回答此问题可获得 20 贡献值,回答如果被采纳可获得 50 分。
<p>我有一个生成如下数据结构的代码:</p>
<pre><code>{'AttributeId': '4192',
'AttributeList': '',
'ClassId': '1014 (AP)',
'InstanceId': '0',
'MessageType': '81 (GetAttributesResponse)',
'ObjectInstance': '',
'Protocol': 'BSMIS Rx',
'RDN': '',
'TransactionId': '66',
'Sequences': [[],
[1,'2013-02-26T15:01:11Z'],
[],
[10564,13,388,0,-321,83,'272','05',67,67,708,896,31,128,-12,-109,0,-20,-111,-1,-1,0],
[10564,13,108,0,-11,83,'272','05',67,67,708,1796,31,128,-12,-109,0,-20,-111,-1,-1,0],
[10589,16,388,0,-15,79,'272','05',67,67,708,8680,31,125,-16,-110,0,-20,-111,-1,-1,0],
[10589,15,108,0,-16,81,'272','05',67,67,708,8105,31,126,-14,-109,0,-20,-111,-1,-1,0],
[10637,40,233,0,-11,89,'272','03',30052,1,5,54013,33,103,-6,-76,1,-20,-111,-1,-1,0],
[10662,46,234,0,-15,85,'272','03',30052,1,5,54016,33,97,-10,-74,1,-20,-111,-1,-1,0],
[10712,51,12,0,-24,91,'272','01',4013,254,200,2973,3,62,-4,-63,0,-20,-111,-1,-1,0],
[10737,15,224,0,-16,82,'272','01',3020,21,21,40770,33,128,-13,-108,0,-20,-111,-1,-1,0],
[10762,14,450,0,-7,78,'272','01',3020,21,21,53215,29,125,-17,-113,0,-20,-111,-1,-1,0],
[10762,15,224,0,-7,85,'272','01',3020,21,21,50770,33,128,-10,-105,0,-20,-111,-1,-1,0],
[10762,14,124,0,-7,78,'272','01',3020,10,10,56880,32,128,-17,-113,0,-20,-111,-1,-1,0],
[10812,11,135,0,-14,81,'272','02',36002,1,11,43159,31,130,-14,-113,1,-20,-111,-1,-1,0],
[10837,42,23,0,-9,89,'272','02',36002,1,11,53529,31,99,-6,-74,1,-20,-111,-1,-1,0,54],
[13,'2013-02-26T15:02:09Z'],
[],
[2,12,7,0,9,70,'272','02',20003,0,0,15535,0,0,0,0,1,100,100,-1,-1,0],
[5,15,44,0,-205,77,'272','02',20003,0,0,15632,0,0,0,0,1,100,100,-1,-1,0],
[7,25,9,0,0,84,'272','02',20002,0,0,50883,0,0,0,0,1,100,100,-1,-1,0]]
}
</code></pre>
<p>然后我把它过滤下来,形成一个相关值的列表,如果长度大于等于22,我只需要序列的前2个元素。我是这样做的:</p>
<pre><code> len22seqs = filter(lambda s: len(s)>=22, data['Sequences'])
UARFCNRSSI = []
for i in range(len(len22seqs)):
UARFCNRSSI.append([len22seqs[i][0], len22seqs[i][1]])
</code></pre>
<p>过滤列表的示例如下:</p>
<pre><code> [[10564, 15], [10564, 13], [10589, 18], [10637, 39], [10662, 38], [10712, 50], [10737, 15], [10762, 14], [10787, 9], [10812, 12], [10837, 45], [3, 17], [7, 21], [46, 26], [48, 12], [49, 24], [64, 14], [66, 17], [976, 27], [981, 22], [982, 22], [983, 17], [985, 13], [517, 9], [521, 15], [525, 11], [526, 13], [528, 14], [698, 14], [788, 24], [792, 19]]
</code></pre>
<p>不过,我现在注意到,在这些子列表中,我需要第三个元素。
就是这样:</p>
<pre><code>[1,'2013-02-26T15:01:11Z'],
</code></pre>
<p>我需要每个长度为2的列表的第一个元素作为第三个元素附加到这个过滤列表中,作为后面的元素。但是当有一个长度为2的新列表时,我需要将这个新值附加到后面的条目中。你知道吗</p>
<p>因此,我的最后一个列表示例可以如下所示,请注意,在找到另一个长度为2的列表时,第三个元素更改为13:</p>
<pre><code>[[10564, 15, 1], [10564, 13, 1], [10589, 18, 1], [10637, 39, 1], [10662, 38, 1], [10837, 45, 1], [3, 17, 13], [7, 21, 13], [46, 26, 13], etc]
</code></pre>
<p>我该怎么做?我是否必须使用len>;=22和len=2进行两次筛选,并且仅使用len>;=22进行单独筛选,因为对于长度为2的列表,我不想将元素0或1附加到最终列表中。你知道吗</p>