I'm new to python. my question is how to pass parameter to date.strftime() or a workaround
Below is the code
from datetime import date
dl_date = date.today()
p = '%d%b%Y' # the format may vary %d%B%Y or %d%m%Y or % d%M%Y etc
file_date_format = "{0}/{1}/{2}".format(str(dl_date.strftime('%r')),str(dl_date.strftime('%r').upper())
, str(dl_date.strftime('%r'))) % (p[:2], p[2:4], p[4:6])
print(file_date_format)
Help is much appreciated.
No need to use percent style string formatting here. Just stick the p slices directly in the strftime calls.
from datetime import date
dl_date = date.today()
p = '%d%b%Y' # the format may vary %d%B%Y or %d%m%Y or % d%M%Y etc
file_date_format = "{0}/{1}/{2}".format(
str(dl_date.strftime(p[:2])),
str(dl_date.strftime(p[2:4]).upper()),
str(dl_date.strftime(p[4:6]))
)
print(file_date_format)
Result:
14/NOV/2014
Related
import datetime as dt
from dateutil.tz import gettz
import time
timezone_a = "Japan"
timezone_b = "Europe/London"
unix_time = 1619238722
result = dt.datetime.fromtimestamp(unix_time, gettz(timezone_a)).strftime("%Y-%m-%d-%H-%M-%S")
print(result, timezone_a)
result = dt.datetime.fromtimestamp(unix_time, gettz(timezone_b)).strftime("%Y-%m-%d-%H-%M-%S")
print(result, timezone_b)
# This code prints
"""
2021-04-24-13-32-02 Japan
2021-04-24-05-32-02 Europe/London
I am trying to reverse it backwards so that input is
2021-04-24-13-32-02 Japan
2021-04-24-05-32-02 Europe/London
And output is 1619238722
"""
Hello, I am trying to figure out how to convert a string with a timezone into Unix time. Any help would be apreciated. Thanks!
afaik, there is no built-in method in the standard lib to parse IANA time zone names. But you can do it yourself like
from datetime import datetime
from zoneinfo import ZoneInfo # Python 3.9+
t = ["2021-04-24-13-32-02 Japan", "2021-04-24-05-32-02 Europe/London"]
# split strings into tuples of date/time + time zone
t = [elem.rsplit(' ', 1) for elem in t]
# parse first element of the resulting tuples to datetime
# add time zone (second element from tuple)
# and take unix time
unix_t = [datetime.strptime(elem[0], "%Y-%m-%d-%H-%M-%S")
.replace(tzinfo=ZoneInfo(elem[1]))
.timestamp()
for elem in t]
# unix_t
# [1619238722.0, 1619238722.0]
See if this code works.
# convert string to datetimeformat
date = datetime.datetime.strptime(date, "%Y-%m-%d-%H-%M-%S %Z")
# convert datetime to timestamp
unixtime = datetime.datetime.timestamp(date)
I have a date variable calls today_date as below. I need to get the 1st calendar day of the current and next month.
In my case, today_date is 4/17/2021, I need to create two more variables calls first_day_current which should be 4/1/2021, and first_day_next which should be 5/1/2021.
Any suggestions are greatly appreciated
import datetime as dt
today_date
'2021-04-17'
Getting just the first date of a month is quite simple - since it equals 1 all the time. You can even do this without needing the datetime module to simplify calculations for you, if today_date is always a string "Year-Month-Day" (or any consistent format - parse it accordingly)
today_date = '2021-04-17'
y, m, d = today_date.split('-')
first_day_current = f"{y}-{m}-01"
y, m = int(y), int(m)
first_day_next = f"{y+(m==12)}-{m%12+1}-01"
If you want to use datetime.date(), then you'll anyway have to convert the string to (year, month, date) ints to give as arguments (or do today_date = datetime.date.today().
Then .replace(day=1) to get first_day_current.
datetime.timedelta can't add months (only upto weeks), so you'll need to use other libraries for this. But it's more imports and calculations to do the same thing in effect.
I found out pd.offsets could accomplish this task as below -
import datetime as dt
import pandas as pd
today_date #'2021-04-17' this is a variable that is being created in the program
first_day_current = today_date.replace(day=1) # this will be 2021-04-01
next_month = first_day_current + pd.offsets.MonthBegin(n=1)
first_day_next = next_month.strftime('%Y-%m-%d') # this will be 2021-05-01
I was given a question to get two time input from user as a function(opening_time,closing_time) and I had to determine the difference between the time,but if one of these values are not in time format,the returned value should be -1.I have computed the time difference but,I am unable to fix a condition that if any one of the variable is not in time format,return -1.
Please I am new to coding,so apologize for any mistake and be kind to write simple solution,not so complex one.
from datetime import datetime
def compute_opening_duration(opening_time, closing_time):
str_format = "%H:%M:%S"
if opening_time or closing_time != datetime.time.str_format:
print(-1)
else:
tdelta = datetime.strptime(closing_time,str_format)
- datetime.strptime(opening_time,str_format)
print(tdelta)
Try this - it will try to cast the inputs to a datetime using your provided string format. If it fails on either one, it will print -1.
from datetime import datetime
def compute_opening_duration(opening_time, closing_time):
str_format = "%H:%M:%S"
try:
t_open_time = datetime.strptime(closing_time,str_format)
t_closing_time = datetime.strptime(opening_time,str_format)
tdelta = datetime.strptime(closing_time,str_format) - datetime.strptime(opening_time,str_format)
print(tdelta)
except:
print(-1)
compute_opening_duration("04:10:21", "08:22:12")
Looking for help on how to sort a python3 dictonary by a datetime object (as shown below, a value in the dictionary) using the timestamp below.
datetime: "2018-05-08T14:06:54-04:00"
Any help would be appreciated, spent a bit of time on this and know that to create the object I can do:
format = "%Y-%m-%dT%H:%M:%S"
# Make strptime obj from string minus the crap at the end
strpTime = datetime.datetime.strptime(ts[:-6], format)
# Create string of the pieces I want from obj
convertedTime = strpTime.strftime("%B %d %Y, %-I:%m %p")
But I'm unsure how to go about comparing that to the other values where it accounts for both day and time correctly, and cleanly.
Again, any nudges in the right direction would be greatly appreciated!
Thanks ahead of time.
Datetime instances support the usual ordering operators (< etc), so you should order in the datetime domain directly, not with strings.
Use a callable to convert your strings to timezone-aware datetime instances:
from datetime import datetime
def key(s):
fmt = "%Y-%m-%dT%H:%M:%S%z"
s = ''.join(s.rsplit(':', 1)) # remove colon from offset
return datetime.strptime(s, fmt)
This key func can be used to correctly sort values:
>>> data = {'s1': "2018-05-08T14:06:54-04:00", 's2': "2018-05-08T14:05:54-04:00"}
>>> sorted(data.values(), key=key)
['2018-05-08T14:05:54-04:00', '2018-05-08T14:06:54-04:00']
>>> sorted(data.items(), key=lambda item: key(item[1]))
[('s2', '2018-05-08T14:05:54-04:00'), ('s1', '2018-05-08T14:06:54-04:00')]
I have a question very similar to this one and this one but I'm stuck on some rounding issue.
I have a time series from a netCDF file and I'm trying to convert them to a datetime format. The format of the time series is in 'days since 1990-01-01 00:00:00'. Eventually I want output in the format .strftime('%Y%m%d.%H%M'). So for example I read my netCDF file as follows
import netCDF4
nc = netCDF4.Dataset(file_name)
time = np.array(nc['time'][:])
I then have
In [180]: time[0]
Out[180]: 365
In [181]: time[1]
Out[181]: 365.04166666651145
I then did
In [182]: start = datetime.datetime(1990,1,1)
In [183]: delta = datetime.timedelta(time[1])
In [184]: new_time = start + delta
In [185]: print(new_time.strftime('%Y%m%d.%H%M'))
19910101.0059
Is there a a way to "round" to the nearest hour so I get 19910101.0100?
You can round down with datetime.replace(), and round up by adding an hour to the rounded down value using datetime.timedelta(hours=1).
import datetime
def round_to_hour(dt):
dt_start_of_hour = dt.replace(minute=0, second=0, microsecond=0)
dt_half_hour = dt.replace(minute=30, second=0, microsecond=0)
if dt >= dt_half_hour:
# round up
dt = dt_start_of_hour + datetime.timedelta(hours=1)
else:
# round down
dt = dt_start_of_hour
return dt
Note that since we're using replace the values we're not replacing (like the timezone - tzinfo) will be preserved.
I don't think datetime provides a way to round times, you'll have to provide the code to do that yourself. Something like this should work:
def round_to_hour(dt):
round_delta = 60 * 30
round_timestamp = datetime.datetime.fromtimestamp(dt.timestamp() + round_delta)
round_dt = datetime.datetime.fromtimestamp(round_timestamp)
return round_dt.replace(microsecond=0, second=0, minute=0)