Query Unix time by day in Python - python

I currently store a timestamp like this:
datetime = int(time.mktime(_scheduled_datetime.timetuple()))
> 1172969203.1
I need to then find all objects with a datetime of now().
Unfortunately, my application does not allow me to query ranges, only whole values. Is it possible to store only the days part of a timestamp i.e. 01/01/2016 So I could get all datetime values for today().

you can try:
import time
import datetime
timestamp = int(time.time())
# date timestamp
print int(
time.mktime(
datetime.date.fromtimestamp(timestamp).timetuple()
)
)
# iso format date string
print datetime.date.fromtimestamp(timestamp).isoformat()

Related

Unix timestamp conversion to time with microsecond accuracy of seconds

I have 1000 of UYC timestamps in csv file, I want to convert it into date and time but I am only interested in second like
Timestamps= 1666181576.26295,
1666181609.54292
19/10/2022 15:45:25.34568
from that I only have interest in 25.34568 seconds, also the numbers after points. How can I get this type of conversion in python? Mostly the search on the internet is interested in conversation from UTC to time and date but I also want precision in seconds.
from datetime import datetime
from decimal import Decimal
ts = 1666181576.26295
timestamp = datetime.fromtimestamp(ts)
result = timestamp.second + Decimal(timestamp.microsecond)/1000000
print(result)
Will result in 56.26295
You can use datetime,
from datetime import datetime
ts = 1666181576.26295
mseconds = datetime.utcfromtimestamp(ts).microsecond
Simplest way I can see to do this is by splitting the timestamp to output everything from seconds onwards
timestamp = 1666181609.54292
temp = datetime.utcfromtimestamp(timestamp)
output = str(temp)
print(output[17:])

Get previous day's date from date passed as a string in "YYYY-MM-DD" format

I'm trying to create a helper function getPreviousDay to be used in the backend (Flask).
From the front end, I'm receiving the date in "YYYY-MM-DD" format as a string.
From this, I want to get the date of the previous day in the same format as a string.
Here's a sample code of what I want to achieve.
def getPreviousDay(date):
'''
todo: previousDay should also be a string in "YYYY-MM-DD" format
'''
return previousDay
current_day = "2022-09-29" #YYYY-MM-DD
yesterday = getPreviousDay(current_day)
The datetime module provides date and timedelta types that can be used for this kind of thing. A "time delta" is a difference between two dates or times, in this case, 1 day. Subtracting one day from today's date gives yesterday's date.
import datetime
def getPreviousDay(date):
today = datetime.date.fromisoformat(date)
yesterday = today - datetime.timedelta(days=1)
return yesterday.isoformat()
This returns:
>>> getPreviousDay('2022-09-29')
'2022-09-28'
The reference documentation for the datetime module has more details.

How to add a certain time to a datetime?

I want to add hours to a datetime and use:
date = date_object + datetime.timedelta(hours=6)
Now I want to add a time:
time='-7:00' (string) plus 4 hours.
I tried hours=time+4 but this doesn't work. I think I have to int the string like int(time) but this doesn't work either.
Better you parse your time like below and access datetime attributes for getting time components from the parsed datetime object
input_time = datetime.strptime(yourtimestring,'yourtimeformat')
input_seconds = input_time.second # for seconds
input_minutes = input_time.minute # for minutes
input_hours = input_time.hour # for hours
# Usage: input_time = datetime.strptime("07:00","%M:%S")
Rest you have datetime.timedelta method to compose the duration.
new_time = initial_datetime + datetime.timedelta(hours=input_hours,minutes=input_minutes,seconds=input_seconds)
See docs strptime
and datetime format
You need to convert to a datetime object in order to add timedelta to your current time, then return it back to just the time portion.
Using date.today() just uses the arbitrary current date and sets the time to the time you supply. This allows you to add over days and reset the clock to 00:00.
dt.time() prints out the result you were looking for.
from datetime import date, datetime, time, timedelta
dt = datetime.combine(date.today(), time(7, 00)) + timedelta(hours=4)
print dt.time()
Edit:
To get from a string time='7:00' to what you could split on the colon and then reference each.
this_time = this_time.split(':') # make it a list split at :
this_hour = this_time[0]
this_min = this_time[1]
Edit 2:
To put it all back together then:
from datetime import date, datetime, time, timedelta
this_time = '7:00'
this_time = this_time.split(':') # make it a list split at :
this_hour = int(this_time[0])
this_min = int(this_time[1])
dt = datetime.combine(date.today(), time(this_hour, this_min)) + timedelta(hours=4)
print dt.time()
If you already have a full date to use, as mentioned in the comments, you should convert it to a datetime using strptime. I think another answer walks through how to use it so I'm not going to put an example.

Check date is within one year python

I have a date string formatted like this: "2017-05-31T06:44:13Z".
I need to check whether this date is within a one year span from today's date.
Which is the best method to do it: convert it into a timestamp and check, or convert into a date format?
Convert the timestamp to a datetime object so it can be compared with other datetime objects using <, >, =.
from datetime import datetime
from dateutil.relativedelta import relativedelta
# NOTE this format basically ignores the timezone. This may or may not be what you want
date_to_check = datetime.strptime('2017-05-31T06:44:13Z', '%Y-%m-%dT%H:%M:%SZ')
today = datetime.today()
one_year_from_now = today + relativedelta(years=1)
if today <= date_to_check <= one_year_from_now:
# do whatever
Use the datetime package together with timedelta:
import datetime
then = datetime.datetime.strptime("2017-05-31T06:44:13Z".replace('T',' ')[:-1],'%Y-%m-%d %H:%M:%S')
now = datetime.datetime.now()
d = datetime.timedelta(days = 365)
and simply check if now-d > then.

Extract Date from excel and append it in a list using python

I have an column in excel which has dates in the format ''17-12-2015 19:35". How can I extract the first 2 digits as integers and append it to a list? In this case I need to extract 17 and append it to a list. Can it be done using pandas also?
Code thus far:
import pandas as pd
Location = r'F:\Analytics Materials\files\paymenttransactions.csv'
df = pd.read_csv(Location)
time = df['Creation Date'].tolist()
print (time)
You could extract the day of each timestamp like
from datetime import datetime
import pandas as pd
location = r'F:\Analytics Materials\files\paymenttransactions.csv'
df = pd.read_csv(location)
timestamps = df['Creation Date'].tolist()
dates = [datetime.strptime(timestamp, '%d-%m-%Y %H:%M') for timestamp in timestamps]
days = [date.strftime('%d') for date in dates]
print(days)
The '%d-%m-%Y %H:%M'and '%d' bits are format specififers, that describe how your timestamp is formatted. See e.g. here for a complete list of directives.
datetime.strptime parses a string into a datetimeobject using such a specifier. dateswill thus hold a list of datetime instances instead of strings.
datetime.strftime does the opposite: It turns a datetime object into string, again using a format specifier. %d simply instructs strftime to only output the day of a date.

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