My project involves reading CSV files concering city data and then computing various statistics. I am trying to create a function that will take city data from a CSV and the create a copy of the data except with a date filter applied. The user is prompted to enter a starting and ending date.
I am currently just testing and trying to get the data to print with the correct date filters applied. My biggest problem so far is that I can't seem to convert the strings entered by the user into a datetime format. Is there a way for the user to directly enter the data in datetime format?
def date_filtering(city_file):
date1 = input('Please enter a valid start date that you wish to explore?\n'
date2 = input('Please enter a valid end date.')
print(city_file.loc[date1:date2])
Thanks in advance!
I am working of the assumption you only need date, because you make no mention of time in the code. What you need to do is parse the input somehow to get the year, month, and day. I am sure it could be done using regex, but it is simpler to do it just with python builtins.
A normal way of representing date in the US is mm/dd/yyyy, so I will use that.
def strtodatetime(str):
month, day, year = [int(d) for d in str.split("/")]
return datetime.date(year, month, day)
date1 = strtodatetime(input('Please enter a valid start date that you wish to explore?\n'))
date2 = strtodatetime(input('Please enter a valid end date.'))
The strtodatetime splits the input string by the "/" character, then casts each element in the resultant list to integers. Then it returns a new datetime.date object with the corresponding dates.
If you are worried about the user inputting something of the wrong format, you can use a try block to make sure they don't. There are three possible sources of exceptions I can see, the user can input too many or too few /'s (like 10/2/1999/2), the user can input a date that doesn't exist (like 13/32/-1), or the user can input something that can't be cast to an int (like March/22/1999). All three raise ValueErrors, so that is the only exception we need to catch here. To catch that, we would change our code to
def strtodatetime(str):
month, day, year = [int(d) for d in str.split("/")]
return datetime.date(year, month, day)
date_filtering(city_file):
try:
date1 = strtodatetime(input('Please enter a valid start date that you wish to explore?\n'))
date2 = strtodatetime(input('Please enter a valid end date.'))
except ValueError:
print('Please enter a valid date in the format mm/dd/yyyy')
return date_filtering(city_file)
return city_file.loc[date1:date2]
print(data_filtering(city_file))
I want to check if the format of the date input by user matches the below:
Jan 5 2018 6:10 PM
Month: First letter should be caps, followed 2 more in small. (total 3 letters)
<Space>: single space, must exist
Date: For single digit it should not be 05, but 5
<Space>: single space, must exist
Hour: 0-12, for single digit it should not be 06, but 6
Minute: 00-59
AM/PM
I'm using the below regex and trying to match:
import re,sys
usr_date = str(input("Please enter the older date until which you want to scan ? \n[Date Format Example: Jan 5 2018 6:10 PM] : "))
valid_usr_date = re.search("^(\s+)*[A-Z]{1}[a-z]{2}\s{1}[1-31]{1}\s{1}[1-2]{1}[0-9]{1}[0-9]{1}[0-9]{1}\s{1}[0-12]{1}:[0-5]{1}[0-9]{1}\s{1}(A|P)M$",usr_date,re.M)
if not valid_usr_date:
print ("The date format is incorrect. Please follow the exact date format as shown in example. Exiting Program!")
sys.exit()
But, even for the correct format it gives a syntax wrong error. What am I doing wrong.
I would not use regex for that, as you have no way to actually validate the date itself (eg, a regex will happily accept Abc 99 9876 9:99 PM).
Instead, use strptime:
from datetime import datetime
string = 'Jan 5 2018 6:10 PM'
datetime.strptime(string, '%b %d %Y %I:%M %p')
If the string would be in the "wrong" format you'd get a ValueError.
The only apparent "problem" with this approach is that for some reason you require the day and hour not to be zero-padded and strptime doesn't seem to have such directives.
A table with all available directives is here.
You could use a function which parses the input string and tries to return a datetime object, if it can't it raises an ValueError:
from datetime import datetime
def valid_date(s):
try:
return datetime.strptime(s, '%Y-%m-%d %H:%M')
except ValueError:
msg = "Not a valid date: '{0}'.".format(s)
raise argparse.ArgumentTypeError(msg)
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.