Parsing Dates In Python - python

I have a list of dates from input like the ones below.
I am working on a project and only want to accept the dates that follow the format April 1, 1990 or the January 13, 2003 (taking in any month) format from user input, and any other date format I do not want to accept. I am struggling on how I would use the replace or find function to obtain these goals? Once I receive that format I want to print out the date in this format 7/19/22. If I have the dates in the right format I used the replace function to replace the space, and comma but how would I take that month and replace it with its numerical value? Sorry for all these questions I am just stuck and have been working on this for a while now.
April 1, 1990
November 2 1995
7/19/22
January 13, 2003
userinput = input("Please enter date")
parsed_date = userinput.replace(" ", "/", 2)
new_parsed_date = parsed_date.replace(',',"")
print(new_parsed_date)
March/1/2019 Here is my output when I parse the date. Is there also any easier way to do this task?

You should take a look at the strptime method of the Python datetime object.
Basically you would write code that looks like this:
>>> from datetime import datetime
>>> datetime.strptime("January 13, 2003", "%B %d, %Y")
datetime.datetime(2003, 1, 13, 0, 0)
Documentation for strptime: https://docs.python.org/3/library/datetime.html#datetime.datetime.strptime

I can't really understand what you're going to do. You said that you only want to accept the certain dates formats from user input and ignore other dates formats.
April 1, 1990 # Accept
January 13, 2003 # Accept
November 2 1995 # Ignore (lack of comma)
7/19/22 # Ignore (use numerical value in month field)
May I just think that you would like to accept the format like January 13, 2003 and print or save them in the format 01/13/2003?
Then you should consider strptime() and strftime() methods for datetime object.
# get the date string from user input
date_input = input("Please Enter Date: ")
input_format = "%B %d, %Y"
output_format = "%m/%d/%Y"
try:
parsered_date = datetime.datetime.strptime(date_input, input_format)
.strftime(output_format)
print(parsered_date)
except ValueError:
print("This is the incorrect date string format.")

Related

Python3.7 Convert comma delimited datetime [duplicate]

This is my code:
import datetime
today = datetime.date.today()
print(today)
This prints: 2008-11-22 which is exactly what I want.
But, I have a list I'm appending this to and then suddenly everything goes "wonky". Here is the code:
import datetime
mylist = []
today = datetime.date.today()
mylist.append(today)
print(mylist)
This prints the following:
[datetime.date(2008, 11, 22)]
How can I get just a simple date like 2008-11-22?
The WHY: dates are objects
In Python, dates are objects. Therefore, when you manipulate them, you manipulate objects, not strings or timestamps.
Any object in Python has TWO string representations:
The regular representation that is used by print can be get using the str() function. It is most of the time the most common human readable format and is used to ease display. So str(datetime.datetime(2008, 11, 22, 19, 53, 42)) gives you '2008-11-22 19:53:42'.
The alternative representation that is used to represent the object nature (as a data). It can be get using the repr() function and is handy to know what kind of data your manipulating while you are developing or debugging. repr(datetime.datetime(2008, 11, 22, 19, 53, 42)) gives you 'datetime.datetime(2008, 11, 22, 19, 53, 42)'.
What happened is that when you have printed the date using print, it used str() so you could see a nice date string. But when you have printed mylist, you have printed a list of objects and Python tried to represent the set of data, using repr().
The How: what do you want to do with that?
Well, when you manipulate dates, keep using the date objects all long the way. They got thousand of useful methods and most of the Python API expect dates to be objects.
When you want to display them, just use str(). In Python, the good practice is to explicitly cast everything. So just when it's time to print, get a string representation of your date using str(date).
One last thing. When you tried to print the dates, you printed mylist. If you want to print a date, you must print the date objects, not their container (the list).
E.G, you want to print all the date in a list :
for date in mylist :
print str(date)
Note that in that specific case, you can even omit str() because print will use it for you. But it should not become a habit :-)
Practical case, using your code
import datetime
mylist = []
today = datetime.date.today()
mylist.append(today)
print mylist[0] # print the date object, not the container ;-)
2008-11-22
# It's better to always use str() because :
print "This is a new day : ", mylist[0] # will work
>>> This is a new day : 2008-11-22
print "This is a new day : " + mylist[0] # will crash
>>> cannot concatenate 'str' and 'datetime.date' objects
print "This is a new day : " + str(mylist[0])
>>> This is a new day : 2008-11-22
Advanced date formatting
Dates have a default representation, but you may want to print them in a specific format. In that case, you can get a custom string representation using the strftime() method.
strftime() expects a string pattern explaining how you want to format your date.
E.G :
print today.strftime('We are the %d, %b %Y')
>>> 'We are the 22, Nov 2008'
All the letter after a "%" represent a format for something:
%d is the day number (2 digits, prefixed with leading zero's if necessary)
%m is the month number (2 digits, prefixed with leading zero's if necessary)
%b is the month abbreviation (3 letters)
%B is the month name in full (letters)
%y is the year number abbreviated (last 2 digits)
%Y is the year number full (4 digits)
etc.
Have a look at the official documentation, or McCutchen's quick reference you can't know them all.
Since PEP3101, every object can have its own format used automatically by the method format of any string. In the case of the datetime, the format is the same used in
strftime. So you can do the same as above like this:
print "We are the {:%d, %b %Y}".format(today)
>>> 'We are the 22, Nov 2008'
The advantage of this form is that you can also convert other objects at the same time.
With the introduction of Formatted string literals (since Python 3.6, 2016-12-23) this can be written as
import datetime
f"{datetime.datetime.now():%Y-%m-%d}"
>>> '2017-06-15'
Localization
Dates can automatically adapt to the local language and culture if you use them the right way, but it's a bit complicated. Maybe for another question on SO(Stack Overflow) ;-)
import datetime
print datetime.datetime.now().strftime("%Y-%m-%d %H:%M")
Edit:
After Cees' suggestion, I have started using time as well:
import time
print time.strftime("%Y-%m-%d %H:%M")
The date, datetime, and time objects all support a strftime(format) method,
to create a string representing the time under the control of an explicit format
string.
Here is a list of the format codes with their directive and meaning.
%a Locale’s abbreviated weekday name.
%A Locale’s full weekday name.
%b Locale’s abbreviated month name.
%B Locale’s full month name.
%c Locale’s appropriate date and time representation.
%d Day of the month as a decimal number [01,31].
%f Microsecond as a decimal number [0,999999], zero-padded on the left
%H Hour (24-hour clock) as a decimal number [00,23].
%I Hour (12-hour clock) as a decimal number [01,12].
%j Day of the year as a decimal number [001,366].
%m Month as a decimal number [01,12].
%M Minute as a decimal number [00,59].
%p Locale’s equivalent of either AM or PM.
%S Second as a decimal number [00,61].
%U Week number of the year (Sunday as the first day of the week)
%w Weekday as a decimal number [0(Sunday),6].
%W Week number of the year (Monday as the first day of the week)
%x Locale’s appropriate date representation.
%X Locale’s appropriate time representation.
%y Year without century as a decimal number [00,99].
%Y Year with century as a decimal number.
%z UTC offset in the form +HHMM or -HHMM.
%Z Time zone name (empty string if the object is naive).
%% A literal '%' character.
This is what we can do with the datetime and time modules in Python
import time
import datetime
print "Time in seconds since the epoch: %s" %time.time()
print "Current date and time: ", datetime.datetime.now()
print "Or like this: ", datetime.datetime.now().strftime("%y-%m-%d-%H-%M")
print "Current year: ", datetime.date.today().strftime("%Y")
print "Month of year: ", datetime.date.today().strftime("%B")
print "Week number of the year: ", datetime.date.today().strftime("%W")
print "Weekday of the week: ", datetime.date.today().strftime("%w")
print "Day of year: ", datetime.date.today().strftime("%j")
print "Day of the month : ", datetime.date.today().strftime("%d")
print "Day of week: ", datetime.date.today().strftime("%A")
That will print out something like this:
Time in seconds since the epoch: 1349271346.46
Current date and time: 2012-10-03 15:35:46.461491
Or like this: 12-10-03-15-35
Current year: 2012
Month of year: October
Week number of the year: 40
Weekday of the week: 3
Day of year: 277
Day of the month : 03
Day of week: Wednesday
Use date.strftime. The formatting arguments are described in the documentation.
This one is what you wanted:
some_date.strftime('%Y-%m-%d')
This one takes Locale into account. (do this)
some_date.strftime('%c')
This is shorter:
>>> import time
>>> time.strftime("%Y-%m-%d %H:%M")
'2013-11-19 09:38'
# convert date time to regular format.
d_date = datetime.datetime.now()
reg_format_date = d_date.strftime("%Y-%m-%d %I:%M:%S %p")
print(reg_format_date)
# some other date formats.
reg_format_date = d_date.strftime("%d %B %Y %I:%M:%S %p")
print(reg_format_date)
reg_format_date = d_date.strftime("%Y-%m-%d %H:%M:%S")
print(reg_format_date)
OUTPUT
2016-10-06 01:21:34 PM
06 October 2016 01:21:34 PM
2016-10-06 13:21:34
Or even
from datetime import datetime, date
"{:%d.%m.%Y}".format(datetime.now())
Out: '25.12.2013
or
"{} - {:%d.%m.%Y}".format("Today", datetime.now())
Out: 'Today - 25.12.2013'
"{:%A}".format(date.today())
Out: 'Wednesday'
'{}__{:%Y.%m.%d__%H-%M}.log'.format(__name__, datetime.now())
Out: '__main____2014.06.09__16-56.log'
Simple answer -
datetime.date.today().isoformat()
With type-specific datetime string formatting (see nk9's answer using str.format().) in a Formatted string literal (since Python 3.6, 2016-12-23):
>>> import datetime
>>> f"{datetime.datetime.now():%Y-%m-%d}"
'2017-06-15'
The date/time format directives are not documented as part of the Format String Syntax but rather in date, datetime, and time's strftime() documentation. The are based on the 1989 C Standard, but include some ISO 8601 directives since Python 3.6.
I hate the idea of importing too many modules for convenience. I would rather work with available module which in this case is datetime rather than calling a new module time.
>>> a = datetime.datetime(2015, 04, 01, 11, 23, 22)
>>> a.strftime('%Y-%m-%d %H:%M')
'2015-04-01 11:23'
You need to convert the datetime object to a str.
The following code worked for me:
import datetime
collection = []
dateTimeString = str(datetime.date.today())
collection.append(dateTimeString)
print(collection)
Let me know if you need any more help.
In Python you can format a datetime using the strftime() method from the date, time and datetime classes in the datetime module.
In your specific case, you are using the date class from datetime. You can use the following snippet to format the today variable into a string with the format yyyy-MM-dd:
import datetime
today = datetime.date.today()
print("formatted datetime: %s" % today.strftime("%Y-%m-%d"))
In the following a more complete example:
import datetime
today = datetime.date.today()
# datetime in d/m/Y H:M:S format
date_time = today.strftime("%d/%m/%Y, %H:%M:%S")
print("datetime: %s" % date_time)
# datetime in Y-m-d H:M:S format
date_time = today.strftime("%Y-%m-%d, %H:%M:%S")
print("datetime: %s" % date_time)
# format date
date = today.strftime("%d/%m/%Y")
print("date: %s" % time)
# format time
time = today.strftime("%H:%M:%S")
print("time: %s" % time)
# day
day = today.strftime("%d")
print("day: %s" % day)
# month
month = today.strftime("%m")
print("month: %s" % month)
# year
year = today.strftime("%Y")
print("year: %s" % year)
More directives:
Sources:
Format DateTime in Python
strftime
You can do:
mylist.append(str(today))
Considering the fact you asked for something simple to do what you wanted, you could just:
import datetime
str(datetime.date.today())
For those wanting locale-based date and not including time, use:
>>> some_date.strftime('%x')
07/11/2019
Since the print today returns what you want this means that the today object's __str__ function returns the string you are looking for.
So you can do mylist.append(today.__str__()) as well.
from datetime import date
def today_in_str_format():
return str(date.today())
print (today_in_str_format())
This will print 2018-06-23 if that's what you want :)
You may want to append it as a string?
import datetime
mylist = []
today = str(datetime.date.today())
mylist.append(today)
print(mylist)
For pandas.Timestamps, strftime() can be used e.g.:
utc_now = datetime.now()
For isoformat:
utc_now.isoformat()
For any format e.g.:
utc_now.strftime("%m/%d/%Y, %H:%M:%S")
You can use easy_date to make it easy:
import date_converter
my_date = date_converter.date_to_string(today, '%Y-%m-%d')
A quick disclaimer for my answer - I've only been learning Python for about 2 weeks, so I am by no means an expert; therefore, my explanation may not be the best and I may use incorrect terminology. Anyway, here it goes.
I noticed in your code that when you declared your variable today = datetime.date.today() you chose to name your variable with the name of a built-in function.
When your next line of code mylist.append(today) appended your list, it appended the entire string datetime.date.today(), which you had previously set as the value of your today variable, rather than just appending today().
A simple solution, albeit maybe not one most coders would use when working with the datetime module, is to change the name of your variable.
Here's what I tried:
import datetime
mylist = []
present = datetime.date.today()
mylist.append(present)
print present
and it prints yyyy-mm-dd.
Here is how to display the date as (year/month/day) :
from datetime import datetime
now = datetime.now()
print '%s/%s/%s' % (now.year, now.month, now.day)
import datetime
import time
months = ["Unknown","January","Febuary","Marchh","April","May","June","July","August","September","October","November","December"]
datetimeWrite = (time.strftime("%d-%m-%Y "))
date = time.strftime("%d")
month= time.strftime("%m")
choices = {'01': 'Jan', '02':'Feb','03':'Mar','04':'Apr','05':'May','06': 'Jun','07':'Jul','08':'Aug','09':'Sep','10':'Oct','11':'Nov','12':'Dec'}
result = choices.get(month, 'default')
year = time.strftime("%Y")
Date = date+"-"+result+"-"+year
print Date
In this way you can get Date formatted like this example: 22-Jun-2017
I don't fully understand but, can use pandas for getting times in right format:
>>> import pandas as pd
>>> pd.to_datetime('now')
Timestamp('2018-10-07 06:03:30')
>>> print(pd.to_datetime('now'))
2018-10-07 06:03:47
>>> pd.to_datetime('now').date()
datetime.date(2018, 10, 7)
>>> print(pd.to_datetime('now').date())
2018-10-07
>>>
And:
>>> l=[]
>>> l.append(pd.to_datetime('now').date())
>>> l
[datetime.date(2018, 10, 7)]
>>> map(str,l)
<map object at 0x0000005F67CCDF98>
>>> list(map(str,l))
['2018-10-07']
But it's storing strings but easy to convert:
>>> l=list(map(str,l))
>>> list(map(pd.to_datetime,l))
[Timestamp('2018-10-07 00:00:00')]
maybe the shortest solution, which exactly matches your situation, would be:
mylist.append(str(AnyDate)[:10])
or even shorter, e.g.:
f'{AnyDate}'[:10]
PS: it doesn't need to be today.

What is the difference between strptime and strftime?

I am working on a project , where I have dictionary with column date called "starttime" .. I need to extract month , hour and year , day of the week.
I am stuck for now I have the below code .
{if city =='NYC':
datn = datum[('starttime')]
datn = dt.strptime(datn,"%m/%d/%y %H:%M:%S")
hour = dt.strftime(datn,"%H")
year = dt.strftime(datn,"%y")
elif city == 'Chicago':
datc = datum[('starttime')]
datc = dt.strptime(datc,"%m/%d/%y %H:%M")
month = dt.strftime(datc,"%m")
hour = dt.strftime(datc,"%H")
year = dt.strftime(datc,"%y")
else:
datw = datum[('start date')]
datw = dt.strftime (datw,"%m")
hour = dt.strftime(datw,"%H")
year = dt.strftime(datw,"%y")
return (month, hour, day_of_week)
}
my import statements are on the top of my code , as below:
from datetime import datetime
strptime translates to
"parse (convert) string to datetime object."
strftime translates to
"create formatted string for given time/date/datetime object according to specified format."
Why do you need strftime?
This is what a datetime object looks like: (2015, 7, 19, 22, 7, 44,
377000)
To someone who isn't quite familiar with this format, with the
exception of the year, what's written up there is not immediately
intuitive. So you probably would be better off with something like
Sun, 19 July, 2015. That's what strftime is used for. You simply need
to learn the proper formatting strings.
One Good Link over SO read about this !
strptime converts the string to a datetime object.
strftime creates a formatted string for given time/date/datetime object according to specified format by the user
you would use strftime to convert a datetime object like this: datetime (2018, 10, 20, 10, 9, 22, 120401) to a more readable format like "20-10-2018" or 20th of October 2018.

Convert Alphanumerical Date String to Numerical Date String

I know there are similar questions out there, but most simply convert the alphanumerical date string to a datetime object with the strptime method, which is not what I'm going for. All I'm trying to do is convert string like so.
Format of Possible Input
December 7, 2015
October 24, 2018
Desired Output
2015-12-07
2018-10-24
How I've Gone About It
""" funding_expiration[0] is equal to strings like 'December 7, 2015' """
funding_expiration = funding_expiration[0].text.split()
""" still need to convert Month (%B) to ## (%m) """
funding_expiration[1] = funding_expiration[1].replace(',', '')
# add padding to single digit days
if len(funding_expiration[1]) is 1:
funding_expiration[1] = funding_expiration[1].zfill(1)
# format numerical date string
funding_expiration = funding_expiration[2] + '-' + funding_expiration[0] + '-' + funding_expiration[1]
I'm still trying to figure out an efficient way to convert the full name of months into their corresponding numerals. I'm new to Python, so I'm wondering whether or not there's a more efficient way to accomplish this?
datetime.strptime can work in your case too. You can use the %B directive to parse full month names.
import datetime
s = 'December 7, 2015'
date_string = str(datetime.datetime.strptime(s, '%B %d, %Y').date())
>>> date_string
'2015-12-07'
Here's a solution using 3rd party dateutil:
from dateutil import parser
L = ['December 7, 2015', 'October 24, 2018']
res_str = [parser.parse(x).strftime('%Y-%m-%d') for x in L]
['2015-12-07', '2018-10-24']

Python Error Control Date

I want my program to take user input in form of a date and the control if it is valid. But with the code I have know the program say that it is wrong regardless of what format i give. I don't see the problem with the code:
import datetime
def visit_date():
while True:
date_visit = input("Enter the date you want to visit the Zoo in YYYY-MM-DD format: ")
try:
return datetime.datetime.strptime(date_visit, "%d/%m/%y")
except ValueError:
print("Not a valid format\n")
You're asking the user for a date in the format YYYY-MM-DD but then trying to parse it according to this format %d/%m/%y.
You instead should parse the string in the same way that you requested it, %Y-%m-%d
You're looking for format %d/%m/%y and asking for %Y-%m-%d
> date_visit = '2016-11-23'
> datetime.datetime.strptime(date_visit, "%Y-%m-%d")
datetime.datetime(2016, 11, 23, 0, 0)
Some notes:
%Y : Year in four digits %y : year in two digits.
%d/%m/%y translates to "day of month in one or two digits, /, month of year in one or two digits, / year in two digits".
%Y-%m-%d translates to "four-digit-year, -, month-of-year, - day-of-month"
Your prompt asks you to enter in YYYY-MM-DD, but your strptime is attempting to use the format %d/%m/%y. You need to have the formats to match for strptime to work
import datetime
d = '2016-11-21'
d = datetime.datetime.strptime(d, "%Y-%m-%d")
d = datetime.datetime(2016, 11, 21, 0, 0)
>>> 2016-11-21 00:00:00
d = '11/21/2016'
d = datetime.datetime.strptime(d, "%d/%M/%Y")
d = datetime.datetime(2016, 11, 21, 0, 0)
>>> 2016-11-21 00:00:00
I personally like to use the python-dateutil module for parsing date strings that allows for different formats
pip install python-dateutil
from dateutil.parser import parse
d1 = 'Tuesday, October 21 2003, 12:14 CDT'
d2 = 'Dec. 23rd of 2012 at 12:34pm'
d3 = 'March 4th, 2016'
d4 = '2015-12-09'
print(parse(d1))
>>> 2003-10-21 12:14:00
print(parse(d2))
>>> 2012-12-23 12:34:00
print(parse(d3))
>>> 2016-03-04 00:00:00
print(parse(d4))
>>> 2015-12-09 00:00:00

python convert single digit day and month to two digits

I'm trying to read and convert the following datetime "6/5/2014 00:09:32" to "Mon Mar 09 07:07:18 +0000 2015".
d = datetime.datetime.strptime('6/5/2014 00:09:32', '%d/%m/%y %H:%M:%S')
Gives me a traceback with:
ValueError: time data '6/5/2014 00:00:07' does not match format '%d/%m/%y %H:%M:%S'
I tried using %e for single digit days as suggested here: http://www.gnu.org/software/libc/manual/html_node/Formatting-Calendar-Time.html#index-strftime-2660, but datetime just says it doesn't recognize 'e'.
Unless I am missing something it looks like single digit days and months are my problem, but from the documentation (https://docs.python.org/2/library/time.html#time.strftime) I don't see a way to get python's datetime to read single digit days or months. But this seems silly. What am I missing?
Thank you for your help.
You need 'Y' not 'y':
In [412]:
d = dt.datetime.strptime('6/5/2014 00:09:32', '%d/%m/%Y %H:%M:%S')
d
Out[412]:
datetime.datetime(2014, 5, 6, 0, 9, 32)
From the docs 'Y' corresponds to:
Year with century as a decimal number.
You tried 'y' which is:
Year without century as a decimal number [00,99].

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