我只是学习python不久。我尽力表示我的数据看起来更好，就像以前一样。 现在我有一些元组数据类型，如下所示：
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{"success":true}
我想知道每个人买了多少件物品。
假设不同的名字是不同的人。
那么我怎么做才能获得如下信息：
('John', '5', 'Coke')
('Mary', '1', 'Pie')
('Jack', '3', 'Milk')
('Mary', '2', 'Water')
('John', '3', 'Coke')
我不知道我现在该怎么办。即使是愚蠢的人也无法想出任何方法。
答案 0 :(得分：8)
我建议使用名字和饮料作为collections.Counter的关键：
from collections import Counter
count = Counter()
for name, amount, drink in tuples:
key = name, drink
count.update({key: int(amount)}) # increment the value
# represent the aggregated data
for (name, drink), amount in count.items():
print('{}: {} {}'.format(name, amount, drink))
更新我做了一些简单的测量，并想出了
count[name, drink] += value
不仅更具可读性，而且比调用update快得多，这不应该是一个惊喜。此外，defaultdict(int)甚至更快（大约两倍）（大概是因为Counter还执行了一些排序。）
答案 1 :(得分：2)
重新安排数据顺序可能有所帮助：
John: 8 Coke
Mary: 1 Pie
Mary: 2 Water
Jack: 3 Milk
可能更具洞察力
(John, Coke) : 8
(Mary, Pie) : 1
(Mary, Water): 2
(Jack, Milk) : 3
如果你知道SQL，这或多或少等同于groupby(name, dish)和sum(count)。
因此，在Python中，您可以为该对创建字典：
data = [
('John', '5', 'Coke'),
('Mary', '1', 'Pie'),
('Jack', '3', 'Milk'),
('Mary', '2', 'Water'),
('John', '3', 'Coke'),
]
orders = {}
for name, count, dish in data:
if (name, dish) in orders:
orders[(name, dish)] += int(count)
else:
# first entry
orders[(name, dish)] = int(count)
更加pythonic，使用collections.defaultdict：
orders = defaultdict(int)
for name, count, dish in data:
orders[(name, dish)] += int(count)
或collections.Counter。
根据需要格式化数据。
答案 2 :(得分：1)
假设你有一个元组列表
tuples = [('John', '5', 'Coke'),
('Mary', '1', 'Pie'),
('Jack', '3', 'Milk'),
('Mary', '2', 'Water'),
('John', '3', 'Coke')]
memory = {}
# First, we calculate the amount for each pair
for tuple in tuples:
# I define a generated key through the names. For example John-Cake, Mary-Pie, Jack-Milk,...
key = (tuple[0],tuple[2])
number = int(tuple[1])
if key in memory:
memory[key] += number
else:
memory[key] = number
# After, we format the information
list = []
for key in memory:
list.append((key[0],memory[key],key[1]))