(reformat_date(date_string, old_format, new_format))
print(reformat_date(date_list, old_format, new_format).values)
print(reformat_date(date_array, old_format, new_format).values)
print(date_series.apply(lambda x: reformat_date(x, old_format, new_format)).values)
Output:
15/02/2010
['15/02/2010' '17/02/2010']
['15/02/2010' '17/02/2010']
['15/02/2010' '17/02/2010']
Just for the sake of completion: when parsing a date using strptime() and the date contains the name of a day, month, etc, be aware that you have to account for the locale.
It's mentioned as a footnote in the docs as well.
As an example:
import locale
print(locale.getlocale())
>> ('nl_BE', 'ISO8859-1')
from datetime import datetime
datetime.strptime('6-Mar-2016', '%d-%b-%Y').strftime('%Y-%m-%d')
>> ValueError: time data '6-Mar-2016' does not match format '%d-%b-%Y'
locale.setlocale(locale.LC_ALL, 'en_US')
datetime.strptime('6-Mar-2016', '%d-%b-%Y').strftime('%Y-%m-%d')
>> '2016-03-06'
use datetime library
http://docs.python.org/library/datetime.html look up 9.1.7.
especiall strptime() strftime() Behavior¶
examples
http://pleac.sourceforge.net/pleac_python/datesandtimes.html
If you dont want to define the input date format then, Install dateparser (pip install dateparser) and,
from dateparser import parse
parse("Mon Feb 15 2010").strftime("%d/%m/%Y")
I have a date string with the format 'Mon Feb 15 2010'. I want to change the format to '15/02/2010'. How can I do this?
datetime module could help you with that:
datetime.datetime.strptime(date_string, format1).strftime(format2)
For the specific example you could do
>>> import datetime
>>> datetime.datetime.strptime('Mon Feb 15 2010', '%a %b %d %Y').strftime('%d/%m/%Y')
'15/02/2010'
>>>
You can install the dateutil library. Its parse function can figure out what format a string is in without having to specify the format like you do with datetime.strptime.
from dateutil.parser import parse
dt = parse('Mon Feb 15 2010')
print(dt)
# datetime.datetime(2010, 2, 15, 0, 0)
print(dt.strftime('%d/%m/%Y'))
# 15/02/2010
convert string to datetime object
from datetime import datetime
s = "2016-03-26T09:25:55.000Z"
f = "%Y-%m-%dT%H:%M:%S.%fZ"
out = datetime.strptime(s, f)
print(out)
output:
2016-03-26 09:25:55
>>> from_date="Mon Feb 15 2010"
>>> import time
>>> conv=time.strptime(from_date,"%a %b %d %Y")
>>> time.strftime("%d/%m/%Y",conv)
'15/02/2010'
As this question comes often, here is the simple explanation.
datetime or time module has two important functions.
strftime - creates a string representation of date or time from a datetime or time object.
strptime - creates a datetime or time object from a string.
In both cases, we need a formating string. It is the representation that tells how the date or time is formatted in your string.
Now lets assume we have a date object.
>>> from datetime import datetime
>>> d = datetime(2010, 2, 15)
>>> d
datetime.datetime(2010, 2, 15, 0, 0)
If we want to create a string from this date in the format 'Mon Feb 15 2010'
>>> s = d.strftime('%a %b %d %y')
>>> print s
Mon Feb 15 10
Lets assume we want to convert this s again to a datetime object.
>>> new_date = datetime.strptime(s, '%a %b %d %y')
>>> print new_date
2010-02-15 00:00:00
Refer This document all formatting directives regarding datetime.
#codeling and #user1767754 : The following two lines will work. I saw no one posted the complete solution for the example problem that was asked. Hopefully this is enough explanation.
import datetime
x = datetime.datetime.strptime("Mon Feb 15 2010", "%a %b %d %Y").strftime("%d/%m/%Y")
print(x)
Output:
15/02/2010
You may achieve this using pandas as well:
import pandas as pd
pd.to_datetime('Mon Feb 15 2010', format='%a %b %d %Y').strftime('%d/%m/%Y')
Output:
'15/02/2010'
You may apply pandas approach for different datatypes as:
import pandas as pd
import numpy as np
def reformat_date(date_string, old_format, new_format):
return pd.to_datetime(date_string, format=old_format, errors='ignore').strftime(new_format)
date_string = 'Mon Feb 15 2010'
date_list = ['Mon Feb 15 2010', 'Wed Feb 17 2010']
date_array = np.array(date_list)
date_series = pd.Series(date_list)
old_format = '%a %b %d %Y'
new_format = '%d/%m/%Y'
print(reformat_date(date_string, old_format, new_format))
print(reformat_date(date_list, old_format, new_format).values)
print(reformat_date(date_array, old_format, new_format).values)
print(date_series.apply(lambda x: reformat_date(x, old_format, new_format)).values)
Output:
15/02/2010
['15/02/2010' '17/02/2010']
['15/02/2010' '17/02/2010']
['15/02/2010' '17/02/2010']
Just for the sake of completion: when parsing a date using strptime() and the date contains the name of a day, month, etc, be aware that you have to account for the locale.
It's mentioned as a footnote in the docs as well.
As an example:
import locale
print(locale.getlocale())
>> ('nl_BE', 'ISO8859-1')
from datetime import datetime
datetime.strptime('6-Mar-2016', '%d-%b-%Y').strftime('%Y-%m-%d')
>> ValueError: time data '6-Mar-2016' does not match format '%d-%b-%Y'
locale.setlocale(locale.LC_ALL, 'en_US')
datetime.strptime('6-Mar-2016', '%d-%b-%Y').strftime('%Y-%m-%d')
>> '2016-03-06'
use datetime library
http://docs.python.org/library/datetime.html look up 9.1.7.
especiall strptime() strftime() Behavior¶
examples
http://pleac.sourceforge.net/pleac_python/datesandtimes.html
If you dont want to define the input date format then, Install dateparser (pip install dateparser) and,
from dateparser import parse
parse("Mon Feb 15 2010").strftime("%d/%m/%Y")
I'm trying to make a simple system which would get the current time and get another time after few secs, then see the difference with both of the times, so it's 2 seconds. So what I need is the other format like this > YEAR,MONTH,DAY HOUR:MIN.
This is the code which I use for this purpose, but in brackets there are just an example of the format I need.
a = datetime.datetime.now( %Y, %m %d %H:%M)
time.sleep(2)
b = datetime.datetime.now( %Y, %m %d %H:%M)
print(b-a)
print(a)
print(b)
I thing strftime is what you're looking for
import datetime
import time
a = datetime.datetime.now()
time.sleep(2)
b = datetime.datetime.now()
print(b-a)
print(a.strftime("%Y, %m, %d %H:%M"))
print(b.strftime("%Y, %m, %d %H:%M"))
prints
0:00:02.001719
2019, 09, 04 15:17
2019, 09, 04 15:17
For more formats, you can see strftime reference: https://www.programiz.com/python-programming/datetime/strftime.
You can convert a datetime.datetime instance to a string formatted to your liking using the strftime() function. For instance, to print with your preferred formatting you could do the following:
>>> import datetime
>>> a = datetime.datetime.now()
>>> print(a.strftime("%Y, %m %d %H:%M")
2019, 09 04 17:11
Subtracting two dates will yield a datetime.timedelta object, you can convert this to the number of seconds using the total_seconds() function:
>>> import datetime
>>> from time import sleep
>>> a = datetime.datetime.now()
>>> sleep(2)
>>> b = datetime.datetime.now()
>>> delta = b - a
>>> print(delta.total_seconds())
2.001301
I have a date string with the format 'Mon Feb 15 2010'. I want to change the format to '15/02/2010'. How can I do this?
datetime module could help you with that:
datetime.datetime.strptime(date_string, format1).strftime(format2)
For the specific example you could do
>>> import datetime
>>> datetime.datetime.strptime('Mon Feb 15 2010', '%a %b %d %Y').strftime('%d/%m/%Y')
'15/02/2010'
>>>
You can install the dateutil library. Its parse function can figure out what format a string is in without having to specify the format like you do with datetime.strptime.
from dateutil.parser import parse
dt = parse('Mon Feb 15 2010')
print(dt)
# datetime.datetime(2010, 2, 15, 0, 0)
print(dt.strftime('%d/%m/%Y'))
# 15/02/2010
convert string to datetime object
from datetime import datetime
s = "2016-03-26T09:25:55.000Z"
f = "%Y-%m-%dT%H:%M:%S.%fZ"
out = datetime.strptime(s, f)
print(out)
output:
2016-03-26 09:25:55
>>> from_date="Mon Feb 15 2010"
>>> import time
>>> conv=time.strptime(from_date,"%a %b %d %Y")
>>> time.strftime("%d/%m/%Y",conv)
'15/02/2010'
As this question comes often, here is the simple explanation.
datetime or time module has two important functions.
strftime - creates a string representation of date or time from a datetime or time object.
strptime - creates a datetime or time object from a string.
In both cases, we need a formating string. It is the representation that tells how the date or time is formatted in your string.
Now lets assume we have a date object.
>>> from datetime import datetime
>>> d = datetime(2010, 2, 15)
>>> d
datetime.datetime(2010, 2, 15, 0, 0)
If we want to create a string from this date in the format 'Mon Feb 15 2010'
>>> s = d.strftime('%a %b %d %y')
>>> print s
Mon Feb 15 10
Lets assume we want to convert this s again to a datetime object.
>>> new_date = datetime.strptime(s, '%a %b %d %y')
>>> print new_date
2010-02-15 00:00:00
Refer This document all formatting directives regarding datetime.
#codeling and #user1767754 : The following two lines will work. I saw no one posted the complete solution for the example problem that was asked. Hopefully this is enough explanation.
import datetime
x = datetime.datetime.strptime("Mon Feb 15 2010", "%a %b %d %Y").strftime("%d/%m/%Y")
print(x)
Output:
15/02/2010
You may achieve this using pandas as well:
import pandas as pd
pd.to_datetime('Mon Feb 15 2010', format='%a %b %d %Y').strftime('%d/%m/%Y')
Output:
'15/02/2010'
You may apply pandas approach for different datatypes as:
import pandas as pd
import numpy as np
def reformat_date(date_string, old_format, new_format):
return pd.to_datetime(date_string, format=old_format, errors='ignore').strftime(new_format)
date_string = 'Mon Feb 15 2010'
date_list = ['Mon Feb 15 2010', 'Wed Feb 17 2010']
date_array = np.array(date_list)
date_series = pd.Series(date_list)
old_format = '%a %b %d %Y'
new_format = '%d/%m/%Y'
print(reformat_date(date_string, old_format, new_format))
print(reformat_date(date_list, old_format, new_format).values)
print(reformat_date(date_array, old_format, new_format).values)
print(date_series.apply(lambda x: reformat_date(x, old_format, new_format)).values)
Output:
15/02/2010
['15/02/2010' '17/02/2010']
['15/02/2010' '17/02/2010']
['15/02/2010' '17/02/2010']
Just for the sake of completion: when parsing a date using strptime() and the date contains the name of a day, month, etc, be aware that you have to account for the locale.
It's mentioned as a footnote in the docs as well.
As an example:
import locale
print(locale.getlocale())
>> ('nl_BE', 'ISO8859-1')
from datetime import datetime
datetime.strptime('6-Mar-2016', '%d-%b-%Y').strftime('%Y-%m-%d')
>> ValueError: time data '6-Mar-2016' does not match format '%d-%b-%Y'
locale.setlocale(locale.LC_ALL, 'en_US')
datetime.strptime('6-Mar-2016', '%d-%b-%Y').strftime('%Y-%m-%d')
>> '2016-03-06'
use datetime library
http://docs.python.org/library/datetime.html look up 9.1.7.
especiall strptime() strftime() Behavior¶
examples
http://pleac.sourceforge.net/pleac_python/datesandtimes.html
If you dont want to define the input date format then, Install dateparser (pip install dateparser) and,
from dateparser import parse
parse("Mon Feb 15 2010").strftime("%d/%m/%Y")