Building a list of months by iterating between two dates in a list (Python)

Question

I have an ordered (i.e. sorted) list that contains dates sorted (as datetime objects) in ascending order.

I want to write a function that iterates through this list and generates another list of the first available dates for each month.

For example, suppose my sorted list contains the following data:

A = [
'2001/01/01',
'2001/01/03',
'2001/01/05',
'2001/02/04',
'2001/02/05',
'2001/03/01',
'2001/03/02',
'2001/04/10',
'2001/04/11',
'2001/04/15',
'2001/05/07',
'2001/05/12',
'2001/07/01',
'2001/07/10',
'2002/03/01',
'2002/04/01',
]

The returned list would be

B = [
'2001/01/01',
'2001/02/04',
'2001/03/01',
'2001/04/10',
'2001/05/07',
'2001/07/01',
'2002/03/01',
'2002/04/01',
]

The logic I propose would be something like this:

def extract_month_first_dates(input_list, start_date, end_date):
    #note: start_date and end_date DEFINITELY exist in the passed in list
    prev_dates, output = [],[]  # <- is this even legal?
    for (curr_date in input_list):
        if ((curr_date < start_date) or (curr_date > end_date)):
            continue

        curr_month = curr_date.date.month
        curr_year = curr_date.date.year
        date_key = "{0}-{1}".format(curr_year, curr_month)
        if (date_key in prev_dates):
            continue
        else:
            output.append(curr_date)
            prev_dates.append(date_key)

    return output

Any comments, suggestions? - can this be improved to be more 'Pythonic' ?

Answer 1

>>> import itertools
>>> [min(j) for i, j in itertools.groupby(A, key=lambda x: x[:7])]
['2001/01/01', '2001/02/04', '2001/03/01', '2001/04/10', '2001/05/07', '2001/07/01', '2002/03/01', '2002/04/01']

Answer 2

Searching lists is a O(n) operation. I think you can simply check whether the key is new:

def extract_month_first_dates(input_list):
    output = []
    last_key = None
    for curr_date in input_list:
        date_key = curr_date.date.month, curr_date.date.year  # no string key required
        if date_key != last_key:
            output.append(curr_date)
            last_key = date_key
    return output

Answer 3

Here is a simple solution in classic python i.e. no itertools ;) and self explanatory

visited = {}
B = []
for a in A:
    month = a[:7]
    if month not in visited:
        B.append(a)
    visited[month] = 1

print B

Ouput:

['2001/01/01', '2001/02/04', '2001/03/01', '2001/04/10', '2001/05/07', '2001/07/01', '2002/03/01', '2002/04/01']