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Python check two dates inside date range

I need to check that two dates, not in any date range on the list.
I want to find out can user check-in in dates (check_range_true - can, check_range_false - can't) or this dates already booked (in date_ranges)
I have range looks like:
date_ranges = [
['2020-1-12', '2020-1-13'],
['2020-1-14', '2020-1-15'],
['2020-1-15', '2020-1-16'],
['2020-1-16', '2020-1-18'],
['2020-1-18', '2020-1-19'],
['2020-1-21', '2020-1-23'],
['2020-1-23', '2020-1-27'],
['2020-1-30', '2020-2-1'],
['2020-2-5', '2020-2-7'],
['2020-2-7', '2020-2-9'],
['2020-2-9', '2020-2-11'],
['2020-2-14', '2020-2-18'],
['2020-2-20', '2020-2-26'],
['2020-3-26', '2020-3-30'],
['2020-5-29', '2020-5-30'],
['2020-10-10', '2021-1-15']
]
And two dates (for example)
check_range_true = ['2020-02-02', '2020-02-04']
check_range_false = ['2020-02-02', '2020-02-05']
I know how check one date in range but not understand how to solve it with two dates.
What to the best way to check these dates in a range and got results, True for the first variable (because of 2020-02-02, 2020-02-04 not "touch" range) and False for the second variable (because of 2020-02-05 is in range of ['2020-2-5', '2020-2-7'])?

What you what to do is to check the dates with (start < first_date < end) and (start < end_date < end) logic
date_ranges = [
['2020-1-12', '2020-1-13'],
['2020-1-14', '2020-1-15'],
['2020-1-15', '2020-1-16'],
['2020-1-16', '2020-1-18'],
['2020-1-18', '2020-1-19'],
['2020-1-21', '2020-1-23'],
['2020-1-23', '2020-1-27'],
['2020-1-30', '2020-2-1'],
['2020-2-5', '2020-2-7'],
['2020-2-7', '2020-2-9'],
['2020-2-9', '2020-2-11'],
['2020-2-14', '2020-2-18'],
['2020-2-20', '2020-2-26'],
['2020-3-26', '2020-3-30'],
['2020-5-29', '2020-5-30'],
['2020-10-10', '2021-1-15']
]
#convert to a flat list
date_ranges = [k for i in date_ranges for k in i]
#truncate the start and the end value
date_ranges = date_ranges[1:-1]
#convert values to datetime
import datetime
date_ranges = [datetime.datetime.strptime(i, '%Y-%m-%d') for i in date_ranges]
#create available time slots
date_ranges = [[date_ranges[i],date_ranges[i+1]] for i in range(0,len(date_ranges),2)]
#convert the check date to date time
check_range = ['2020-02-02', '2020-02-04']
check_range = [datetime.datetime.strptime(i, '%Y-%m-%d') for i in check_range]
# apply the logic of start < date < end twice
any([(i[0] < check_range[0] < i[1]) and (i[0] < check_range[1] < i[1]) for i in date_ranges])
True
check_range = ['2020-02-02', '2020-02-05']
check_range = [datetime.datetime.strptime(i, '%Y-%m-%d') for i in check_range]
any([(i[0] < check_range[0] < i[1]) and (i[0] < check_range[1] < i[1]) for i in date_ranges])
False

If I understand this correctly you want to check if given date range (e.g. check_range_true) overlaps (or not) with any other date range specified in the list. To achieve this, I would first transform string values to proper datetime objects for easier dates comparison. This could be achieved with list comprehension and strptime:
from datetime import datetime
booked_date_ranges = [
[datetime.strptime(start_date, '%Y-%m-%d'), datetime.strptime(end_date, '%Y-%m-%d')]
for start_date, end_date in date_ranges
]
Then I would create a function, which would check if provided date range (consisting of start date, and end date) overlaps with any date range from the previously specified list. You need to check if start date is between date range or end date is between date range. It would be something along these lines:
def dates_overlap(date_range_to_check, booked_date_ranges):
dates = [datetime.strptime(date, '%Y-%m-%d') for date in date_range_to_check]
return any(
(start_date <= dates[0] and dates[0] <= end_date) or (start_date <= dates[1] and dates[1] <= end_date)
for start_date, end_date in booked_date_ranges
)
Then if you want to check if given date range DOES NOT overlap, you can just use the dates_overlap function and negate the result:
>>> not dates_overlap(check_range_false, booked_date_ranges)
False
>>> not dates_overlap(check_range_true, booked_date_ranges)
True
I hope this answers your question. Of course this is just a draft and there's definitely some room for improvement, but should be a working solution to given problem.

How to increment date (month and year) and set the day to default based on specific conditions?

My Date should always fall on 8th or 22nd that comes off the input date.
For Example:
If the input date is 20190415 then the output date should be 20190422 as that's the nearest date and if input date is 20190424 then the output date should be 20190508.
Example1:
input_date = 20190415
Expected output_date = 20190422
Example2:
input_date = 20190424
Expected output_date = 20190508
Example3:
input_date = 20190506
Expected output_date = 20190508
Example4:
input_date = 20191223
Expected output_date = 20200108
How do we achieve this using Python?

You can check if the day is greater than 22, and if so you set it to the 8th of the next month. If it's between 8 and 22 you set it to 22 of the same month and if it's below the 8th you set it to the 8th of the month. There's probably more elegant ways to do it using date math, but this will work for your scenario.
Use the datetime module to find out what the "next month" is. One way to do it is to add a timedelta of 1 month to the first of the current month, and then change the date on that datetime object to the 8th. Here's a quick example of how that might look like:
from datetime import date, timedelta
input_date = date(2019, 12, 23)
if input_date.day > 22:
output_date = date(input_date.year, input_date.month) + timedelta(days=31)
output_date = output_date.replace(day = 8)
You can read a lot more about the details of how the datetime module works on the official documentation. It's kind of a long read, but I actually have that page bookmarked because I always have to go back and reference how to actually use the module :)

Considering the input as string, next date can be calculated using timedelta, check out the below code:
if 8<datetime.strptime(input_date, "%Y%m%d").day < 22:
delta = 22 - datetime.strptime(input_date, "%Y%m%d").day
print((datetime.strptime(input_date, "%Y%m%d") +
timedelta(days=delta)).strftime("%Y%m%d"))
elif datetime.strptime(str(input_date), "%Y%m%d").day < 8:
delta = 8 - datetime.strptime(input_date, "%Y%m%d").day
print((datetime.strptime(input_date, "%Y%m%d") +
timedelta(days=delta)).strftime("%Y%m%d"))
else:
delta = (datetime.strptime(input_date, "%Y%m%d")+ relativedelta(months=+1)).day -8
print((datetime.strptime(input_date, "%Y%m%d") + relativedelta(months=+1) -
timedelta(days=delta)).strftime("%Y%m%d") )

How to count days belonging to a given month in the range of two Python datetimes?

I have two Python datetime and I want to count the days between those dates, counting ONLY the days belonging to the month I choose. The range might overlap multiple months/years.
Example:
If I have 2017-10-29 & 2017-11-04 and I chose to count the days in October, I get 3 (29, 30 & 31 Oct.).
I can't find a straightforward way to do this so I think I'm going to iterate over the days using datetime.timedelta(days=1), and increment a count each time the day belongs to the month I chose.
Do you know a more performant method?
I'm using Python 2.7.10 with the Django framework.

Iterating over the days would be the most straightforward way to do it. Otherwise, you would need to know how many days are in a given month and you would need different code for different scenarios:
The given month is the month of the first date
The given month is the month of the second date
The given month is between the first and the second date (if dates span more than two months)
If you want to support dates spanning more than one year then you would need the input to include month and year.
Your example fits scenario #1, which I guess you could do like this:
>>> from datetime import datetime, timedelta
>>>
>>> first_date = datetime(2017, 10, 29)
>>>
>>> first_day_of_next_month = first_date.replace(month=first_date.month + 1, day=1)
>>> last_day_of_this_month = first_day_of_next_month - timedelta(1)
>>> number_of_days_in_this_month = last_day_of_this_month.day
>>> number_of_days_in_this_month - first_date.day + 1
3
This is why I would suggest implementing it the way you originally intended and only turning to this if there's a performance concern.

You can get difference between two datetime objects by simply subtracting them.
So, we start by getting the difference between the two dates.
And then we generate all the dates between the two using
gen = (start_date + datetime.timedelta(days = e) for e in range(diff + 1))
And since we only want the dates between the specified ones, we apply a filter.
filter(lambda x : x==10 , gen)
Then we will sum them over.
And the final code is this:
diff = start_date - end_date
gen = (start_date + datetime.timedelta(days = e) for e in range(diff + 1))
filtered_dates = filter(
lambda x : x.month == 10 ,
gen
)
count = sum(1 for e in filtered_dates)
You can also use reduce but sum() is a lot more readable.

A potential method of achieving this is to first compare whether your start or end dates you are comparing have the same month that you want to choose.
For example:
start = datetime(2017, 10, 29)
end = datetime(2017, 11, 4)
We create a function to compare the dates like so:
def daysofmonth(start, end, monthsel):
if start.month == monthsel:
days = (datetime(start.year, monthsel+1, 1) - start).days
elif end.month == monthsel:
days = (end - datetime(end.year, monthsel, 1)).days
elif not (monthsel > start.month) & (end.month > monthsel):
return 0
else:
days = (datetime(start.year, monthsel+1, 1) - datetime(start.year, monthsel, 1)).days
return days
So, in our example setting monthsel gives:
>>> daysofmonth(start, end, 10)
>>> 3

Using pandas whit your dates:
import pandas as pd
from datetime import datetime
first_date = datetime(2017, 10, 29)
second_date = datetime(2017, 11, 4)
days_count = (second_date - first_date).days
month_date = first_date.strftime("%Y-%m")
values = pd.date_range(start=first_date,periods=days_count,freq='D').to_period('M').value_counts()
print(values)
print(values[month_date])
outputs
2017-10 3
2017-11 3
3

Checking for missing values in CSV

I have a CSV file that counts the data in a timestamp every 15 min.
I need tried to figure out how to see, if there is missing any of the 15 min in the file. But i cant get the code to work 100%.
hope you can help!
First i geathered the data from csv and set it in timestamp. The format is yyyy-mm-dd-hh:mm.
lst = list()
with open("CHFJPY15.csv", "r") as f:
f_r = f.read()
sline = f_r.split()
for line in sline:
parts = line.split(',')
date = parts[0]
time = parts[1]
closeingtime = parts[5]
timestamp = date + ' ' + time + ' ' + closeingtime
lst.append(timestamp)
print(lst, "liste")
(All credits to BillBell for the code below)
Here try creating a consistently formatted list of data items.
from datetime import timedelta
interval = timedelta(minutes=15)
from datetime import datetime
current_time = datetime(2015,12,9,19,30)
data = []
omits = [3,5,9,11,17]
for i in range(20):
current_time += interval
if i in omits:
continue
data.append(current_time.strftime('%y.%m.%d.%H:%M')+' 123.456')
Now I read through the dates subtracting each from it predecessor. I set the first 'predecessor', which I call previous to now because that's bound to differ from the other dates.
I split each datum from the list into two, ignoring the second piece. Using strptime I turn strings into dates. Dates can be subtracted and the differences compared.
previous = datetime.now().strftime('%y.%m.%d.%H:%M')
first = True
for d in data:
date_part, other = d.split(' ')
if datetime.strptime(date_part, '%y.%m.%d.%H:%M') - datetime.strptime(previous, '%y.%m.%d.%H:%M') != interval:
if not first:
'unacceptable gap prior to ', date_part
else:
first = False
previous = date_part
Hope you can see the problem.

Python: get all months in range?

I want to get all months between now and August 2010, as a list formatted like this:
['2010-08-01', '2010-09-01', .... , '2016-02-01']
Right now this is what I have:
months = []
for y in range(2010, 2016):
for m in range(1, 13):
if (y == 2010) and m < 8:
continue
if (y == 2016) and m > 2:
continue
month = '%s-%s-01' % (y, ('0%s' % (m)) if m < 10 else m)
months.append(month)
What would be a better way to do this?

dateutil.relativedelta is handy here.
I've left the formatting out as an exercise.
from dateutil.relativedelta import relativedelta
import datetime
result = []
today = datetime.date.today()
current = datetime.date(2010, 8, 1)
while current <= today:
result.append(current)
current += relativedelta(months=1)

I had a look at the dateutil documentation. Turns out it provides an even more convenient way than using dateutil.relativedelta: recurrence rules (examples)
For the task at hand, it's as easy as
from dateutil.rrule import *
from datetime import date
months = map(
date.isoformat,
rrule(MONTHLY, dtstart=date(2010, 8, 1), until=date.today())
)
The fine print
Note that we're cheating a little bit, here. The elements dateutil.rrule.rrule produces are of type datetime.datetime, even if we pass dtstart and until of type datetime.date, as we do above. I let map feed them to date's isoformat function, which just turns out to convert them to strings as if it were just dates without any time-of-day information.
Therefore, the seemingly equivalent list comprehension
[day.isoformat()
for day in rrule(MONTHLY, dtstart=date(2010, 8, 1), until=date.today())]
would return a list like
['2010-08-01T00:00:00',
'2010-09-01T00:00:00',
'2010-10-01T00:00:00',
'2010-11-01T00:00:00',
⋮
'2015-12-01T00:00:00',
'2016-01-01T00:00:00',
'2016-02-01T00:00:00']
Thus, if we want to use a list comprehension instead of map, we have to do something like
[dt.date().isoformat()
for dt in rrule(MONTHLY, dtstart=date(2010, 8, 1), until=date.today())]

use datetime and timedelta standard Python's modules - without installing any new libraries
from datetime import datetime, timedelta
now = datetime(datetime.now().year, datetime.now().month, 1)
ctr = datetime(2010, 8, 1)
list = [ctr.strftime('%Y-%m-%d')]
while ctr <= now:
ctr += timedelta(days=32)
list.append( datetime(ctr.year, ctr.month, 1).strftime('%Y-%m-%d') )
I'm adding 32 days to enter new month every time (longest months has 31 days)

It's seems like there's a very simple and clean way to do this by generating a list of dates and subsetting to take only the first day of each month, as shown in the example below.
import datetime
import pandas as pd
start_date = datetime.date(2010,8,1)
end_date = datetime.date(2016,2,1)
date_range = pd.date_range(start_date, end_date)
date_range = date_range[date_range.day==1]
print(date_range)

I got another way using datetime, timedelta and calender:
from calendar import monthrange
from datetime import datetime, timedelta
def monthdelta(d1, d2):
delta = 0
while True:
mdays = monthrange(d1.year, d1.month)[1]
d1 += timedelta(days=mdays)
if d1 <= d2:
delta += 1
else:
break
return delta
start_date = datetime(2016, 1, 1)
end_date = datetime(2016, 12, 1)
num_months = [i-12 if i>12 else i for i in range(start_date.month, monthdelta(start_date, end_date)+start_date.month+1)]
monthly_daterange = [datetime(start_date.year,i, start_date.day, start_date.hour) for i in num_months]

You could reduce the number of if statements to two lines instead of four lines because having a second if statement that does the same thing with the previous if statement is a bit redundant.
if (y == 2010 and m < 8) or (y == 2016 and m > 2):
continue

I don't know whether it's better, but an approach like the following might be considered more 'pythonic':
months = [
'{}-{:0>2}-01'.format(year, month)
for year in xrange(2010, 2016 + 1)
for month in xrange(1, 12 + 1)
if not (year <= 2010 and month < 8 or year >= 2016 and month > 2)
]
The main differences here are:
As we want the iteration(s) to produce a list, use a list comprehension instead of aggregating list elements in a for loop.
Instead of explicitly making a distinction between numbers below 10 and numbers 10 and above, use the capabilities of the format specification mini-language for the .format() method of str to specify
a field width (the 2 in the {:0>2} place holder)
right-alignment within the field (the > in the {:0>2} place holder)
zero-padding (the 0 in the {:0>2} place holder)
xrange instead of range returns a generator instead of a list, so that the iteration values can be produced as they're being consumed and don't have to be held in memory. (Doesn't matter for ranges this small, but it's a good idea to get used to this in Python 2.) Note: In Python 3, there is no xrange and the range function already returns a generator instead of a list.
Make the + 1 for the upper bounds explicit. This makes it easier for human readers of the code to recognize that we want to specify an inclusive bound to a method (range or xrange) that treats the upper bound as exclusive. Otherwise, they might wonder what's the deal with the number 13.

A different approach that doesn't require any additional libraries, nor nested or while loops. Simply convert your dates into an absolute number of months from some reference point (it can be any date really, but for simplicity we can use 1st January 0001). For example
a=datetime.date(2010,2,5)
abs_months = a.year * 12 + a.month
Once you have a number representing the month you are in you can simply use range to loop over the months, and then convert back:
Solution to the generalized problem:
import datetime
def range_of_months(start_date, end_date):
months = []
for i in range(start_date.year * 12 + start_date.month, end_date.year*12+end_date.month + 1)
months.append(datetime.date((i-13) // 12 + 1, (i-1) % 12 + 1, 1))
return months
Additional Notes/explanation:
Here // divides rounding down to the nearest whole number, and % 12 gives the remainder when divided by 12, e.g. 13 % 12 is 1.
(Note also that in the above date.year *12 + date.month does not give the number of months since the 1st of January 0001. For example if date = datetime.datetime(1,1,1), then date.year * 12 + date.month gives 13. If I wanted to do the actual number of months I would need to subtract 1 from the year and month, but that would just make the calculations more complicated. All that matters is that we have a consistent way to convert to and from some integer representation of what month we are in.)

fresh pythonic one-liner from me
from dateutil.relativedelta import relativedelta
import datetime
[(start_date + relativedelta(months=+m)).isoformat()
for m in range(0, relativedelta(start_date, end_date).months+1)]

In case you don't have any months duplicates and they are in correct order you can get what you want with this.
from datetime import date, timedelta
first = date.today()
last = first + timedelta(weeks=20)
date_format = "%Y-%m"
results = []
while last >= first:
results.append(last.strftime(date_format))
last -= timedelta(days=last.day)

Similar to #Mattaf, but simpler...
pandas.date_range() has an option frequency freq='m'...
Here I am adding a day (pd.Timedelta('1d')) in order to reach the beginning of each new month:
import pandas as pd
date_range = pd.date_range('2010-07-01','2016-02-01',freq='M')+pd.Timedelta('1d')
print(list(date_range))