I have scraped some data from the yellow pages using scrapy.
The hours of the business provided from scraping are in a 12-hour format and I need to convert it into 24 hours.
The format for the business hours I scraped are:
Mon - Fri:,10:00 am - 7:00 pm.
I need to extract the two values for opening and closing time, convert them both into 24-hour format and then concatenate the string back together again.
As a result, I need to devise a regex that will extract the time and then change it into a 24 hour format.
The final string should (as per previous example) should be:
Mon - Fri:,10:00 - 19:00
I have tried different regex. I tried the following:
import re
txt = 'Mon - Fri:,10:00 am - 7:00 pm'
data = re.findall(r'\s(\d{2}\:\d{2}\s?(?:AM|PM|am|pm))', txt)
print(data)
i am not python developer but we can in this way in javascript. you can convert logic into python
in this way you can convert this time to miltary time (24 hour)
https://jsfiddle.net/1hxojLdf/2/
let text='Mon - Fri:,10:00 am - 7:00 pm';
const regex=/(\w+\s-\s\w+:.)(\d{1,2}:\d{1,2}\s(am|pm))\s-\s(\d{1,2}:\d{1,2}\s(am|pm))/;
const result=text.match(regex);
let timeone=result[2];
let timetwo=result[4];
timeone= moment(timeone,"h:mm A").format('HH:mm');
timetwo= moment(timetwo,"h:mm A").format('HH:mm');
text=result[1]+timeone+"-"+timetwo;
alert(text)
Related
I need to calculate the date based on the number of minutes from the start of the year. I just need the day and month.
I have tried a few different classes within Python datetime but I can not get the desired result.
for example, at 1700 today, minutes since the 01/01/2022 is 263100
print(datetime.timedelta(0,0,0,0,263100))
This returns
182 days, 17:00:00
Before I explicitly write out the months and how many days they have and work it out that way, is there something already built in that I am missing within datetime?
I think this is what you are looking for. We add the minutes timedelta to the start datetime, and return it formatted.
from datetime import datetime, timedelta
def datesince_min(min:int, start:tuple=(2022,1,1)) -> str:
date = datetime(*start) + timedelta(minutes=min)
return date.strftime("%A, %B %d, %Y %I:%M:%S")
print(datesince_min(263100)) #Saturday, July 02, 2022 06:00:00
For more information regarding strftime formatting, refer to this.
Before I explicitly write out the months and how many days they have and work it out that way...
You shouldn't ever have to do that, and if you did it would almost certainly have to be for a system or language that is in development.
I found this code that should do the trick:
number_of_days = ((datetime.timedelta(0,0,0,0,263100)).split())[0]
months = (number_of_days - years * 365) // 30
days = (number_of_days - years * 365 - months*30)
print(months, days)
Where you replace 263100 with whatever minutes you wish ofc
(source: https://www.codesansar.com/python-programming-examples/convert-number-days-years-months-days.htm#:~:text=Python%20Source%20Code%3A%20Days%20to%20Years%2C%20Months%20%26,print%28%22Months%20%3D%20%22%2C%20months%29%20print%28%22Days%20%3D%20%22%2C%20days%29)
:)
Can below piece of shell date format be converted to python date format?
date_used = $(date --date="$(date +%Y-%m-15) - 1 month" "+%Y_%m")
As per my understanding this above format is just taking day as 15 of current month and it simply subtracts 1 month and results in giving output in the format of year_month.
Output of the above shell date format -->
echo $date_used = 2022_05
Can this particular scenario be done using python?
Any update about it would be really appreciable.
An equivalent would be:
from datetime import datetime,timedelta
# Current date, replace date with 15, subtract 30 days and format
date_used = (datetime.now().replace(day=15) - timedelta(days=30)).strftime('%Y_%m')
print(date_used)
Output:
2022_05
You can use python's datetime module to do this
it has a function named strptime you can use to read date and time data with format code very similar to unix date format (or maybe its even same i'm not sure)
and strftime to output in a certain format as well
you can see the functions and the format code to see if there's any different on
https://docs.python.org/3/library/datetime.html#strftime-and-strptime-format-codes
Example code
from datetime import datetime, timedelta
date = datetime.strptime((datetime.now() - timedelta(days=30)).strftime("%Y-%m") + "-15", "%Y-%m-%d")
print(date)
print(date.strftime("%Y_%m"))
output
2022-05-15 00:00:00
2022_05
I need to parse a few dates that are roughly in the format (1 or 2-digit year)-(Month abbreviation), for example:
5-Jun (June 2005)
13-Jan (January 2013)
I tried using strptime with the format %b-%y but it did not consistently produce the desired date. Per the documentation, this is because some years in my dataset are not zero-padded.
Further, when I tested the datetime module (please see below for my code) on the string "5-Jun", I got "2019-06-05", instead of the desired result (June 2005), even if I set yearfirst=True when calling parse.
from dateutil.parser import parse
parsed = parse("5-Jun",yearfirst=True)
print(parsed)
It will be easier if 0 is padded to single digit years, as it can be directly converted to time using format. Regular expression is used here to replace any instance of single digit number with it's '0 padded in front' value. I've used regex from here.
Sample code:
import re
match_condn = r'\b([0-9])\b'
replace_str = r'0\1'
datetime.strptime(re.sub(match_condn, replace_str, '15-Jun'), '%y-%b').strftime("%B %Y")
Output:
June 2015
One approach is to use str.zfill
Ex:
import datetime
d = ["5-Jun", "13-Jan"]
for date in d:
date, month = date.split("-")
date = date.zfill(2)
print(datetime.datetime.strptime(date+"-"+month, "%y-%b").strftime("%B %Y"))
Output:
June 2005
January 2013
Ah. I see from #Rakesh's answer what your data is about. I thought you needed to parse the full name of the month. So you had your two terms %b and %y backwards, but then you had the problem with the single-digit years. I get it now. Here's a much simpler way to get what you want if you can assume your dates are always in one of the two formats you indicate:
inp = "5-Jun"
t = time.strptime(("0" + inp)[-6:], "%y-%b")
I have exported a list of AD Users out of AD and need to validate their login times.
The output from the powershell script give lastlogin as LDAP/FILE time
EXAMPLE 130305048577611542
I am having trouble converting this to readable time in pandas
Im using the following code:
df['date of login'] = pd.to_datetime(df['FileTime'], unit='ns')
The column FileTime contains time formatted like the EXAMPLE above.
Im getting the following output in my new column date of login
EXAMPLE 1974-02-17 03:50:48.577611542
I know this is being parsed incorrectly as when i input this date time on a online converter i get this output
EXAMPLE:
Epoch/Unix time: 1386031258
GMT: Tuesday, December 3, 2013 12:40:58 AM
Your time zone: Monday, December 2, 2013 4:40:58 PM GMT-08:00
Anyone have an idea of what occuring here why are all my dates in the 1970'
I know this answer is very late to the party, but for anyone else looking in the future.
The 18-digit Active Directory timestamps (LDAP), also named 'Windows NT time format','Win32 FILETIME or SYSTEMTIME' or NTFS file time. These are used in Microsoft Active Directory for pwdLastSet, accountExpires, LastLogon, LastLogonTimestamp and LastPwdSet. The timestamp is the number of 100-nanoseconds intervals (1 nanosecond = one billionth of a second) since Jan 1, 1601 UTC.
Therefore, 130305048577611542 does indeed relate to December 3, 2013.
When putting this value through the date time function in Python, it is truncating the value to nine digits. Therefore the timestamp becomes 130305048 and goes from 1.1.1970 which does result in a 1974 date!
In order to get the correct Unix timestamp you need to do:
(130305048577611542 / 10000000) - 11644473600
Here's a solution I did in Python that worked well for me:
import datetime
def ad_timestamp(timestamp):
if timestamp != 0:
return datetime.datetime(1601, 1, 1) + datetime.timedelta(seconds=timestamp/10000000)
return np.nan
So then if you need to convert a Pandas column:
df.lastLogonTimestamp = df.lastLogonTimestamp.fillna(0).apply(ad_timestamp)
Note: I needed to use fillna before using apply. Also, since I filled with 0's, I checked for that in the conversion function about, if timestamp != 0. Hope that makes sense. It's extra stuff but you may need it to convert the column in question.
I've been stuck on this for couple of days. But now i am ready to share really working solution in more easy to use form:
import datetime
timestamp = 132375402928051110
value = datetime.datetime (1601, 1, 1) +
datetime.timedelta(seconds=timestamp/10000000) ### combine str 3 and 4
print(value.strftime('%Y-%m-%d %H:%M:%S'))
Hi everyone I have a code like this to calculate the exact day on six_months back but unfortunately it prints the yy-mm-dd format and I want the dd/mm/yy format how do I do it(I tried to convert but it doesn't work)?What's wrong with my code?
import datetime
six_months = str(datetime.date.today() - datetime.timedelta(6*365/12-1)
datetime.datetime.strptime(six_months, '%Y-%m-%d').strftime('%d/%m/%y')
expected output=04/02/2017
current output=2017-02-04
The code is fine, you just forgot to save to result of
datetime.datetime.strptime(six_months, '%Y-%m-%d').strftime('%d/%m/%y')
to any variable. strptime doesn't change the object it is called on in any way, it returns a string.
I reckon your algorithm for computing 6 months back doesn't correspond to the real-world understanding of that phrase. Six months back from 4 August is 4 February and your computation gives the right answer for that. But six months back from 4 September is 4 March, and your computation gives the answer 7 March.
Your code also unnecessarily formats the computed date to a string, and then has to parse the string back into a date to get the dd/mm/yy format you want.
import datetime
from dateutil.relativedelta import *
six_months = datetime.date.today() + relativedelta(months=-6)
print (f"{six_months:%d/%m/%y}")
Output (until tomorrow) is
04/02/17