I have spent some time trying to figure out how to get a time delta between time values. The only issue is that one of the times was stored in a file. So I have one string which is in essence str(datetime.datetime.now()) and datetime.datetime.now().
Specifically, I am having issues getting a delta because one of the objects is a datetime object and the other is a string.
I think the answer is that I need to get the string back in a datetime object for the delta to work.
I have looked at some of the other Stack Overflow questions relating to this including the following:
Python - Date & Time Comparison using timestamps, timedelta
Comparing a time delta in python
Convert string into datetime.time object
Converting string into datetime
Example code is as follows:
f = open('date.txt', 'r+')
line = f.readline()
date = line[:26]
now = datetime.datetime.now()
then = time.strptime(date)
delta = now - then # This does not work
Can anyone tell me where I am going wrong?
For reference, the first 26 characters are acquired from the first line of the file because this is how I am storing time e.g.
f.write(str(datetime.datetime.now())
Which would write the following:
2014-01-05 13:09:42.348000
time.strptime returns a struct_time.
datetime.datetime.now() returns a datetime object.
The two can not be subtracted directly.
Instead of time.strptime you could use datetime.datetime.strptime, which returns a datetime object. Then you could subtract now and then.
For example,
import datetime as DT
now = DT.datetime.now()
then = DT.datetime.strptime('2014-1-2', '%Y-%m-%d')
delta = now - then
print(delta)
# 3 days, 8:17:14.428035
By the way, you need to supply a date format string to time.strptime or DT.datetime.strptime.
time.strptime(date)
should have raised a ValueError.
It looks like your date string is 26 characters long. That might mean you have a date string like 'Fri, 10 Jun 2011 11:04:17 '.
If that is true, you may want to parse it like this:
then = DT.datetime.strptime('Fri, 10 Jun 2011 11:04:17 '.strip(), "%a, %d %b %Y %H:%M:%S")
print(then)
# 2011-06-10 11:04:17
There is a table describing the available directives (like %Y, %m, etc.) here.
Try this:
import time
import datetime
d = datetime.datetime.now()
now = time.mktime(d.timetuple())
And then apply the delta
if you have the year,month,day of 'then' you may use:
year = 2013
month = 1
day = 1
now_date = datetime.datetime.now()
then_date = now_date.replace(year = year, month = month, day = day)
delta = now_date - then_date
Related
Does anyone know how I can extract the end 6 characters in a absoloute URL e.g
/es/ideas-de-trading-y-noticias/el-ibex-35-insiste-en-buscar-los-7900-puntos-a-la-espera-de-las--221104
This is not a typical URL sometimetimes it ends -221104
Also, is there a way to turn 221104 into the date 04 11 2022 easily?
Thanks in advance
Mark
You should use the datetime module for parsing strings into datetimes, like so.
from datetime import datetime
url = 'https://www.ig.com/es/ideas-de-trading-y-noticias/el-ibex-35-insiste-en-buscar-los-7900-puntos-a-la-espera-de-las--221104'
datetime_string = url.split('--')[1]
date = datetime.strptime(datetime_string, '%y%m%d')
print(f"{date.day} {date.month} {date.year}")
the %y%m%d text tells the strptime method that the string of '221104' is formatted in the way that the first two letters are the year, the next two are the month, and the final two are the day.
Here is a link to the documentation on using this method:
https://docs.python.org/3/library/datetime.html#strftime-and-strptime-behavior
If the url always has this structure (that is it has the date at the end after a -- and only has -- once), you can get the date with:
str_date = str(url).split("--")[1]
Relaxing the assumption to have only one --, we can have the code working by just taking the last element of the splitted list (again assuming the date is always at the end):
str_date = str(url).split("--")[-1]
(Thanks to #The Myth for pointing that out)
To convert the obtained date into a datetime.date object and get it in the format you want:
from datetime import datetime
datetime_date = datetime.strptime(str_date, "%y%m%d")
formatted_date = datetime_date.strftime("%d %m %Y")
print(formatted_date) # 04 11 2022
Docs:
strftime
strptime
behaviour of the above two functions and format codes
Taking into consideration the date is constant in the format yy-mm-dd. You can split the URL by:
url = "https://www.ig.com/es/ideas-de-trading-y-noticias/el-ibex-35-insiste-en-buscar-los-7900-puntos-a-la-espera-de-las--221104"
time = url[-6:] # Gets last 6 values
To convert yy-mm-dd into dd mm yy we will use the DateTime module:
import datetime as dt
new_time = dt.datetime.strptime(time, '%y%m%d') # Converts your date into datetime using the format
format_time = dt.datetime.strftime(new_time, '%d-%m-%Y') # Format
print(format_time)
The whole code looks like this:
url = "https://www.ig.com/es/ideas-de-trading-y-noticias/el-ibex-35-insiste-en-buscar-los-7900-puntos-a-la-espera-de-las--221104"
time = url[-6:] # Gets last 6 values
import datetime as dt
new_time = dt.datetime.strptime(time, '%y%m%d') # Converts your date into datetime using the format
format_time = dt.datetime.strftime(new_time, '%d %m %Y') # Format
print(format_time)
Learn more about datetime
You can use python built-in split function.
date = url.split("--")[1]
It gives us 221104
then you can modify the string by rearranging it
date_string = f"{date[4:6]} {date[2:4]} {date[0:2]}"
this gives us 04 11 22
Assuming that -- will only be there as it is in the url you posted, you can do something as follows:
You can split the URL at -- & extract the element
a = 'https://www.ig.com/es/ideas-de-trading-y-noticias/el-ibex-35-insiste-en-buscar-los-7900-puntos-a-la-espera-de-las--221104'
desired_value = a.split('--')[1]
& to convert:
from datetime import datetime
converted_date = datetime.strptime(desired_value , "%y%m%d")
formatted_date = datetime.strftime(converted_date, "%d %m %Y")
My code is the following:
date = datetime.datetime.now()- datetime.datetime.now()
print date
h, m , s = str(date).split(':')
When I print h the result is:
-1 day, 23
How do I get only the hour (the 23) from the substract using datetime?
Thanks.
If you subtract the current date from a past date, you would get a negative timedelta value.
You can get the seconds with td.seconds and corresponding hour value via just dividing by 3600.
from datetime import datetime
import time
date1 = datetime.now()
time.sleep(3)
date2 = datetime.now()
# timedelta object
td = date2 - date1
print(td.days, td.seconds // 3600, td.seconds)
# 0 0 3
You're not too far off but you should just ask your question as opposed to a question with a "real scenario" later as those are often two very different questions. That way you get an answer to your actual question.
All that said, rather than going through a lot of hoop-jumping with splitting the datetime object, assigning it to a variable which you then later use look for what you need in, it's better to just know what DateTime can do since that can be such a common part of your coding. You would also do well to look at timedelta (which is part of datetime) and if you use pandas, timestamp.
from datetime import datetime
date = datetime.now()
print(date)
print(date.hour)
I can get you the hour of datetime.datetime.now()
You could try indexing a list of a string of datetime.datetime.now():
print(list(str(datetime.datetime.now()))[11] + list(str(datetime.datetime.now()))[12])
Output (in my case when tested):
09
Hope I am of help!
I want to add a time to a datetime. My initial datetime is: initial_datetime='2015-11-03 08:05:22' and is a string and this_hour and this_min are strings too. I use:
time='-7:00'
time = time.split(':')
this_hour = time[0]
this_min = time[1]
initial_datetime='2015-11-03 08:05:22'
new_date = datetime.combine(initial_datetime, time(this_hour, this_min))
+ timedelta(hours=4)
But there comes an error:
'str' object is not callable.
My desired output is the initial_datetime plus my time (in this case -7 hours ) and then add 4 hours. So, in my example, the new date should be '2015-11-03 05:05:22'.
datetime.combine is typically used to combine a date object with a time object rather than incrementing or decrementing a datetime object. In your case, you need to convert your datetime string to a datetime object and convert the parts of your time string to integers so you can add them to your datetime with timedelta. As an aside, be careful about using variable names, like time, that conflict with your imports.
from datetime import datetime, timedelta
dtstr = '2015-11-03 08:05:22'
tstr = '-7:00'
hours, minutes = [int(t) for t in tstr.split(':')]
dt = datetime.strptime(dtstr, '%Y-%m-%d %H:%M:%S') + timedelta(hours=hours+4, minutes=minutes)
print(dt)
# 2015-11-03 05:05:22
I tried:
df["datetime_obj"] = df["datetime"].apply(lambda dt: datetime.strptime(dt, "%d/%m/%Y %H:%M"))
but got this error:
ValueError: time data '10/11/2006 24:00' does not match format
'%d/%m/%Y %H:%M'
How to solve it correctly?
The reason why this does not work is because the %H parameter only accepts values in the range of 00 to 23 (both inclusive). This thus means that 24:00 is - like the error says - not a valid time string.
I think therefore we have not much other options than convert the string to a valid format. We can do this by first replacing 24:00 with 00:00, and then later increment the day for these timestamps.
Like:
from datetime import timedelta
import pandas as pd
df['datetime_zero'] = df['datetime'].str.replace('24:00', '0:00')
df['datetime_er'] = pd.to_datetime(df['datetime_zero'], format='%d/%m/%Y %H:%M')
selrow = df['datetime'].str.contains('24:00')
df['datetime_obj'] = df['datetime_er'] + selrow * timedelta(days=1)
The last line thus adds one day to the rows that contain 24:00, such that '10/11/2006 24:00' gets converted to '11/11/2006 24:00'. Note however that the above is rather unsafe since depending on the format of the timestamp this will/will not work. For the above it will (probably) work, since there is only one colon. But if for example the datetimes have seconds as well, the filter could get triggered for 00:24:00, so it might require some extra work to get it working.
Your data doesn't follow the conventions used by Python / Pandas datetime objects. There should be only one way of storing a particular datetime, i.e. '10/11/2006 24:00' should be rewritten as '11/11/2006 00:00'.
Here's one way to approach the problem:
# find datetimes which have '24:00' and rewrite
twenty_fours = df['strings'].str[-5:] == '24:00'
df.loc[twenty_fours, 'strings'] = df['strings'].str[:-5] + '00:00'
# construct datetime series
df['datetime'] = pd.to_datetime(df['strings'], format='%d/%m/%Y %H:%M')
# add one day where applicable
df.loc[twenty_fours, 'datetime'] += pd.DateOffset(1)
Here's some data to test:
dateList = ['10/11/2006 24:00', '11/11/2006 00:00', '12/11/2006 15:00']
df = pd.DataFrame({'strings': dateList})
Result after transformations described above:
print(df['datetime'])
0 2006-11-11 00:00:00
1 2006-11-11 00:00:00
2 2006-11-12 15:00:00
Name: datetime, dtype: datetime64[ns]
As indicated in the documentation (https://docs.python.org/2/library/datetime.html#strftime-strptime-behavior), hours go from 00 to 23. 24:00 is then an error.
I run a sql query that returns a date in the format '2015-03-01T17:09:00.000+0000' I want to subtract this from today's date.
I am getting today's date with the following:
import datetime
now = datetime.datetime.now()
The formats don't seem to line up and I can't figure out a standardize format.
You can use strptime from datetime module to get python compatible date time from your query result using a format string. (You might have to play with the format string a bit to suit your case)
ts = '2015-03-01T17:09:00.000+0000' to a format string like
f = '%Y-%m-%dT%H:%M:%S.%f%z'
date_from_sql = datetime.datetime.strptime(ts, f)
now = datetime.datetime.now()
delta = date_from_sql - now
The .000 is probably microseconds (denoted by %f in the format string) and the +0000 is the utc offset (denoted by %z in the format string). Check this out for more formatting options: https://docs.python.org/2/library/datetime.html#strftime-strptime-behavior
Check out this thread for an example: what is the proper way to convert between mysql datetime and python timestamp?
Checkout this for more on strptime https://docs.python.org/2/library/datetime.html#datetime.datetime.strptime
Getting the delta between two datetime objects in Python is really simple, you simply subtract them.
import datetime
d1 = datetime.datetime.now()
d2 = datetime.datetime.now()
delta = d2 - d1
print delta.total_seconds()
d2 - d1 returns a datetime.timedelta object, from which you can get the total second difference between the two dates.
As for formatting the dates, you can read about formatting strings into datetime objects, and datetime objects into string here
You'll read about the strftime() and strptime() functions, and with them you can get yourself two datetime objects which you can subtract from each other.