I am trying the following code to log using the timezone 'America/New_York'
the time is shown correct, but the %z and %Z are not showing the right time Zone
from datetime import datetime
from pytz import timezone
import logging
def timetz(*args):
tz = timezone('America/New_York')
print(tz)
dt = datetime.now(tz)
print(dt)
return dt.timetuple()
logger = logging.getLogger()
formatter = logging.Formatter(fmt='%(asctime)s::%(levelname)-8s\n%(message)s',datefmt='%Y-%m-%d %H:%M:%S %p %Z %z')
formatter.converter = timetz
formatter.default_msec_format = '%s.%03d'
handler= logging.StreamHandler()
handler.setFormatter(formatter)
handler.setLevel(logging.INFO)
logger.addHandler(handler)
logger.error('test')
I get
2021-03-05 14:53:04 PM UTC +0000::ERROR
test
But i should get
2021-03-05 14:53:04 PM EST -0500::ERROR
test
The %z and %Z are not working
I'm doing a parse from string ISO8601 to a datetime and it's working. Now I want to return datetime on localtime but my code is returning same timestamp from input:
def format_string_to_timestamp(dt, defaultTimezone='America/Sao_Paulo'):
origin_dt = datetime.strptime(dt, '%Y-%m-%dT%H:%M:%S.%f')
tz_local = pytz.timezone (defaultTimezone)
dt_local = origin_dt.astimezone(tz_local).replace(tzinfo=None)
print(dt)
print(dt_local)
print(origin_dt)
return dt_local.strftime('%Y-%m-%d %H:%M:%S')
# example input: 2019-02-25T17:58:53.753
What is missing to return dt_local as America/Sao_Paulo timezone?
I am trying to find the average time taken. The value 'x' will allow me to get the time taken for every row there is, but how am I able to find the average time taken for all the rows. I will think it is something like x divided by count, but I am not able to find a solution to this... any pros out there can help me??
import datetime,time,csv
from itertools import islice
from Tkinter import Tk #python GUI programming
from tkFileDialog import askopenfilename
from collections import Counter
from datetime import datetime
import pandas
Tk().withdraw()
category_list=[]
description_list=[]
reported_date=[]
acknowledged_date=[]
count = 0
# hard code all possible date formats
date_formats = ['%m/%d/%Y %H:%M', '%-d-%b-%y', '%d/%m/%Y %h:%M %p', '%d/%m/%Y %H:%M', '%A, %d %B %Y %H:%M','%A, %d %B %Y %H:%M','%A %d %B %Y %H%M',"%d/%m/%Y %H:%M %p"," %d/%m/%Y %H:%M %p", '%d-%b-%y' ,
'%d.%m.%Y', '%d %b %Y %H%M hrs', '%d %b %Y %H%M', '%d-%m-%y', '%d-%b-%y', '%b-%d-%y', '%d-%a-%y','%e-%a-%y','%b %d %Y %H%M hrs','%d/%b/%Y %m:%M %p','%A, %e %B %Y %H:%M',' %d/%m/%Y %h:%M','%d-%b-%y','%m/%d/%Y %H:%M:%S %p']
#file = askopenfilename() #ask user which file to open
#f = open(file,'r')
with open('Feedback and Complaints_Sample Dataset.csv', 'rb') as f:
reader = csv.reader(f)
header = next(reader) #read 2nd line onwards
data= [] #make a list called data
for row in reader: #assign data in every column and name them respectively
for format in date_formats:
try:
reported_on = datetime.strptime(row[0], format) #try and get the dates
acknowledged_on = datetime.strptime(row[12], format) #try and get the dates
x= acknowledged_on-reported_on #time taken to acknowledge
#acknowledged_date.append(acknowledged_on)
#reported_date.append(reported_on)
count += 1
break # if correct format, dont test any other formats
except ValueError:
pass # if incorrect format, try other formats`enter code here`
Subtracting two datetime objects creates a timedelta object. You need to keep a total time, so create a timedelta object, and for each x add it to your total.
At the end, you can then divide your total_time by your count:
import csv
from itertools import islice
from datetime import datetime, timedelta
count = 0
total_time = timedelta()
# hard code all possible date formats
date_formats = ['%m/%d/%Y %H:%M', '%-d-%b-%y', '%d/%m/%Y %h:%M %p', '%d/%m/%Y %H:%M', '%A, %d %B %Y %H:%M','%A, %d %B %Y %H:%M','%A %d %B %Y %H%M',"%d/%m/%Y %H:%M %p"," %d/%m/%Y %H:%M %p", '%d-%b-%y' ,
'%d.%m.%Y', '%d %b %Y %H%M hrs', '%d %b %Y %H%M', '%d-%m-%y', '%d-%b-%y', '%b-%d-%y', '%d-%a-%y','%e-%a-%y','%b %d %Y %H%M hrs','%d/%b/%Y %m:%M %p','%A, %e %B %Y %H:%M',' %d/%m/%Y %h:%M','%d-%b-%y','%m/%d/%Y %H:%M:%S %p']
with open('Feedback and Complaints_Sample Dataset.csv', 'rb') as f:
reader = csv.reader(f)
header = next(reader) #read 2nd line onwards
for row in reader:
for format in date_formats:
try:
reported_on = datetime.strptime(row[0], format) #try and get the dates
acknowledged_on = datetime.strptime(row[12], format) #try and get the dates
x = acknowledged_on - reported_on #time taken to acknowledge
total_time += x
count += 1
break # if correct format, don't test any other formats
except ValueError:
pass # if incorrect format, try other formats`enter code here`
print "Total time taken:", total_time
print "Average time taken:", total_time / count
Note: your logic for the date_formats implies that both dates in a single row will always share the same date format.
I have the input date of 2017-08-22T11:32:31+10:00
I wish to convert this to UTC which would be 2017-08-22+01:32:31
Code so far
from datetime import datetime, timedelta
from pytz import timezone
import pytz
fmt = "%Y-%m-%d+%H:%M:%S"
now_time = datetime('2017-08-22T11:32:31+10:00')
zone = 'UTC'
now_time = now_time.timezone(zone)
print now_time.strftime(fmt)
Error
now_time = datetime('2017-08-22T11:32:31+10:00')
TypeError: an integer is required
You can use dateutil.parser to infer the datetime format when creating your datetime object.
import dateutil.parser
your_date = dateutil.parser.parse('2017-08-22T11:32:31+10:00')
Next, you can use the .astimezone function to convert your_date to UTC:
utc_date = your_date.astimezone(pytz.utc)
print(utc_date)
Output:
2017-08-22 01:32:31+00:00
I am looking for a comparison of two times in Python. One time is the real time from computer and the other time is stored in a string formatted like "01:23:00".
import time
ctime = time.strptime("%H:%M:%S") # this always takes system time
time2 = "08:00:00"
if (ctime > time2):
print("foo")
import datetime
now = datetime.datetime.now()
my_time_string = "01:20:33"
my_datetime = datetime.datetime.strptime(my_time_string, "%H:%M:%S")
# I am supposing that the date must be the same as now
my_datetime = now.replace(hour=my_datetime.time().hour, minute=my_datetime.time().minute, second=my_datetime.time().second, microsecond=0)
if (now > my_datetime):
print("Hello")
EDIT:
The above solution was not taking into account leap second days (23:59:60). Below is an updated version that deals with such cases:
import datetime
import calendar
import time
now = datetime.datetime.now()
my_time_string = "23:59:60" # leap second
my_time_string = now.strftime("%Y-%m-%d") + " " + my_time_string # I am supposing the date must be the same as now
my_time = time.strptime(my_time_string, "%Y-%m-%d %H:%M:%S")
my_datetime = datetime.datetime(1970, 1, 1) + datetime.timedelta(seconds=calendar.timegm(my_time))
if (now > my_datetime):
print("Foo")
https://docs.python.org/2/library/datetime.html
The datetime module will parse dates, times, or combined date-time values into objects that can be compared.
from datetime import datetime
current_time = datetime.strftime(datetime.utcnow(),"%H:%M:%S") #output: 11:12:12
mytime = "10:12:34"
if current_time > mytime:
print "Time has passed."