1

I have a list that looks like this:

relationShipArray = []

relationShipArray.append([340859419124453377, 340853571828469762])
relationShipArray.append([340859419124453377, 340854579195432961])
relationShipArray.append([340770796777660416, 340824159120654336])
relationShipArray.append([340509588065513473, 340764841658703872])
relationShipArray.append([340478540048916480, 340671891540934656])
relationShipArray.append([340853571828469762, 340854579195432961])
relationShipArray.append([340842710057492480, 340825411573399553])
relationShipArray.append([340825411573399553, 340770796777660416])
relationShipArray.append([340825411573399553, 340824159120654336])
relationShipArray.append([340824159120654336, 340770796777660416])
relationShipArray.append([340804620295221249, 340825411573399553])
relationShipArray.append([340684236191313923, 340663388122279937])
relationShipArray.append([340663388122279937, 340684236191313923])
relationShipArray.append([340859507280318464, 340859419124453377])
relationShipArray.append([340859507280318464, 340853571828469762])
relationShipArray.append([340859507280318464, 340854579195432961])
relationShipArray.append([340854599697178624, 340845885439229952])
relationShipArray.append([340836561937641472, 340851694759972864])
relationShipArray.append([340854579195432961, 340853571828469762])
relationShipArray.append([340844519832580096, 340854599697178624])
relationShipArray.append([340814054610305024, 340748443670683648])
relationShipArray.append([340851694759972864, 340836561937641472])
relationShipArray.append([340748443670683648, 340814054610305024])
relationShipArray.append([340739498356912128, 340825992832638977])

As you can see there are cases that are duplicated. e.g. 

[340853571828469762, 340854579195432961]

is the same as (but inverted)

[340854579195432961, 340853571828469762]

What is the best way (with some efficiency but can live without it if need be) to remove the duplicates from this list? So in this case I would keep [340853571828469762, 340854579195432961], but remove the [340854579195432961, 340853571828469762].

4

3 に答える 3

1

順序を維持する必要がある場合は、OrderedDict を使用します。

from collections import OrderedDict

>>> L = [[1, 2], [4, 5], [1,2], [2, 1]]
>>> [[x, y] for x, y in OrderedDict.fromkeys(frozenset(x) for x in L)]
[[1, 2], [4, 5]]

編集1

順序が重要でない場合は、セットを使用できます。

>>> [[x, y] for x, y in set(frozenset(x) for x in L)]
[[1, 2], [4, 5]]

編集2

2 つの要素だけでなく、さまざまな長さのリストに対して機能する、より一般的なソリューション:

[list(entry) for entry in set(frozenset(x) for x in L)]
[list(entry) for entry in OrderedDict.fromkeys(frozenset(x) for x in L)]
于 2013-06-02T16:40:57.687 に答える
0

ワンライナーソリューション

relationShipArray = []

relationShipArray.append([340859419124453377, 340853571828469762])
relationShipArray.append([340859419124453377, 340854579195432961])
relationShipArray.append([340770796777660416, 340824159120654336])
relationShipArray.append([340509588065513473, 340764841658703872])
relationShipArray.append([340478540048916480, 340671891540934656])
relationShipArray.append([340853571828469762, 340854579195432961])
relationShipArray.append([340842710057492480, 340825411573399553])
relationShipArray.append([340825411573399553, 340770796777660416])
relationShipArray.append([340825411573399553, 340824159120654336])
relationShipArray.append([340824159120654336, 340770796777660416])
relationShipArray.append([340804620295221249, 340825411573399553])
relationShipArray.append([340684236191313923, 340663388122279937])
relationShipArray.append([340663388122279937, 340684236191313923])
relationShipArray.append([340859507280318464, 340859419124453377])
relationShipArray.append([340859507280318464, 340853571828469762])
relationShipArray.append([340859507280318464, 340854579195432961])
relationShipArray.append([340854599697178624, 340845885439229952])
relationShipArray.append([340836561937641472, 340851694759972864])
relationShipArray.append([340854579195432961, 340853571828469762])
relationShipArray.append([340844519832580096, 340854599697178624])
relationShipArray.append([340814054610305024, 340748443670683648])
relationShipArray.append([340851694759972864, 340836561937641472])
relationShipArray.append([340748443670683648, 340814054610305024])
relationShipArray.append([340739498356912128, 340825992832638977])

すべてのリストrelationShipArrayとそれらの逆ピアを含む配列を作成します。使用してnp.uniqueください。

import numpy as np
Y = list(np.unique(np.array(relationShipArray + 
                       [X[::-1] for X in relationShipArray])))
于 2013-06-02T16:47:47.617 に答える
0

の順序がrelationShipArray重要でない場合:

result = {tuple(sorted(item)) for item in relationShipArray}
于 2013-06-02T16:36:55.087 に答える