如果第一个元素是重复项，则从列表中删除元组的大多数pythonic方法 [英] Most pythonic way to remove tuples from a list if first element is a duplicate

问题描述

到目前为止，我的代码非常丑陋:

The code I have so far is pretty ugly:

orig = [(1,2),(1,3),(2,3),(3,3)]
previous_elem = []
unique_tuples = []
for tuple in orig:
    if tuple[0] not in previous_elem:
        unique_tuples += [tuple]
    previous_elem += [tuple[0]]
assert unique_tuples == [(1,2),(2,3),(3,3)]

必须有一个更加Python化的解决方案.

There must be a more pythonic solution.

推荐答案

如果您不在乎返回的元组是否重复，则可以始终将列表转换为字典并返回:

If you don't care which tuple round you return for duplicates, you could always convert your list to a dictionary and back:

>>> orig = [(1,2),(1,3),(2,3),(3,3)]
>>> list(dict(orig).items())
[(1, 3), (2, 3), (3, 3)]

如果您想返回 第一 元组，则可以将列表反转两次，并使用

If you want to return the first tuple round, you could reverse your list twice and use an OrderedDict, like this:

>>> from collections import OrderedDict
>>> orig = [(1,2),(1,3),(2,3),(3,3)]
>>> new = list(OrderedDict(orig[::-1]).items())[::-1]
[(1, 2), (2, 3), (3, 3)]

这些不是最有效的解决方案(如果非常重要，则 )，但是它们确实构成了惯用的单行代码.

These are not the most efficient solutions (if that's of great importance) , but they do make for nice idiomatic one-liners.

请注意速度的差异，并且如果您不关心返回的元组，则第一种选择的效率会更高:

Note the difference in speed, and that should you not care which tuple round you return, the first option is much more efficient:

>>> import timeit
>>> setup = '''
orig = [(1,2),(1,3),(2,3),(3,3)]
'''
>>> print (min(timeit.Timer('(list(dict(orig).items()))', setup=setup).repeat(7, 1000)))
0.0015771419037069459

相比

>>>setup = '''
orig = [(1,2),(1,3),(2,3),(3,3)]
from collections import OrderedDict
'''
>>> print (min(timeit.Timer('(list(OrderedDict(orig[::-1]).items())[::-1])', 
             setup=setup).repeat(7, 1000)))
0.024554947372323

根据这些速度测试，第一种选择的速度快将近15倍.

The first option is nearly 15 times faster according to these speed tests.

话虽这么说， Saksham的答案也是O(n)，并明智地粉碎了这些字典方法:

That being said however, Saksham's answer is also O(n) and smashes these dictionary methods efficiency wise:

>>> setup = '''
orig = [(1,2),(1,3),(2,3),(3,3)]
newlist = []
seen = set()
def fun():
    for (a, b) in orig:
        if not a in seen:
            newlist.append((a, b))
            seen.add(a)
    return newlist
'''
>>> print (min(timeit.Timer('fun()', setup=setup).repeat(7, 1000)))
0.0004833390384996095

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