Python - 对元组列表进行分组和求和 [英] Python - Group by and sum a list of tuples

问题描述

鉴于以下列表:

<预><代码>[('A', '', Decimal('4.0000000000'), 1330, datetime.datetime(2012, 6, 8, 0, 0)),('B', '', Decimal('31.0000000000'), 1330, datetime.datetime(2012, 6, 4, 0, 0)),('AA', 'C', Decimal('31.0000000000'), 1330, datetime.datetime(2012, 5, 31, 0, 0)),('B', '', Decimal('7.0000000000'), 1330, datetime.datetime(2012, 5, 24, 0, 0)),('A', '', Decimal('21.0000000000'), 1330, datetime.datetime(2012, 5, 14, 0, 0))]

我想按元组中的第一、第二、第四和第五列对它们进行分组，并对第三列求和.在本例中，我将列命名为 col1、col2、col3、col4、col5.

在 SQL 中，我会做这样的事情:

从我的表中选择 col1, col2, sum(col3), col4, col5按 col1、col2、col4、col5 分组

是否有一种很酷"的方法可以做到这一点，还是全部是手动循环?

解决方案

>>>[(x[0:2] + (sum(z[2] for z in y),) + x[2:5]) for (x, y) initertools.groupby(sorted(L, key=operator.itemgetter(0, 1, 3, 4)),键=operator.itemgetter(0, 1, 3, 4))][('A', '', Decimal('21.0000000000'), 1330, datetime.datetime(2012, 5, 14, 0, 0)),('A', '', Decimal('4.0000000000'), 1330, datetime.datetime(2012, 6, 8, 0, 0)),('AA', 'C', Decimal('31.0000000000'), 1330, datetime.datetime(2012, 5, 31, 0, 0)),('B', '', Decimal('7.0000000000'), 1330, datetime.datetime(2012, 5, 24, 0, 0)),('B', '', Decimal('31.0000000000'), 1330, datetime.datetime(2012, 6, 4, 0, 0))]

(注意:输出重新格式化)

Given the following list:

[
    ('A', '', Decimal('4.0000000000'), 1330, datetime.datetime(2012, 6, 8, 0, 0)),
    ('B', '', Decimal('31.0000000000'), 1330, datetime.datetime(2012, 6, 4, 0, 0)),
    ('AA', 'C', Decimal('31.0000000000'), 1330, datetime.datetime(2012, 5, 31, 0, 0)),
    ('B', '', Decimal('7.0000000000'), 1330, datetime.datetime(2012, 5, 24, 0, 0)),
    ('A', '', Decimal('21.0000000000'), 1330, datetime.datetime(2012, 5, 14, 0, 0))
]

I would like to group these by the first, second, fourth and fifth columns in the tuple and sum the 3rd. For this example I'll name the columns as col1, col2, col3, col4, col5.

In SQL I would do something like this:

select col1, col2, sum(col3), col4, col5 from my table
group by col1, col2, col4, col5

Is there a "cool" way to do this or is it all a manual loop?

解决方案

>>> [(x[0:2] + (sum(z[2] for z in y),) + x[2:5]) for (x, y) in
      itertools.groupby(sorted(L, key=operator.itemgetter(0, 1, 3, 4)),
      key=operator.itemgetter(0, 1, 3, 4))]
[
  ('A', '', Decimal('21.0000000000'), 1330, datetime.datetime(2012, 5, 14, 0, 0)),
  ('A', '', Decimal('4.0000000000'), 1330, datetime.datetime(2012, 6, 8, 0, 0)),
  ('AA', 'C', Decimal('31.0000000000'), 1330, datetime.datetime(2012, 5, 31, 0, 0)),
  ('B', '', Decimal('7.0000000000'), 1330, datetime.datetime(2012, 5, 24, 0, 0)),
  ('B', '', Decimal('31.0000000000'), 1330, datetime.datetime(2012, 6, 4, 0, 0))
]

(NOTE: output reformatted)

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