Invalid Dates Python - python-3.x

I am new to Python. I was just wondering, how can you write code that makes beyond a certain date an invalid input. For example, if the user inputs anything after 12/02/2013, it will produce an error. Everything after that date will work perfectly

As glibdud suggested, use datetime objects.
date = datetime.date(YYYY, MM, DD)
where (YYYY, MM, DD) are integers representing years, months, and days. The condition can then be checked in your script with
inputDate > maxDate
for example:
import datetime
maxDate = datetime.date(2013, 12, 2)
y = int(input('Enter year:'))
m = int(input('Enter numerical month (1-12):'))
d = int(input('Enter numerical day (1-31):'))
inputDate = datetime.date(y, m, d)
if inputDate > maxDate:
print('Error - date after 02 December 2013')
else:
print('Success!')
Gives:
Enter year:2018
Enter numerical month (1-12):1
Enter numerical day (1-31):1
Error - date after 02 December 2013
and
Enter year:2000
Enter numerical month (1-12):1
Enter numerical day (1-31):1
Success!

Related

Group by the dates to weeks

I have time in ms in epoch format, I need to translate this into a date and group it by a week number.
I tried the following procedure:
df.loc[0, 'seconds'] = df['seconds'].iloc[0]
for _, grp in df.groupby(pd.TimeGrouper(key='seconds', freq='7D')):x
print (grp)
df["week"].to_period(freq='w')
For example, if my 'seconds' column is presented like 1557499095332, then I want the 'dates' column to be 10-05-2019 20:08:15 and the 'Week' column to present W19 or 19.
How do I go about this?
Try using strftime method:
from datetime import datetime as dt
x = 1557499095332
dt.fromtimestamp(x/1000).strftime("%A, %B %d, %Y %I:%M:%S")
dt.fromtimestamp(x/1000).strftime("%W")
3rd line will return 'Friday, May 10, 2019 03:38:15'
4th line will return '18' (it's because 1st of January 2019 will return '0' as it's first week)

Pandas to recognize current date, and filter a date column relative to today's date

Having a lot of trouble translating the logic below in pandas/python, so I do not even have sample code or a df to work with :x
I run a daily report, that essentially filters for data from Monday thru the day before what 'Today' is. I have a Date column [ in dt.strftime('%#m/%#d/%Y') format] . It will never be longer than a Monday-Sunday scope.
1) Recognize the day it is 'today' when running the report, and recognize what day the closet Monday prior was. Filter the "Date" Column for the Monday-day before today's date [ in dt.strftime('%#m/%#d/%Y') format ]
2) Once the df is filtered for that, take this group of rows that have dates in the logic above, have it check for dates in a new column "Date2". If any dates are before the Monday Date, in Date2, change all of those earlier dates in 'Date2' to the Monday date it the 'Date' column.
3) If 'Today' is a Monday, then filter the scope from the Prior Monday through - Sunday in the "Date" Column. While this is filtered, do the step above [step 2] but also, for any dates in the "Date2" column that are Saturday and Sunday Dates - changes those to the Friday date.
Does this make sense?
Here're the steps:
from datetime import datetime
today = pd.to_datetime(datetime.now().date())
day_of_week = today.dayofweek
last_monday = today - pd.to_timedelta(day_of_week, unit='d')
# if today is Monday, we need to step back another week
if day_of_week == 0:
last_monday -= pd.to_timedelta(7, unit='d')
# filter for last Monday
last_monday_flags = (df.Date == last_mon)
# filter for Date2 < last Monday
date2_flags = (df.Date2 < last_monday)
# update where both flags are true
flags = last_monday_flags & date2_flags
df.loc[flags, 'Date2'] = last_monday
# if today is Monday
if day_of_week == 0:
last_sunday = last_monday + pd.to_timedelta(6, unit='d')
last_sat = last_sunday - pd.to_timedelta(1, unit='d')
last_week_flags = (df.Date >= last_monday) & (df.Date <= next_sunday)
last_sat_flags = (df.Date2 == last_sat)
last_sun_flags = (df.Date2 == last_sun)
# I'm just too lazy and not sure how Sat and Sun relates to Fri
# but i guess just subtract 1 day or 2 depending on which day
...

How to extract or validate date format from a text using python?

I'm trying to execute this code:
import datefinder
string_with_dates = 'The stock has a 04/30/2009 great record of positive Sept 1st, 2005 earnings surprises, having beaten the trade Consensus EPS estimate in each of the last four quarters. In its last earnings report on May 8, 2018, Triple-S Management reported EPS of $0.6 vs.the trade Consensus of $0.24 while it beat the consensus revenue estimate by 4.93%.'
matches = datefinder.find_dates(string_with_dates)
for match in matches:
print(match)
The output is:
2009-04-30 00:00:00
2005-09-01 00:00:00
2018-05-08 00:00:00
2019-02-04 00:00:00
The last date has come due to the percentage value 4.93% ... How to overcome this situation?
I cannot fix the datefinder module issue. You stated that you needed a solution, so I put this together for you. It's a work in progress, which means that you can adjusted it as needed. Also, some of the regex could have been consolidated, but I wanted to break them out for you. Hopefully, this answer helps you until you find another solution that works better for your needs.
import re
string_with_dates = 'The stock has a 04/30/2009 great record of positive Sept 1st, 2005 earnings surprises having beaten the trade Consensus EPS estimate in each of the last ' \
'four quarters In its last earnings report on March 8, 2018, Triple-S Management reported EPS of $0.6 vs.the trade Consensus of $0.24 while it beat the ' \
'consensus revenue estimate by 4.93%. The next trading day will occur at 2019-02-15T12:00:00-06:30'
def find_dates(input):
'''
This function is used to extract date strings from provide text.
Symbol references:
YYYY = four-digit year
MM = two-digit month (01=January, etc.)
DD = two-digit day of month (01 through 31)
hh = two digits of hour (00 through 23) (am/pm NOT allowed)
mm = two digits of minute (00 through 59)
ss = two digits of second (00 through 59)
s = one or more digits representing a decimal fraction of a second
TZD = time zone designator (Z or +hh:mm or -hh:mm)
:param input: text
:return: date string
'''
date_formats = [
# Matches date format MM/DD/YYYY
'(\d{2}\/\d{2}\/\d{4})',
# Matches date format MM-DD-YYYY
'(\d{2}-\d{2}-\d{4})',
# Matches date format YYYY/MM/DD
'(\d{4}\/\d{1,2}\/\d{1,2})',
# Matches ISO 8601 format (YYYY-MM-DD)
'(\d{4}-\d{1,2}-\d{1,2})',
# Matches ISO 8601 format YYYYMMDD
'(\d{4}\d{2}\d{2})',
# Matches full_month_name dd, YYYY or full_month_name dd[suffixes], YYYY
'(January|February|March|April|May|June|July|August|September|October|November|December)(\s\d{1,2}\W\s\d{4}|\s\d(st|nd|rd|th)\W\s\d{4})',
# Matches abbreviated_month_name dd, YYYY or abbreviated_month_name dd[suffixes], YYYY
'(Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sept|Oct|Nov|Dec)(\s\d{1,2}\W\s\d{4}|\s\d(st|nd|rd|th)\W\s\d{4})',
# Matches ISO 8601 format with time and time zone
# yyyy-mm-ddThh:mm:ss.nnnnnn+|-hh:mm
'\d{4}-\d{2}-\d{2}T\d{2}:\d{2}:\d{2}(\+|-)\d{2}:\d{2}',
# Matches ISO 8601 format Datetime with timezone
# yyyymmddThhmmssZ
'\d{8}T\d{6}Z',
# Matches ISO 8601 format Datetime with timezone
# yyyymmddThhmmss+|-hhmm
'\d{8}T\d{6}(\+|-)\d{4}'
]
for item in date_formats:
date_format = re.compile(r'\b{}\b'.format(item), re.IGNORECASE|re.MULTILINE)
find_date = re.search(date_format, input)
if find_date:
print (find_date.group(0))
find_dates(string_with_dates)
# outputs
04/30/2009
March 8, 2018
Sept 1st, 2005
2019-02-15T12:00:00-06:30

Python date range based on current date

I'm looking to achieve a search of files based on user input on a date. For example, I'm attempting to write the script to ask user for a range in which to search (month to date, last full week, or specific day).
last full week needs to go backward, to the last full week - so if today is Wednesday, the script should go back to the previous (2)Sunday(s) as a start range to the Saturday that just past, while also accounting for what day it is currently:
Sun(start)---Mon---Tue---Wed---Thu---Fri---Sat(end)---Sun---Mon---Tue---Wed (today)
Howevver, it needs to also account for what day it is in relation to the above, meaning that regardless of what "today" is, the search criteria is always one full week behind (if its Sunday, it just goes to last sunday to 'yesterday, Saturday')
From some examples attempting similar things I've seen here and here, I've attempted to join, modify, and add over the last couple of days:
import datetime
import os
import dateutil.relativedelta
import timedelta
class AuditFileCheck():
"""File Compliance Checker."""
def datechoice(self):
"""Select date."""
print("Checking the Audit Files for compliance.")
print("Today is", datetime.date.today().strftime(" %A."))
print("\nI will check either for file compliances."
"\nSearch criteria is either by MONTH to date, last full WEEK, "
"or individual DAY: [M/W/D]")
print(now.strftime('Week of %Y%m%d'))
weekbefore = now - timedelta(days=6)
print(
"Week of {weekbefore:%A, %m-%d-%Y} to {now:%A, %m-%d-%Y}").format(**vars())
input_search = input(
"Enter search range: Month to date, Prior Week, or by day [M/W/D]")
def search_month(d, w, m, weekday, month):
"""Establish search from month start, or prior month if today is first of current month."""
if input_search.lower() == "m":
current_month = datetime(today.month, 1, today.year)
if current_month == datetime.today():
current_month == dateutil.relativedelta.relativedelta(
months=-1)
return m - datetime.timedelta(current_month)
m = current_month()
print(current_month)
# TODO ensure the prior month is used if 'today' is before the end of
# first full week in current month
if input_search.lower() == "w":
prior_week = weekday + d.weekday()
if prior_week >= 0: # Target day already happened this week
prior_week -= 6
return d - datetime.timedelta(prior_week)
d = datetime.date.today()
# 6 = Sunday, 0 = Monday, 1 = Tuesday...
previous_monday = previous_weekday(d, 6)
print(previous_monday)
# TODO search files
if input_search.lower() == "d":
day_search = input(
"Enter a specific date within to search [YYYYMMDD]")
return d
print("Searching through...")
# TODO search files from set_day
This bit:
previous_sunday = previous_weekday(d, 8)
adjusting the integer adjusts how far back it looks.
I'm having some trouble with getting this to function properly. What am I doing wrong here? The more I attempt to play with it, the more confused I become and less it works...

taking date input and grouping pandas datfarme as per time

I have a pandas dataframe with date and time values as follows.
Date Time Pattern
0 06/01/13 0:00:01 A
1 06/02/13 1:00:01 B
2 06/03/13 2:00:01 A
3 06/04/13 3:00:01 C
Now i intend to take date input from user as follows:
date = str(input('Input date in mm-dd-yy format'))
Now how should i find/group by all the rows with input date by user and copy it to a new dataframe. I tried many things but got confused with datatime conversion.
How should i go about it?
First make sure your Date column is datetime
df.Date = pd.to_datetime(df.Date)
Then use query
date = pd.to_datetime(input('Input date in mm-dd-yyyy format'))
df.query('Date == #date')
response to #learningprogramming
You can include other criteria in query
date = pd.to_datetime(input('Input date in mm-dd-yyyy format: '))
df.query('Date == #date & Pattern == "B"')
loc works as well
date = pd.to_datetime(input('Input date in mm-dd-yyyy format: '))
df.loc[(df.Date == date) & (df.Pattern == 'B')]
putting all in the inputs
date = pd.to_datetime(input('Input date in mm-dd-yyyy format: '))
pattern = str(input('Input pattern type: '))
df.query('Date == #date & Pattern == #pattern')
Is the column named 'Date' a string? If so, you can try something like:
subset = df[df['Date'] == date]

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