Python时间序列：将字典中的每日数据合并为每周数据

Question

I have a dictionary as below.

my_dict.keys() = 
dict_keys([20160101, 20160102, 20160103, 20160104, 20160105, 20160106,
       20160107, 20160108, 20160109, 20160110, 20160111, 20160112,
       20160113, 20160114, 20160115, 20160116, 20160117, 20160118,
       20160119, 20160120, 20160121, 20160122, 20160123, 20160124,
       ......    
       20171203, 20171204, 20171213, 20171215, 20171216, 20171217,
       20171218, 20171219, 20171220, 20171221, 20171222, 20171223,
       20171224, 20171225, 20171226, 20171227, 20171228, 20171229,
       20171230, 20171231])

my_dict[20160101] = 
array([[ 0.,  0.,  1.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  2.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.],
       [ 1.,  0.,  0.,  0.,  0.,  2.],
       [ 0.,  0.,  4.,  0.,  0.,  0.]])

So, as you already notice that my keys are indicating dates and each date i have array with 6 by 6 floats. In every keys in my_dict, all the indexes are same.

**Important thing to notice is that my_dict does not have every day. For example, after 20171204, its 20171213 and 20171215. So dates can be skipped.

Now my task is to get daily data (not every single day) to weekly data and add all the values within a week. In other words, starting from the first week of 2016 to last week of 2017, add every values within a week and provide weekly data. Also, since first week of 2016 starts with 20160103 (Sun), I can disregard 20160101 and 20160102 data in my_dict as well as end week of 2017. Can you guys help me with this problem? Thanks in advance!

-------edit--------- It seems like my question is not clear enough. So I will provide a quick example. Since I want to follow the standard of pandas datatime week, so each week starts with Sunday. So first week of 2016 will be 20160103,20160104,20160105,20160106,20160107,20160108,201601‌​09.

So my new dictionary, weekly_dict[201601] <- where 201601 indicates the first week of 2016, all the values in key 20160103,20160104,20160105,20160106,20160107,20160108,201601‌​09 will be added and input as values.

weekly_dict = {}
weekly_dict[201601] = my_dict[20160103] + my_dict[20160104] + my_dict[20160105] + my_dict[20160106] + my_dict[20160107] + my_dict[20160108] + my_dict[20160109]

And continues. Hope this makes sense. Thanks!

Answer 1

This is probably a job for pandas:

import pandas as pd

# First, get a list of keys
date_ints = list(my_dict)
# Turn them into a pandas Series object
date_int_series = pd.Series(date_ints)
# Cast them to a string, then format them into a full datetime-type with the proper
# format specification
datetime_series = pd.to_datetime(date_int_series.astype('str'), format='%Y%m%d')
# Create a dictionary mapping each date integer -> week of the year
date_int_to_week = dict(zip(date_int_series, datetime_series.dt.week))

This dictionary has each key of my_dict as a key, with its corresponding week of the year as its value.

Edit:

If what you're looking for is to sum each entry of your original dictionary based on week, you can do something like this:

week_to_date_list = {}
for date_int, week in date_int_to_week.items():
    if week not in week_to_date_list:
        week_to_date_list[week] = []
    week_to_date_list[week].append(date_int)

my_dict_weekly = {}
for week in week_to_date_list:
    arrays_in_week = [my_dict[day_int] for day_int in week_to_date_list[week]]
    my_dict_weekly[week] = reduce(sum, arrays_in_week)

my_dict_weekly should now be a dictionary that has weeks of the year as its key, then the sum of all of the arrays corresponding to that week. If you're using python 3, you'll need to import reduce from functools .

Answer 2

If i did understand well your question, i think that you can solve it using datetime and timedelta from datetime module like this example:

from datetime import datetime, timedelta

def get_days_of_week(year, week=1):
    # number of the days
    days = {'Monday': 1, 'Tuesday': 2, 'Wednesday': 3, 
            'Thursday': 4, 'Friday': 5, 'Saturday': 6, 'Sunday': 7}
    # construct the datetime object with the year and the desired week
    a = datetime.strptime('{0}'.format(year), '%Y') + timedelta(days=7*(week-1))
    # Every week should start by Sunday .. So escaping days untill the first Sunday
    a += timedelta(days=7-days.get(a.strftime('%A'), 0))
    for k in range(0, 7):
        yield (a + timedelta(days=k)).strftime('%Y%m%d')

days = list(get_days_of_week(2016, week=1))
print('2016 / week = 1:', days)

days = list(get_days_of_week(2016, week=22))
print('2016 / week = 22:', days)

Output:

2016 / week = 1: 
 ['20160103',
 '20160104',
 '20160105',
 '20160106',
 '20160107',
 '20160108',
 '20160109']

2016 / week = 22: 
 ['20160529',
 '20160530',
 '20160531',
 '20160601',
 '20160602',
 '20160603',
 '20160604']

Edit:

According to your last edit, this code may fulfill your needs:

from datetime import datetime, timedelta

def get_days_of_week(data):
    # number of the days
    days = {'Monday': 1, 'Tuesday': 2, 'Wednesday': 3,
            'Thursday': 4, 'Friday': 5, 'Saturday': 6, 'Sunday': 7}
    date = datetime.strptime('{}'.format(data), '%Y%m%d')
    # get week number
    week = int(date.strftime('%U'))
    # get year
    year = date.strftime('%Y')
    # construct the datetime object with the year and the desired week
    a = datetime.strptime(year, '%Y') + timedelta(days=7*week)
    # Every week should start by Synday .. So escaping days untill the first Sunday
    a += timedelta(days=7-days.get(a.strftime('%A'), 0))

    return {int(str(data)[:-2]): [int((a + timedelta(days=k)).strftime('%Y%m%d')) for k in range(0, 7)]}

week_dict = {}
week_dict.update(get_days_of_week(20160101))
week_dict.update(get_days_of_week(20160623))
print(week_dict[201601])
print(week_dict[201606])

print(week_dict)

Output:

[20160103, 20160104, 20160105, 20160106, 20160107, 20160108, 20160109]
[20160626, 20160627, 20160628, 20160629, 20160630, 20160701, 20160702]
{ 201601: [ 20160103,
            20160104,
            20160105,
            20160106,
            20160107,
            20160108,
            20160109],
  201606: [ 20160626,
            20160627,
            20160628,
            20160629,
            20160630,
            20160701,
            20160702]}