вы можете использовать collection.defaultdict с statistics.mean :
from collections import defaultdict
from statistics import mean
l = [('2017-12-01', ['5', '6', '0', False]),
('2017-12-02', ['5', '7', '0', False]),
('2017-12-03', ['6', '7', '0.5', True]),
('2017-12-04', ['6', '7', '0.5', True]),
('2017-12-05', ['5', '6', '0.4', True]),
('2018-01-01', ['5', '6', '0', False]),
('2018-01-02', ['5', '6', '0', False])]
my_dict = defaultdict(lambda : [[], []])
for d, v in l:
y = int(d[:4])
my_dict[y][0].append(float(v[0]))
my_dict[y][1].append(float(v[1]))
result = {k: [mean(e) for e in v] for k, v in my_dict.items()}
result
вывод:
{2017: [5.4, 6.6], 2018: [5.0, 6.0]}
также , вы можете сначала использовать pandas
1) , вам необходимо преобразовать данные в pandas.DataFrame:
import pandas as pd
df = pd.DataFrame([[f, *map(float, s[:2])] for f, s in l], columns=['date', 'col0', 'col1'])
df['date']= pd.to_datetime(df['date'])
df
:
2 ) теперь вы можете манипулировать своим pd.DataFrame, используя pandas .Dataframe.groupby чтобы получить желаемый результат:
df.groupby(df.date.dt.year).mean().transpose().to_dict('l')
выход:
{2017: [5.4, 6.6], 2018: [5.0, 6.0]}
, поскольку вам требуется более простой подход, который вы можете использовать:
# group col0 and col1 values base on the year
year_cols = {}
for date, cols in l:
# the year is in the first 4 characters so using a slice will get the year
# then convert to integer
year = int(date[:4])
col0 = cols[0]
col1 = cols[1]
# store the values from column 0 and column 1 base on the year
if year in year_cols: # check if new element/year
# if not a new elemnt
year_cols[year]['col0'].append(float(col0)) # convert to float to be able to compute the average
year_cols[year]['col1'].append(float(col1)) # convert to float to be able to compute the average
else: # in case of a new element/year
col01_data = {'col0': [float(col0)], 'col1': [float(col1)]}
year_cols[year] = col01_data
# get the average for each year on each column
result = {}
for year, col0_col1 in year_cols.items():
col0 = col0_col1['col0']
col1 = col0_col1['col1']
# compute the average for each column
# average formula: sum of all elements divided by the number of elemetns
result[year] = [sum(col0) / len(col0), sum(col1) / len(col1)]
result
вывод:
{2017: [5.4, 6.6], 2018: [5.0, 6.0]}