Question

我是元组列表：

mylist = [ (user1, 23, 32), (user1, 23, 34), (user3, 34, 34), (user2, 34, 45), (user1, 45,23),(user2, 56, 56), (user5, 4,4)]

我需要找出每个用户的平均时间。我的问题是我不知道有多少用户和价值观。所以我做不到

for item in mylist:
    if item[0] == 'user1':
       new_list.append( ( item1, item2))

如何获得值

user1 = [ (23,32), (23,34), (45,23)]
user2 = [(34, 45), (56,56)]
usre3 = [(34,34)]

感谢

Answer 1

>>> from collections import defaultdict    
>>> d = defaultdict(list)
>>> for user, *items in mylist:
    d[user] += items,


>>> d
defaultdict(<class 'list'>, {'user5': [[4, 4]], 'user2': [[34, 45], [56, 56]], 'user3': [[34, 34]], 'user1': [[23, 32], [23, 34], [45, 23]]})

Answer 2

你显然需要一本字典，对吗？一种方法是：

results = {}
for user, val1, val2 in mylist:
    if user not in results:
        results[user] = []
    results[user].append((val1, val2))

但是感谢collections.defaultdict（从2.5开始），我们可以简化这一点：

import collections

results = collections.defaultdict(list)
for users, val1, val2 in mylist:
    results[user].append((val1, val2))

Answer 3

我猜你实际上有这样一个列表：

mylist = [ ('user1', 23, 32), ('user1', 23, 34), ('user3', 34, 34), ('user2', 34, 45), ('user1', 45,23), ('user2', 56, 56), ('user5', 4,4)]

你不想制作一个新的清单，你想要一本字典：

from collections import defaultdict

users = defaultdict(list)
for u, t1, t2 in mylist:
    users[u].append((t1, t2))

然后你可以，例如：

print users['user1']

产生

[(23, 32), (23, 34), (45, 23)]

Answer 4

使用字典：

avgtimes = {}
for item in mylist:
    avgtimes.setdefault(item[0], []).append(item[1:])

Answer 5

from collections import defaultdict
h=defaultdict(list)
mylist = [ ("user1", 23, 32), ("user1", 23, 34), ("user3", 34, 34), ("user2", 34, 45), ("user1", 45,23),("user2", 56, 56), ("user5", 4,4)]
for item in mylist:
   h[item[0]].append(item[1:])
print h

用于查找值的列表内部元组

5 个答案: