比较双列表中的元素并返回双列表

Question

我有这个双重名单：

doubleL = [['123', 'user1', 'its.zip', '05-19-17'],
['123', 'user1', 'zippedtto.zip', '05-24-17'],
['123', 'user1', 'zippedtto.zip', '05-19-17'],
['123', 'user2', 'Iam.zip', '05-19-17'], 
['abcd', 'Letsee.zip', '05-22-17']]

screenshot of my double list

我想使用python并检查每个子列表中的前两个元素是否相同，并返回带有最后一个元素的双列表。例如，在这种情况下，我的输出将是：

output = [ ['05-19-17', '05-24-17', '05-19-17' ], ['05-19-17'], ['05-22-17']]

获得此输出的最pythonic方式是什么？

谢谢！

Answer 1

没有defaultdict

tmp = {} # A dictionary

for x in doubleL: # Iterate over outer list
    # Set default as empty list if key not present
    # Else/And just append the last value
    tmp.setdefault((x[0], x[1]), []).append(x[-1]) 

print(list(tmp.values()) # Print values as list

使用defaultdict

from collections import defaultdict

for x in doubleL: # Iterate over outer list
    # Just append the last value
    tmp[(x[0], x[1])].append(x[-1]) 

print(list(tmp.values()) # Print values as list

Answer 2

您可以使用字典，但如果您想保留订单，则应使用模块集合和OrderedDict，如下所示：

LET date = 1495616898128

FOR invoice IN data_invoices
    FILTER invoice.tenantId IN ['xxx', 'yyy']
    FILTER invoice.createdOn<=date
    LET invoiceData = (invoice.data[* FILTER CURRENT.createdOn<=date LIMIT 1])[0]
    FILTER invoiceData.values.amount>1000
    LET jobs = (
        FOR job, link IN 1 OUTBOUND invoice data_links
            FILTER link.createdOn<=date AND (link.replacedOn == NULL 
                OR link.replacedOn>date)
            LET jobData = (job.data[* FILTER CURRENT.createdOn<=date LIMIT 1])[0]
            FILTER CONTAINS(jobData.values.name, 'a')
            COLLECT WITH COUNT INTO count
            RETURN {
                count 
            } 
    )[0]
    FILTER jobs.count>0
    SORT invoiceData.values.amount ASC
    LIMIT 0,8
    RETURN invoice

Answer 3

非词典路线将使用itertools.groupby和sorted以正确的顺序获取所有项目

from itertools import groupby

groups = []
for g, l in groupby(sorted(doubleL, key=lambda o: o[:2]), key=lambda o: o[:2]):
    groups.append([i[-1] for i in l])

print groups
>>> [['05-19-17', '05-24-17', '05-19-17'], ['05-19-17'], ['05-22-17']]