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")
This is the data that is being returned from my API:
"Jun 02, 2021, 2 PMEST"
If I'm within 7 days of the current date which I'm getting by doing this:
from datetime import date
today = date.today()
print("Today's date:", today)
Just need to convert Jun to a number and 02 and compare to see if it's within 7 days in the future of the current date, then return True
APPROACH 0:
Given the format of your example data, you should be able to convert it to a datetime using this code:
datetime.strptime("Jun 02, 2021, 2 PMEST", "%b %d, %Y, %I %p%Z")
The details about this format string are here: https://docs.python.org/3/library/datetime.html#strftime-strptime-behavior
However, when I tested this locally, it worked for this input:
"Jun 02, 2021, 2 PMUTC"
but not for your input (which has different timezone):
"Jun 02, 2021, 2 PMEST"
I have investigated this some more and "read the docs" (https://docs.python.org/3/library/time.html).
To get EST parsing to work, you would have to change your OS timezone and reset the time module's timezones like this:
from datetime import datetime
import os
import time
os.environ["TZ"] = "US/Eastern". # change timezone
time.tzset(). # reset time.tzname tuple
datetime.strptime("Jun 02, 2021, 2 PMEST", "%b %d, %Y, %I %p%Z")
When you're done, be safe and delete the "hacked" environment variable:
del os.environ["TZ"]
Note - Since your system timezone is presumably still UTC, it can still parse UTC timezone too.
See this thread for detailed discussion: https://bugs.python.org/issue22377
Also note that the timestamp is not actually captured. The result you get with EST and UTC is a naive datetime object.
APPROACH 1
So, it seems like there is a better way to approach this.
First, you need to pip install dateutils if you don't already have it.
THen do something like this:
from dateutil import parser
from dateutil.tz import gettz
tzinfos = {"EST": gettz("US/Eastern")}
my_datetime = parser.parse("Jun 02, 2021, 2 PM EST", tzinfos=tzinfos)
What's happening here is we use gettz to get timezone information from the timezones listed in usr/share/zoneinfo. Then the parse function can (fuzzy) parse your string (no format needs to be specified!) and returns my_datetime which has timezone information on it. Here are the parser docs: https://dateutil.readthedocs.io/en/stable/parser.html
I don't know how many different timezones you need to deal with so the rest is up to you. Good luck.
Convert the date to a datetime structure and take the direct difference. Note that today must be a datetime, too.
import datetime
date_string = "Jun 02, 2021, 2 PMEST"
today = datetime.datetime.today()
date = datetime.datetime.strptime(date_string,
"%b %d, %Y, %I %p%Z") # Corrected
(date - today).days
#340
I want to check if the format of the date input by user matches the below:
Jan 5 2018 6:10 PM
Month: First letter should be caps, followed 2 more in small. (total 3 letters)
<Space>: single space, must exist
Date: For single digit it should not be 05, but 5
<Space>: single space, must exist
Hour: 0-12, for single digit it should not be 06, but 6
Minute: 00-59
AM/PM
I'm using the below regex and trying to match:
import re,sys
usr_date = str(input("Please enter the older date until which you want to scan ? \n[Date Format Example: Jan 5 2018 6:10 PM] : "))
valid_usr_date = re.search("^(\s+)*[A-Z]{1}[a-z]{2}\s{1}[1-31]{1}\s{1}[1-2]{1}[0-9]{1}[0-9]{1}[0-9]{1}\s{1}[0-12]{1}:[0-5]{1}[0-9]{1}\s{1}(A|P)M$",usr_date,re.M)
if not valid_usr_date:
print ("The date format is incorrect. Please follow the exact date format as shown in example. Exiting Program!")
sys.exit()
But, even for the correct format it gives a syntax wrong error. What am I doing wrong.
I would not use regex for that, as you have no way to actually validate the date itself (eg, a regex will happily accept Abc 99 9876 9:99 PM).
Instead, use strptime:
from datetime import datetime
string = 'Jan 5 2018 6:10 PM'
datetime.strptime(string, '%b %d %Y %I:%M %p')
If the string would be in the "wrong" format you'd get a ValueError.
The only apparent "problem" with this approach is that for some reason you require the day and hour not to be zero-padded and strptime doesn't seem to have such directives.
A table with all available directives is here.
You could use a function which parses the input string and tries to return a datetime object, if it can't it raises an ValueError:
from datetime import datetime
def valid_date(s):
try:
return datetime.strptime(s, '%Y-%m-%d %H:%M')
except ValueError:
msg = "Not a valid date: '{0}'.".format(s)
raise argparse.ArgumentTypeError(msg)
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
This is my code:
import datetime
today = datetime.date.today()
print(today)
This prints: 2008-11-22 which is exactly what I want.
But, I have a list I'm appending this to and then suddenly everything goes "wonky". Here is the code:
import datetime
mylist = []
today = datetime.date.today()
mylist.append(today)
print(mylist)
This prints the following:
[datetime.date(2008, 11, 22)]
How can I get just a simple date like 2008-11-22?
The WHY: dates are objects
In Python, dates are objects. Therefore, when you manipulate them, you manipulate objects, not strings or timestamps.
Any object in Python has TWO string representations:
The regular representation that is used by print can be get using the str() function. It is most of the time the most common human readable format and is used to ease display. So str(datetime.datetime(2008, 11, 22, 19, 53, 42)) gives you '2008-11-22 19:53:42'.
The alternative representation that is used to represent the object nature (as a data). It can be get using the repr() function and is handy to know what kind of data your manipulating while you are developing or debugging. repr(datetime.datetime(2008, 11, 22, 19, 53, 42)) gives you 'datetime.datetime(2008, 11, 22, 19, 53, 42)'.
What happened is that when you have printed the date using print, it used str() so you could see a nice date string. But when you have printed mylist, you have printed a list of objects and Python tried to represent the set of data, using repr().
The How: what do you want to do with that?
Well, when you manipulate dates, keep using the date objects all long the way. They got thousand of useful methods and most of the Python API expect dates to be objects.
When you want to display them, just use str(). In Python, the good practice is to explicitly cast everything. So just when it's time to print, get a string representation of your date using str(date).
One last thing. When you tried to print the dates, you printed mylist. If you want to print a date, you must print the date objects, not their container (the list).
E.G, you want to print all the date in a list :
for date in mylist :
print str(date)
Note that in that specific case, you can even omit str() because print will use it for you. But it should not become a habit :-)
Practical case, using your code
import datetime
mylist = []
today = datetime.date.today()
mylist.append(today)
print mylist[0] # print the date object, not the container ;-)
2008-11-22
# It's better to always use str() because :
print "This is a new day : ", mylist[0] # will work
>>> This is a new day : 2008-11-22
print "This is a new day : " + mylist[0] # will crash
>>> cannot concatenate 'str' and 'datetime.date' objects
print "This is a new day : " + str(mylist[0])
>>> This is a new day : 2008-11-22
Advanced date formatting
Dates have a default representation, but you may want to print them in a specific format. In that case, you can get a custom string representation using the strftime() method.
strftime() expects a string pattern explaining how you want to format your date.
E.G :
print today.strftime('We are the %d, %b %Y')
>>> 'We are the 22, Nov 2008'
All the letter after a "%" represent a format for something:
%d is the day number (2 digits, prefixed with leading zero's if necessary)
%m is the month number (2 digits, prefixed with leading zero's if necessary)
%b is the month abbreviation (3 letters)
%B is the month name in full (letters)
%y is the year number abbreviated (last 2 digits)
%Y is the year number full (4 digits)
etc.
Have a look at the official documentation, or McCutchen's quick reference you can't know them all.
Since PEP3101, every object can have its own format used automatically by the method format of any string. In the case of the datetime, the format is the same used in
strftime. So you can do the same as above like this:
print "We are the {:%d, %b %Y}".format(today)
>>> 'We are the 22, Nov 2008'
The advantage of this form is that you can also convert other objects at the same time.
With the introduction of Formatted string literals (since Python 3.6, 2016-12-23) this can be written as
import datetime
f"{datetime.datetime.now():%Y-%m-%d}"
>>> '2017-06-15'
Localization
Dates can automatically adapt to the local language and culture if you use them the right way, but it's a bit complicated. Maybe for another question on SO(Stack Overflow) ;-)
import datetime
print datetime.datetime.now().strftime("%Y-%m-%d %H:%M")
Edit:
After Cees' suggestion, I have started using time as well:
import time
print time.strftime("%Y-%m-%d %H:%M")
The date, datetime, and time objects all support a strftime(format) method,
to create a string representing the time under the control of an explicit format
string.
Here is a list of the format codes with their directive and meaning.
%a Locale’s abbreviated weekday name.
%A Locale’s full weekday name.
%b Locale’s abbreviated month name.
%B Locale’s full month name.
%c Locale’s appropriate date and time representation.
%d Day of the month as a decimal number [01,31].
%f Microsecond as a decimal number [0,999999], zero-padded on the left
%H Hour (24-hour clock) as a decimal number [00,23].
%I Hour (12-hour clock) as a decimal number [01,12].
%j Day of the year as a decimal number [001,366].
%m Month as a decimal number [01,12].
%M Minute as a decimal number [00,59].
%p Locale’s equivalent of either AM or PM.
%S Second as a decimal number [00,61].
%U Week number of the year (Sunday as the first day of the week)
%w Weekday as a decimal number [0(Sunday),6].
%W Week number of the year (Monday as the first day of the week)
%x Locale’s appropriate date representation.
%X Locale’s appropriate time representation.
%y Year without century as a decimal number [00,99].
%Y Year with century as a decimal number.
%z UTC offset in the form +HHMM or -HHMM.
%Z Time zone name (empty string if the object is naive).
%% A literal '%' character.
This is what we can do with the datetime and time modules in Python
import time
import datetime
print "Time in seconds since the epoch: %s" %time.time()
print "Current date and time: ", datetime.datetime.now()
print "Or like this: ", datetime.datetime.now().strftime("%y-%m-%d-%H-%M")
print "Current year: ", datetime.date.today().strftime("%Y")
print "Month of year: ", datetime.date.today().strftime("%B")
print "Week number of the year: ", datetime.date.today().strftime("%W")
print "Weekday of the week: ", datetime.date.today().strftime("%w")
print "Day of year: ", datetime.date.today().strftime("%j")
print "Day of the month : ", datetime.date.today().strftime("%d")
print "Day of week: ", datetime.date.today().strftime("%A")
That will print out something like this:
Time in seconds since the epoch: 1349271346.46
Current date and time: 2012-10-03 15:35:46.461491
Or like this: 12-10-03-15-35
Current year: 2012
Month of year: October
Week number of the year: 40
Weekday of the week: 3
Day of year: 277
Day of the month : 03
Day of week: Wednesday
Use date.strftime. The formatting arguments are described in the documentation.
This one is what you wanted:
some_date.strftime('%Y-%m-%d')
This one takes Locale into account. (do this)
some_date.strftime('%c')
This is shorter:
>>> import time
>>> time.strftime("%Y-%m-%d %H:%M")
'2013-11-19 09:38'
# convert date time to regular format.
d_date = datetime.datetime.now()
reg_format_date = d_date.strftime("%Y-%m-%d %I:%M:%S %p")
print(reg_format_date)
# some other date formats.
reg_format_date = d_date.strftime("%d %B %Y %I:%M:%S %p")
print(reg_format_date)
reg_format_date = d_date.strftime("%Y-%m-%d %H:%M:%S")
print(reg_format_date)
OUTPUT
2016-10-06 01:21:34 PM
06 October 2016 01:21:34 PM
2016-10-06 13:21:34
Or even
from datetime import datetime, date
"{:%d.%m.%Y}".format(datetime.now())
Out: '25.12.2013
or
"{} - {:%d.%m.%Y}".format("Today", datetime.now())
Out: 'Today - 25.12.2013'
"{:%A}".format(date.today())
Out: 'Wednesday'
'{}__{:%Y.%m.%d__%H-%M}.log'.format(__name__, datetime.now())
Out: '__main____2014.06.09__16-56.log'
Simple answer -
datetime.date.today().isoformat()
With type-specific datetime string formatting (see nk9's answer using str.format().) in a Formatted string literal (since Python 3.6, 2016-12-23):
>>> import datetime
>>> f"{datetime.datetime.now():%Y-%m-%d}"
'2017-06-15'
The date/time format directives are not documented as part of the Format String Syntax but rather in date, datetime, and time's strftime() documentation. The are based on the 1989 C Standard, but include some ISO 8601 directives since Python 3.6.
I hate the idea of importing too many modules for convenience. I would rather work with available module which in this case is datetime rather than calling a new module time.
>>> a = datetime.datetime(2015, 04, 01, 11, 23, 22)
>>> a.strftime('%Y-%m-%d %H:%M')
'2015-04-01 11:23'
You need to convert the datetime object to a str.
The following code worked for me:
import datetime
collection = []
dateTimeString = str(datetime.date.today())
collection.append(dateTimeString)
print(collection)
Let me know if you need any more help.
In Python you can format a datetime using the strftime() method from the date, time and datetime classes in the datetime module.
In your specific case, you are using the date class from datetime. You can use the following snippet to format the today variable into a string with the format yyyy-MM-dd:
import datetime
today = datetime.date.today()
print("formatted datetime: %s" % today.strftime("%Y-%m-%d"))
In the following a more complete example:
import datetime
today = datetime.date.today()
# datetime in d/m/Y H:M:S format
date_time = today.strftime("%d/%m/%Y, %H:%M:%S")
print("datetime: %s" % date_time)
# datetime in Y-m-d H:M:S format
date_time = today.strftime("%Y-%m-%d, %H:%M:%S")
print("datetime: %s" % date_time)
# format date
date = today.strftime("%d/%m/%Y")
print("date: %s" % time)
# format time
time = today.strftime("%H:%M:%S")
print("time: %s" % time)
# day
day = today.strftime("%d")
print("day: %s" % day)
# month
month = today.strftime("%m")
print("month: %s" % month)
# year
year = today.strftime("%Y")
print("year: %s" % year)
More directives:
Sources:
Format DateTime in Python
strftime
You can do:
mylist.append(str(today))
Considering the fact you asked for something simple to do what you wanted, you could just:
import datetime
str(datetime.date.today())
For those wanting locale-based date and not including time, use:
>>> some_date.strftime('%x')
07/11/2019
Since the print today returns what you want this means that the today object's __str__ function returns the string you are looking for.
So you can do mylist.append(today.__str__()) as well.
from datetime import date
def today_in_str_format():
return str(date.today())
print (today_in_str_format())
This will print 2018-06-23 if that's what you want :)
You may want to append it as a string?
import datetime
mylist = []
today = str(datetime.date.today())
mylist.append(today)
print(mylist)
For pandas.Timestamps, strftime() can be used e.g.:
utc_now = datetime.now()
For isoformat:
utc_now.isoformat()
For any format e.g.:
utc_now.strftime("%m/%d/%Y, %H:%M:%S")
You can use easy_date to make it easy:
import date_converter
my_date = date_converter.date_to_string(today, '%Y-%m-%d')
A quick disclaimer for my answer - I've only been learning Python for about 2 weeks, so I am by no means an expert; therefore, my explanation may not be the best and I may use incorrect terminology. Anyway, here it goes.
I noticed in your code that when you declared your variable today = datetime.date.today() you chose to name your variable with the name of a built-in function.
When your next line of code mylist.append(today) appended your list, it appended the entire string datetime.date.today(), which you had previously set as the value of your today variable, rather than just appending today().
A simple solution, albeit maybe not one most coders would use when working with the datetime module, is to change the name of your variable.
Here's what I tried:
import datetime
mylist = []
present = datetime.date.today()
mylist.append(present)
print present
and it prints yyyy-mm-dd.
Here is how to display the date as (year/month/day) :
from datetime import datetime
now = datetime.now()
print '%s/%s/%s' % (now.year, now.month, now.day)
import datetime
import time
months = ["Unknown","January","Febuary","Marchh","April","May","June","July","August","September","October","November","December"]
datetimeWrite = (time.strftime("%d-%m-%Y "))
date = time.strftime("%d")
month= time.strftime("%m")
choices = {'01': 'Jan', '02':'Feb','03':'Mar','04':'Apr','05':'May','06': 'Jun','07':'Jul','08':'Aug','09':'Sep','10':'Oct','11':'Nov','12':'Dec'}
result = choices.get(month, 'default')
year = time.strftime("%Y")
Date = date+"-"+result+"-"+year
print Date
In this way you can get Date formatted like this example: 22-Jun-2017
I don't fully understand but, can use pandas for getting times in right format:
>>> import pandas as pd
>>> pd.to_datetime('now')
Timestamp('2018-10-07 06:03:30')
>>> print(pd.to_datetime('now'))
2018-10-07 06:03:47
>>> pd.to_datetime('now').date()
datetime.date(2018, 10, 7)
>>> print(pd.to_datetime('now').date())
2018-10-07
>>>
And:
>>> l=[]
>>> l.append(pd.to_datetime('now').date())
>>> l
[datetime.date(2018, 10, 7)]
>>> map(str,l)
<map object at 0x0000005F67CCDF98>
>>> list(map(str,l))
['2018-10-07']
But it's storing strings but easy to convert:
>>> l=list(map(str,l))
>>> list(map(pd.to_datetime,l))
[Timestamp('2018-10-07 00:00:00')]
maybe the shortest solution, which exactly matches your situation, would be:
mylist.append(str(AnyDate)[:10])
or even shorter, e.g.:
f'{AnyDate}'[:10]
PS: it doesn't need to be today.