A function returns date and time in unicode format.
u'2014-03-06T04:38:51Z'
I wish to convert this to date and time format and subtract it with current datetime to get the number of days in between.
Thanks in advance
Check string is unicode
>>> import types
>>> type(u'2014-03-06T04:38:51Z') is types.UnicodeType
True
Converting strings to datetime:
>>> import datetime
>>> datetime.datetime.strptime(u'2014-03-06T04:38:51Z', '%Y-%m-%dT%H:%M:%SZ')
datetime.datetime(2014, 3, 6, 4, 38, 51)
Subtract from today to
>>> import datetime
>>> today = datetime.datetime.today()
>>> yourdate = datetime.datetime.strptime(u'2014-03-06T04:38:51Z', '%Y-%m-%dT%H:%M:%SZ')
>>> difference = today - yourdate
print str(difference)
First you have to convert your string to a datetime.datetime object.
import datetime
then = datetime.datetime.strptime(u'2014-03-06T04:38:51Z', "%Y-%m-%dT%H:%M:%SZ")
then represents itself as datetime.datetime(2014, 3, 6, 4, 38, 51), which looks about right. Then you have to get today's date as a datetime.datetime.
now = datetime.datetime.now()
Finally subtract it from your date (or vice versa - the question didn't make it clear).delta is a datetime.timedelta object that stores increments in days, seconds and microseconds. The latter two are always positive, the first can be negative.
for delta in (now-then, then-now):
print(delta, "::", delta.days, delta.seconds, delta.microseconds)
This prints out:
-1 day, 20:18:14.250142 :: -1 73094 250142
3:41:45.749858 :: 0 13305 749858
Best try it with a few examples to convince yourself it's correct.
Related
I am given a number like: 737556.5965277777.
>>> datetime.timedelta(days=737556.5965277777)
datetime.timedelta(days=737556, seconds=51539, microseconds=999996)
>>> datetime.datetime.now()
datetime.datetime(2020, 5, 11, 15, 36, 41, 711686)
How can I compare this with the current datetime to check whether it is after? Either get the current timedelta and compare those or convert it to a timestamp first, then compare.
One way would be to convert the current datetime to an ordinal, and then compare the two. If the ordinal datetime is greater than the now() ordinal datetime, then it is after.
You can use datetimes toordinal()
from datetime import datetime as dt
dt_now = dt.now()
ordinal_date = dt_now.toordinal()
print(ordinal_date)
The ordinal datetime is just a number of day starting with date 0001-01-01 (as day 1). So you can just add the timedelta (minus one day, to compensate for 0001-01-01 not being day 0) to that date:
yourdate = datetime.datetime(1,1,1) + datetime.timedelta(days=737556.5965277777 - 1)
and compare it to now:
if yourdate > datetime.datetime.now():
Note that yourdate.toordinal() returns the integer part of your initial number: 737556.
I have a JSON object with a date that returns
print row['ApplicationReceivedDateTime']
/Date(1454475600000)/
how do I process this using the pythons datetime module?
print type(row['ApplicationReceivedDateTime'])
returns <type 'unicode'>
print repr(row['ApplicationReceivedDateTime'])
returns u'/Date(1454475600000)/'
That looks like milliseconds. Try dividing by 1000.
import datetime as dt
>>> dt.datetime.fromtimestamp(1454475600000 / 1000)
datetime.datetime(2016, 2, 2, 21, 0)
If the date is in the string format per your question, extract the numeric portion using re.
date = '/Date(1454475600000)/'
>>> dt.datetime.fromtimestamp(int(re.findall(r"\d+", date)[0]) / 1000)
datetime.datetime(2016, 2, 2, 21, 0)
You probably want
datetime.datetime.strptime(string_date, "%Y-%m-%d %H:%M:%S.%f")
And the values of Year, Month, Day, Hour, Minute, Second and F, for that you can write a manual function for that like this
def generate_date_time_str(date_str):
"""Login to parse the date str"""
return date_str
the date_str will look link this
"%Y-%m-%d %H:%M:%S.%f"
There is no python module directly convert any random date str to DateTime object
You can use re to get the integer value and then use datetime.datetime.fromtimestamp to get the date value:
from datetime import datetime
import re
string_time = row['ApplicationReceivedDateTime']
parsed_time = int(re.search('\((\d+)\)', string_time)[1]) / 1e3 #1e3 == 1000
rcvd_date = datetime.fromtimestamp(parsed_time)
print(rcvd_date.strftime('%Y-%m-%d %H:%M:%S'))
Prints:
'2016-02-03 05:00:00'
I have strings like:
first = '2018-09-16 15:00:00'
second = '1900-01-01 09:45:55.500597'
I want to compare them.
All methods I found like Convert string date to timestamp in Python requires to know the format of the string.
I don't know the format of the strings (see differences between first and second) all I know is that they can be converted to timestamps.
How can I convert them in order to compare them?
Edit:
The "largest" string that I can get is:
1900-01-01 09:45:55.500597
but I can also get:
1900-01-01
1900-01-01 09:45
1900-01-01 09:45:55
etc..
It's always YYYY-MM-DD HH-MM....
You can use pandas.to_datetime. It offers a lot of flexibility in the string timestamp format, and you can use it on single strings or list/series/tuples of timestamps.
>>> import pandas as pd
>>> day = pd.to_datetime('1900-01-01')
>>> minute = pd.to_datetime('1900-01-01 09:45')
>>> second = pd.to_datetime('1900-01-01 09:45:55')
>>> subsecond = pd.to_datetime('1900-01-01 09:45:55.500597')
>>> assert subsecond > second
>>> assert minute < second
>>> assert day < minute
You can use the dateutil module (pip install python-dateutil):
>>> from dateutil.parser import parse
>>> parse('2018-09-16 15:00:00')
datetime.datetime(2018, 9, 16, 15, 0)
>>> parse('1900-01-01 09:45:55.500597')
datetime.datetime(1900, 1, 1, 9, 45, 55, 500597)
From the list of its features:
Generic parsing of dates in almost any string format;
Once you have the datetime objects, you can compare them directly, there's no need to calculate the timestamps.
Given a string that looks like "Hours:5 Minutes:34 Seconds:28" or "Minutes:34 Seconds:28", is there any pythonic way to convert it to a datetime object? I do not want to use a regex if there's an easier way.
Yes, there is. You can do it like this:
import time
datetime_string = "Hours:5 Minutes:34 Seconds:28"
if "Hours" in datetime_string:
datetime_object = time.strptime(datetime_string, "Hours:%H Minutes:%M Seconds:%S")
elif "Minutes" in datetime_string:
datetime_object = time.strptime(datetime_string, "Minutes:%M Seconds:%S")
else:
datetime_object = time.strptime(datetime_string, "Seconds:%S")
Note: When You create datetime object, values that You do not provide will be filled with default values.So, in case datetime_string contains only seconds, hours and minutes will be set to 0.
You may use datetime.strptime() to convert string into datetime object as:
>>> from datetime import datetime
>>> date_object = datetime.strptime('Hours:5 Minutes:34 Seconds:28', 'Hours:%H Minutes:%M Seconds:%S')
>>> date_object
datetime.datetime(1900, 1, 1, 5, 34, 28)
# ^ ^ ^
# Hour Min Seconds
Since you do not have date in the string, it will keep the default date of 1 Jan 1990. I think what you need is datetime.time() which return time object with same hour, minute, second and microsecond as in you datetime object. (tzinfo is None). For example:
>>> date_object.time()
datetime.time(5, 34, 28)
# ^ ^ ^
# Hour Min Seconds
where date_object is of datetime type created earlier.
I have a time which is 13:11:06 and i want to -GMT (i.e -0530). I can minus it by simply doing -5 by splitting the string taking the first digit (convert to int) and then minus it and then re-join. But then i get it in a format which is 8:11:06 which is not right as it should be 08:11:06, secondly its a lengthy process. Is there a easy way to get my time in -GMT format (08:11:06)
This is what i did to get -GMT time after getting the datetime
timesplithour = int(timesplit[1]) + -5
timesplitminute = timesplit[2]
timesplitseconds = timesplit[3]
print timesplithour
print timesplitminute
print timesplitseconds
print timesplithour + ":" + timesplitminute + ":" + timesplitseconds
You could use Python's datatime library to help you as follows:
import datetime
my_time = "13:11:06"
new_time = datetime.datetime.strptime("2016 " + my_time, "%Y %H:%M:%S") - datetime.timedelta(hours=5, minutes=30)
print new_time.strftime("%H:%M:%S")
This would print:
07:41:06
First it converts your string into a datetime object. It then creates a timedelta object allowing you to subtract 5 hours 30 minutes from the datetime object. Finally it uses strftime to format the resulting datetime into a string in the same format.
Use the datetime module:
from datetime import datetime, timedelta
dt = datetime.strptime('13:11:06', '%H:%M:%S')
time_gmt = (dt - timedelta(hours=5, minutes=30)).time()
print(time_gmt.hour)
print(time_gmt.minute)
print(time_gmt.second)
s = time_gmt.strftime('%H:%M:%S')
print(s)
Output
7
41
6
07:41:06
Note that this subtracts 5 hours and 30 minutes as initially mentioned in the question. If you really only want to subtract 5 hours, use timedelta(hours=5).
You can use datetimes timedelta.
print datetime.datetime.today()
>>> datetime.datetime(2016, 3, 3, 10, 45, 6, 270711)
print datetime.datetime.today() - datetime.timedelta(days=3)
>>> datetime.datetime(2016, 2, 29, 10, 45, 8, 559073)
This way you can subtract easily
Assuming the time is a datetime instance
import datetime as dt
t = datetime(2015,12,31,13,11,06)
#t.time() # gives time object. ie no date information
offset = dt.timedelta(hours=5,minutes=30) # or hours=5.5
t -= offset
t.strftime(("%H:%M:%S") # output as your desired string
#'18:41:06'
If the object is datetime and you don't care about DST, the simplest thing you can do is,
In [1]: from datetime import datetime
In [2]: curr = datetime.now()
In [3]: curr
Out[3]: datetime.datetime(2016, 3, 3, 9, 57, 31, 302231)
In [4]: curr.utcnow()
Out[4]: datetime.datetime(2016, 3, 3, 8, 57, 57, 286956)