1

它有一种内置（或天真的）方式来处理（求和、计数）由 itertools.groupby 生成的聚合吗？

例如，给定示例代码中的表格，折扣为 10%...

我想要：

# Select each city...
for city,city_purchases_d in itertools.groupby(transaction_l, 
                                               lambda d: d["city"]):
  print Aggregate( city,sum(|qty|),sum(|qty * price|)*(1-discount) ) *
              city_purchases_d

输入数据：

discount=0.10 # 10%

transaction_l=(
  dict(trans=201, name="Anne",  city="LAX", item="Apple", qty=10, price=1.33),
  dict(trans=202, name="Betty", city="LAX", item="Banana",qty=20, price=2.33),
  dict(trans=203, name="Carol", city="LAX", item="Cherry",qty=30, price=3.33),
  dict(trans=101, name="Andy",  city="NYC", item="Avodado",qty=1, price=1.32),
  dict(trans=102, name="Andy",  city="NYC", item=u"Açaí",  qty=1, price=1.70),
  dict(trans=103, name="Bob",   city="NYC", item="Bacuri", qty=3, price=2.10),
  dict(trans=104, name="Cliff", city="NYC", item="Carrot", qty=4, price=2.22),
  dict(trans=105, name="David", city="NYC", item="Donut",  qty=5, price=3.00)
)

输出将是：

('LAX',60,143.82)
('NYC',14,29.88)

IE

In LAX purchased 60 fruit at the total price of $143.82   
In NYC purchased 14 fruit at the total price of $29.88

附言。我注意到有很多类似的问题......但没有一个只是简单地采用（类似）一个天真city,sum(|qty|),sum(|qty * price|)*(1-discount)的聚合表达式。

编辑：（以使用生成器理解为代价）几乎可以达到如下效果：

discount=0.10 # 10%
desc_f="In %s purchased %s fruit at the total price of $%.2f"
for city,city_purchases_d in itertools.groupby(transaction_l, lambda d: d["city"]):
# alternatively - Plan B: manually creating aggregation DOES also work:
  qty_x_price=list(trans["qty"]*trans["price"] for trans in list(city_purchases_d))
  qty=(trans["qty"] for trans in city_purchases_d)
  print desc_f%(city,sum(qty),sum(qty_x_price)*(1-discount))
4

3 回答 3

1

我假设您在数据的聚合处理中需要一些灵活性，也许是用户输入的？否则很容易做到这一点itertools.groupby：

from itertools import groupby

discount=0.10
transaction_l=(
  dict(trans=201, name="Anne",  city="LAX", item="Apple", qty=10, price=1.33),
  dict(trans=202, name="Betty", city="LAX", item="Banana",qty=20, price=2.33),
  dict(trans=203, name="Carol", city="LAX", item="Cherry",qty=30, price=3.33),
  dict(trans=101, name="Andy",  city="NYC", item="Avodado",qty=1, price=1.32),
  dict(trans=102, name="Andy",  city="NYC", item=u"Açaí",  qty=1, price=1.70),
  dict(trans=103, name="Bob",   city="NYC", item="Bacuri", qty=3, price=2.10),
  dict(trans=104, name="Cliff", city="NYC", item="Carrot", qty=4, price=2.22),
  dict(trans=105, name="David", city="NYC", item="Donut",  qty=5, price=3.00)
)
desc_f = 'In %s purchased %s fruit at the total price of $%.2f'

for city, transactions in groupby(transaction_l, key=lambda d: d['city']):
    transactions = list(transactions)
    print desc_f % (city,
                    sum(t['qty'] for t in transactions),
                    sum( (t['qty']*t['price'])*(1-discount)
                        for t in transactions))

输出

在洛杉矶以 143.82 美元的总价购买了 60 个水果
在纽约以 29.88 美元的总价购买了 14 个水果

如果您需要对数据执行任意“查询”的灵活性，这可能是一个幼稚（甚至是奇怪）的建议，但是对内存中的 SQLite 数据库使用 SQL 查询？

import sqlite3

discount=0.10 # 10%

transaction_l=(
  dict(trans=201, name="Anne",  city="LAX", item="Apple", qty=10, price=1.33),
  dict(trans=202, name="Betty", city="LAX", item="Banana",qty=20, price=2.33),
  dict(trans=203, name="Carol", city="LAX", item="Cherry",qty=30, price=3.33),
  dict(trans=101, name="Andy",  city="NYC", item="Avodado",qty=1, price=1.32),
  dict(trans=102, name="Andy",  city="NYC", item=u"Açaí",  qty=1, price=1.70),
  dict(trans=103, name="Bob",   city="NYC", item="Bacuri", qty=3, price=2.10),
  dict(trans=104, name="Cliff", city="NYC", item="Carrot", qty=4, price=2.22),
  dict(trans=105, name="David", city="NYC", item="Donut",  qty=5, price=3.00)
)

memdb = sqlite3.connect(':memory:')
cursor = memdb.cursor()
# create an in-memory table
r = cursor.execute('create table transactions (trans int, name varchar(30), city char(3), item varchar(20), qty int, price numeric)')
result = cursor.executemany('insert into transactions (trans, name, city, item, qty, price) values (:trans, :name, :city, :item, :qty, :price)', transaction_l)
assert result.rowcount == len(transaction_l)

result = cursor.execute('select city, sum(qty), sum(qty*price)*(1-{}) from transactions group by city'.format(discount))

desc_f = 'In {} purchased {} fruit at the total price of ${:.2f}'
for row in result:
    print desc_f.format(*row)

memdb.close()

输出

在洛杉矶以 143.82 美元的总价购买了 60 个水果
在纽约以 29.88 美元的总价购买了 14 个水果

所以，现在你的任务是创建一个这样的 SQL 查询：

select city, sum(qty), sum(qty*price)*(1-0.1) from transactions group by city

由此：

city,sum(|qty|),sum(|qty * price|)*(1-discount)

这似乎很可行。

于 2015-03-15T12:34:02.143 回答
0

我仅出于兴趣添加以下代码示例...

def eval_iter(expr_str, global_d, local_d, dict_iter, sep="|"):
  expr_l=expr_str.split(sep)
  if isinstance(dict_iter, dict): dict_iter=dict_iter.itervalues()

  aggregation_l=[]
  for eval_locals in dict_iter:
    locals().update(eval_locals)
    aggregation_l.append(
      eval(",".join(expr_l[1::2]).join("[]"), globals(), locals()))
    for key in eval_locals: del locals()[key] # cleanup a bit

  expr_l[1::2]=["aggregation_l[%d]"%enum for enum in range(len(expr_l)/2)]

  local_d["aggregation_l"]=zip(*aggregation_l)
  return eval("".join(expr_l), global_d, local_d)


discount=0.10 # 10%
desc_f="In %s purchased %s fruit at the total price of $%.2f"

# The QUERY: -------- 8>< - - - - cut here - - - -
for city,city_purchases_d in itertools.groupby(transaction_l, 
                                               lambda d: d["city"]):
  print desc_f%eval_iter("city,sum(|qty|),sum(|qty * price|)*(1-discount)",
                         globals(), locals(), city_purchases_d)

输出：根据需要...

In LAX purchased 60 fruit at the total price of $143.82 
In NYC purchased 14 fruit at the total price of $29.88
于 2015-03-15T13:26:15.993 回答
0

可以使用 Pandas 模块以更简单的方式归档所需的结果，

import pandas as pd

discount = 0.2

df = pd.DataFrame(list(transaction_l))
df['total_price'] = df.qty*df.price*(1-discount)
res = df.groupby('city').sum()[['qty', 'total_price']]
print(res)
#       qty  total_price
#city 
#LAX    60       127.84
#NYC    14        26.56
于 2015-03-15T12:34:29.253 回答