对列表中的项目进行分组并计算总和。

Question

我有一个包含每周数字的列表，需要获得按月分组的总数。

下面的代码可以完成这项工作，但应该有更多的pythonic方法来使用标准库来完成这项工作，下面代码的缺点是列表需要排序。

#Test data (not sorted)
sum_weekly=[('2020/01/05', 59), ('2020/01/19', 88), ('2020/01/26', 95), ('2020/02/02', 89),
 ('2020/02/09', 113), ('2020/02/16', 90), ('2020/02/23', 68), ('2020/03/01', 74), ('2020/03/08', 85),
  ('2020/04/19', 6), ('2020/04/26', 5), ('2020/05/03', 14),
 ('2020/05/10', 5), ('2020/05/17', 20), ('2020/05/24', 28),('2020/03/15', 56), ('2020/03/29', 5), ('2020/04/12', 2),]

month = sum_weekly[0][0].split('/')[1]
count=0
out=[]
for item in sum_weekly:
    m_sel = item[0].split('/')[1]
    if m_sel!=month:
        out.append((month, count))
        count=item[1]
    else:
        count+=item[1]
    month = m_sel
out.append((month, count))

# monthly sums output as ('01', 242), ('02', 360), ('03', 220), ('04', 13), ('05', 67)
print (out)

Answer 1

你可以使用 defaultdict 来存储结果，而不是一个列表。字典的键是月份，你可以简单地添加相同月份（键）的值。

可能的实现。

# Test Data
from collections import defaultdict

sum_weekly = [('2020/01/05', 59), ('2020/01/19', 88), ('2020/01/26', 95), ('2020/02/02', 89),
              ('2020/02/09', 113), ('2020/02/16', 90), ('2020/02/23', 68), ('2020/03/01', 74), ('2020/03/08', 85),
              ('2020/03/15', 56), ('2020/03/29', 5), ('2020/04/12', 2), ('2020/04/19', 6), ('2020/04/26', 5),
              ('2020/05/03', 14),
              ('2020/05/10', 5), ('2020/05/17', 20), ('2020/05/24', 28)]


results = defaultdict(int)
for date, count in sum_weekly: # used unpacking to make it clearer
    month = date.split('/')[1]
    # because we use a defaultdict if the key does not exist it
    # the entry for the key will be created and initialize at zero
    results[month] += count

print(results)

Answer 2

你可以使用 itertools.groupby (它是标准库的一部分)--它做的事情和你在引擎盖下做的差不多(将关键函数给出相同输出的元素序列分组)。它可以看起来像下面这样。

import itertools

def select_month(item):
    return item[0].split('/')[1]

def get_value(item):
    return item[1]

result = [(month, sum(map(get_value, group))) 
            for month, group in itertools.groupby(sorted(sum_weekly), select_month)]
print(result)

Answer 3

简洁，但也许不是那么pythonic:

import calendar, functools, collections

{calendar.month_name[i]: val for i, val in functools.reduce(lambda a, b: a + b, [collections.Counter({datetime.datetime.strptime(time, '%Y/%m/%d').month: val}) for time, val in sum_weekly]).items()}

Answer 4

一个使用pyspark的方法

from pyspark import SparkContext

sc = SparkContext()

l = sc.parallelize(sum_weekly)

r = l.map(lambda x: (x[0].split("/")[1], x[1])).reduceByKey(lambda p, q: (p + q)).collect()

print(r) #[('04', 13), ('02', 360), ('01', 242), ('03', 220), ('05', 67)]

Answer 5

你可以用Pandas的数据框架来实现。首先，你把月份隔离开来，然后使用groupby.sum()。

import pandas as pd

sum_weekly=[('2020/01/05', 59), ('2020/01/19', 88), ('2020/01/26', 95), ('2020/02/02', 89), ('2020/02/09', 113), ('2020/02/16', 90), ('2020/02/23', 68), ('2020/03/01', 74), ('2020/03/08', 85), ('2020/04/19', 6), ('2020/04/26', 5), ('2020/05/03', 14), ('2020/05/10', 5), ('2020/05/17', 20), ('2020/05/24', 28),('2020/03/15', 56), ('2020/03/29', 5), ('2020/04/12', 2)]

df= pd.DataFrame(sum_weekly)
df.columns=['Date','Sum']
df['Month'] = df['Date'].str.split('/').str[1]
print(df.groupby('Month').sum())