I using:
s = "20200113"
final = datetime.datetime.strptime(s, '%Y%m%d')
I need convert a number in date format (2020-01-13)
but when I print final:
2020-01-13 00:00:00
Tried datetime.date(s, '%Y%m%d') but It's returns a error:
an integer is required (got type str)
Is there any command to get only date without hour?
Once you have a datetime object just use strftime
import datetime
d = datetime.datetime.now() # Some datetime object.
print(d.strftime('%Y-%m-%d'))
which gives
2020-02-20
You can use strftime to convert back in the format you need :
import datetime
s = "20200113"
temp = datetime.datetime.strptime(s, '%Y%m%d')
# 2020-01-13 00:00:00
final = temp.strftime('%Y-%m-%d')
print(final)
# 2020-01-13
Use datetime.date(year, month, day). Slice your string and convert to integers to get the year, month and day. Now it is a datetime.date object, you can use it for other things. Here, however, we use .strftime to convert it back to text in your desired format.
s = "20200113"
year = int(s[:4]) # 2020
month = int(s[4:6]) # 1
day = int(s[6:8]) # 13
>>> datetime.date(year, month, day).strftime('%Y-%m-%d')
'2020-01-13'
You can also convert directly via strings.
>>> f'{s[:4]}-{s[4:6]}-{s[6:8]}'
'2020-01-13'
You can use .date() on datetime objects to 'remove' the time.
my_time_str = str(final.date())
will give you the wanted result
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")
I have this column where the string has date, month, year and also time information. I need to take the date, month and year only.
There is no space in the string.
The string is on this format:
date
Tuesday,August22022-03:30PMWIB
Monday,July252022-09:33PMWIB
Friday,January82022-09:33PMWIB
and I expect to get:
date
2022-08-02
2022-07-25
2022-01-08
How can I get the date, month and year only and change the format into yyyy-mm-dd in python?
thanks in advance
Use strptime from datetime library
var = "Tuesday,August22022-03:30PMWIB"
date = var.split('-')[0]
formatted_date = datetime.strptime(date, "%A,%B%d%Y")
print(formatted_date.date()) #this will get your output
Output:
2022-08-02
You can use the standard datetime library
from datetime import datetime
dates = [
"Tuesday,August22022-03:30PMWIB",
"Monday,July252022-09:33PMWIB",
"Friday,January82022-09:33PMWIB"
]
for text in dates:
text = text.split(",")[1].split("-")[0]
dt = datetime.strptime(text, '%B%d%Y')
print(dt.strftime("%Y-%m-%d"))
An alternative/shorter way would be like this (if you want the other date parts):
for text in dates:
dt = datetime.strptime(text[:-3], '%A,%B%d%Y-%I:%M%p')
print(dt.strftime("%Y-%m-%d"))
The timezone part is tricky and works only for UTC, GMT and local.
You can read more about the format codes here.
strptime() only accepts certain values for %Z:
any value in time.tzname for your machine’s locale
the hard-coded values UTC and GMT
You can convert to datetime object then get string back.
from datetime import datetime
datetime_object = datetime.strptime('Tuesday,August22022-03:30PM', '%A,%B%d%Y-%I:%M%p')
s = datetime_object.strftime("%Y-%m-%d")
print(s)
You can use the datetime library to parse the date and print it in your format. In your examples the day might not be zero padded so I added that and then parsed the date.
import datetime
date = 'Tuesday,August22022-03:30PMWIB'
date = date.split('-')[0]
if not date[-6].isnumeric():
date = date[:-5] + "0" + date[-5:]
newdate = datetime.datetime.strptime(date, '%A,%B%d%Y').strftime('%Y-%m-%d')
print(newdate)
# prints 2022-08-02
I have a date in the format 2012-01-01 06:00:00. I want to get only the date in the format 2012-01-01.
I've tried multiple links such as Converting (YYYY-MM-DD-HH:MM:SS) date time
But, I could not find the solution.
Parse. str -> date.
from datetime import datetime
s = "2012-01-01 06:00:00"
dt = datetime.strptime(s, "%Y-%m-%d %H:%M:%S")
Format. date -> str.
s_ymd = dt.strftime("%Y-%m-%d")
Result:
>>> s_ymd
'2012-01-01'
Assuming your date is a string, the following works fine:
str = "2012-01-01 06:00:00"
print(str[:10])
The notation [:10] basically means "take first 10 characters of the string".
I want to change date format using Python. I don't know How to do that.
I have date format as shown below
2020-10-22 12:14:41.293000+00:00
I want to change above date format into below format:
date(2020, 10, 22)
I want this format because I want to get different between two dates.
with above format I can get difference by using following code,
d0 = date(2017, 8, 18)
d1 = date(2017, 10, 26)
delta = d1 - d0
print(delta.days)
So how to Change date Format as I discussed above.and also let me know if anyone know other way to get difference between these two dates 2020-10-22 12:14:41.293000+00:00 and 2020-10-25 12:14:41.293000+00:00 without changing its formet.I will be thankful if anyone can help me with this issue.
use fromisoformat and date() to get only the date, without the time:
from datetime import datetime
d = datetime.fromisoformat('2020-10-22 12:14:41.293000+00:00').date()
# d
# datetime.date(2020, 10, 22)
To convert a string to a datetime object, we use the datetime module.
from datetime import datetime
def str_to_date(s):
s = s.split()[0] # Separate the str by spaces and get the first item (the date)
s = datetime.strptime(s, "%Y-%m-%d")
return s # You now have a datetime object to do operations on.
If the format is always in the form YYYY-MM-DD ..... then you can strip the first 10 characters of the date and use datetime to convert into datetime object
from datetime import datetime
unformatted_date = "2020-10-22 12:14:41.293000+00:00"
date_obj = datetime.strptime(unformatted_date[0:11], "%Y-%m-%d").date()
print(date_obj)
Change string to datetime format:
from datetime import datetime
date1 = datetime.strptime('2020-10-22 12:14:41.293000+00:00', '%Y-%m-%d %H:%M:%S.%f%z').date()
date2 = datetime.strptime('2020-10-23 12:14:41.293000+00:00', '%Y-%m-%d %H:%M:%S.%f%z').date()
print(date2-date1)
#1 day, 0:00:00
I'm trying to subtract a day from this date 06-30-2019 in order to make it 06-29-2019 but can't figure out any way to achive that.
I've tried with:
import datetime
date = "06-30-2019"
date = datetime.datetime.strptime(date,'%m-%d-%Y').strftime('%m-%d-%Y')
print(date)
It surely gives me back the date I used above.
How can I subtract a day from a date in the above format?
try this
import datetime
date = "06/30/19"
date = datetime.datetime.strptime(date, "%m/%d/%y")
NewDate = date + datetime.timedelta(days=-1)
print(NewDate) # 2019-06-29 00:00:00
Your code:
date = "06-30-2019"
date = datetime.datetime.strptime(date,'%m-%d-%Y').strftime('%m-%d-%Y')
Check type of date variable.
type(date)
Out[]: str
It is in string format. To perform subtraction operation you must convert it into date format first. You can use pd.to_datetime()
# Import packages
import pandas as pd
from datetime import timedelta
# input date
date = "06-30-2019"
# Convert it into pd.to_datetime format
date = pd.to_datetime(date)
print(date)
# Substracting days
number_of_days = 1
new_date = date - timedelta(number_of_days)
print(new_date)
output:
2019-06-29 00:00:00
If you want to get rid of timestamp you can use:
str(new_date.date())
Out[]: '2019-06-29'
use timedelta
import datetime
date = datetime.datetime.strptime("06/30/19" ,"%m/%d/%y")
print( date - datetime.timedelta(days=1))