Python，从元组列表中删除重复项

I have the following list:

我有以下列表：

[('mail', 167, datetime.datetime(2010, 9, 29)) , 
 ('name', 1317, datetime.datetime(2011, 12, 12)), 
 ('mail', 1045, datetime.datetime(2010, 8, 13)), 
 ('name', 3, datetime.datetime(2011, 11, 3))]

And I want to remove items from the list with coinciding first item in a tuple where date is not the latest. In other words I need to get this:

并且我想从列表中删除项目，其中元组中的第一项是重合的，其中日期不是最新的。换句话说，我需要得到这个：

[('mail', 167, datetime.datetime(2010, 9, 29)) , 
 ('name', 1317, datetime.datetime(2011, 12, 12))]

5 个解决方案

#1

You can use a dictionary to store the highest value found for a given key so far:

到目前为止，您可以使用字典存储给定键的最高值：

temp = {}
for key, number, date in input_list:
    if key not in temp: # we see this key for the first time
        temp[key] = (key, number, date)
    else:
        if temp[key][2] < date: # the new date is larger than the old one
            temp[key] = (key, number, date)
result = temp.values()

#2

The following approach uses a dictionary to overwrite entries with the same key. Since the list is sorted by the date, older entries get overwritten by newer ones.

以下方法使用字典覆盖具有相同键的条目。由于列表按日期排序，较旧的条目会被较新的条目覆盖。

temp = {}
for v in sorted(L, key=lambda L: L[2]): # where L is your list
    temp[v[0]] = v
result = temp.values()

Or, for something a lot more compact (but much less readable):

或者，对于更紧凑（但可读性更低）的东西：

result = dict((v[0],v) for v in sorted(L, key=lambda L: L[2])).values()

Update

This method would be reasonably quick if the list is already (or mostly) sorted by date. If it isn't, and especially if it is a large list, then this may not be the best approach.

如果列表已经（或大部分）按日期排序，则此方法将相当快。如果不是，特别是如果它是一个大型列表，那么这可能不是最好的方法。

For unsorted lists, you will likely get a some performance improvement by sorting by the key first, then the date. i.e. sorted(L, key=lambda L: (L[0],L[2])).

对于未排序的列表，您可能会先通过键排序，然后按日期排序，从而获得一些性能提升。即排序（L，key =λL：（L [0]，L [2]））。

Or, better yet, go for Space_C0wb0y's answer.

或者，更好的是，去找Space_C0wb0y的答案。

#3

d = {}

for item in list:
    if (item[0], item[1]) not in d:
        d[(item[0], item[1])] = item[2]
    else:
        if item[2] > d[(item[0], item[1])]:
            d[(item[0], item[1])] = item[2]

item = [(x[0], x[1], d[x] for x in d.keys()]

#4

You can do it via sorting the list and getting the highest values by d[2]:

你可以通过对列表进行排序并通过d [2]获得最高值来实现：

    In [26]: d
    Out[26]: 
    [('mail', 167, datetime.datetime(2010, 9, 29, 0, 0)),
     ('name', 1317, datetime.datetime(2011, 12, 12, 0, 0)),
     ('mail', 1045, datetime.datetime(2010, 8, 13, 0, 0)),
     ('name', 3, datetime.datetime(2011, 11, 3, 0, 0))]

    In [27]: d.sort(key = lambda i: i[2], reverse=True)

    In [28]: d
    Out[28]: 
    [('name', 1317, datetime.datetime(2011, 12, 12, 0, 0)),
     ('name', 3, datetime.datetime(2011, 11, 3, 0, 0)),
     ('mail', 167, datetime.datetime(2010, 9, 29, 0, 0)),
     ('mail', 1045, datetime.datetime(2010, 8, 13, 0, 0))]

    In [29]: [i for pos, i in enumerate(d) if i[0] in [j[0] for j in d[pos+1:]]]
    Out[29]: 
    [('name', 1317, datetime.datetime(2011, 12, 12, 0, 0)),
     ('mail', 167, datetime.datetime(2010, 9, 29, 0, 0))]

#5

Here you go.

干得好。

#!/usr/bin/python2

from pprint import pprint
import datetime

ol = [('mail', 167, datetime.datetime(2010, 9, 29)) , 
     ('name', 1317, datetime.datetime(2011, 12, 12)), 
     ('mail', 1045, datetime.datetime(2010, 8, 13)), 
     ('name', 3, datetime.datetime(2011, 11, 3))]

d = {}

for t in sorted(ol, key=lambda t: (t[0], t[2])):
    d[t[0]] = t
out = d.values()

pprint(out)

That sorts the list using the first and third tuple elements as keys, then removes duplicates by using a hash table.

这使用第一个和第三个元组元素作为键对列表进行排序，然后使用哈希表删除重复项。