我在串列串列中有元組，并且只想提取元組中的一些元素。輸入資料的示例如下。
# input
[[('ab', 0.026412873688749918), ('dc', 0.016451082731822664), ('on', 0.014278088125928066),
('qc', 0.009752817881775656), ('mn', 0.008332886637563352), ('nt', 0.008250535392602258),
('nsw', 0.006874273287824427), ('bar', 0.005878684829852004), ('tor', 0.005741627328513831),
('wds', 0.004119216502907735)],
[('nb', 0.03053649661493629), ('ns', 0.01925207174326825), ('ham', 0.016207228280183325),
('bra', 0.013390785663058102), ('nia', 0.00878166482558038), ('knxr', 0.004648856466085521),
('nwm', 0.004463444159552605), ('md', 0.004377821331080258), ('ut', 0.004165890522922745),
('va', 0.0037484060754341083)]]
我想要做的是獲取元組中的第一項。
# output
[['ab', 'dc', 'on', 'qc', 'mn', 'nt', 'nsw', 'bar', 'tor', 'wds'],
['nb', 'ns', 'ham', 'bra', 'nia', 'knxr', 'nwm', 'md', 'ut', 'va']]
uj5u.com熱心網友回復：
input = [
[('ab', 0.026412873688749918), ('dc', 0.016451082731822664), ('on', 0.014278088125928066),
('qc', 0.009752817881775656), ('mn', 0.008332886637563352), ('nt', 0.008250535392602258),
('nsw', 0.006874273287824427), ('bar', 0.005878684829852004), ('tor', 0.005741627328513831),
('wds', 0.004119216502907735)],
[('nb', 0.03053649661493629), ('ns', 0.01925207174326825), ('ham', 0.016207228280183325),
('bra', 0.013390785663058102), ('nia', 0.00878166482558038), ('knxr', 0.004648856466085521),
('nwm', 0.004463444159552605), ('md', 0.004377821331080258), ('ut', 0.004165890522922745),
('va', 0.0037484060754341083)]
]
如評論中所示，您可以使用串列推導來實作這一點：
[[idx for idx, val in x] for x in input]
# Result
[['ab', 'dc', 'on', 'qc', 'mn', 'nt', 'nsw', 'bar', 'tor', 'wds'],
['nb', 'ns', 'ham', 'bra', 'nia', 'knxr', 'nwm', 'md', 'ut', 'va']]
實作此目的的更復雜方法是??使用zip()將元組的第一個元素與第二個元素分開，如下所示：
[('ab', 'dc', 'on', 'qc', 'mn', 'nt', 'nsw', 'bar', 'tor', 'wds'),
(0.026412873688749918,0.016451082731822664,0.014278088125928066,0.009752817881775656,0.008332886637563352,0.008250535392602258,0.006874273287824427,0.005878684829852004,0.005741627328513831,0.004119216502907735)]
這種方法可以使用：
[list(list(zip(*x))[0]) for x in input]
# Result
[['ab', 'dc', 'on', 'qc', 'mn', 'nt', 'nsw', 'bar', 'tor', 'wds'],
['nb', 'ns', 'ham', 'bra', 'nia', 'knxr', 'nwm', 'md', 'ut', 'va']]
uj5u.com熱心網友回復：
您可以使用回圈或串列理解來執行此操作。
輸入資料是包含元組的串列串列。使用tuple[0]訪問元組的第一個元素并將其保存到一個空串列中，如下所示：-
input_data = [
[('ab', 0.026412873688749918), ('dc', 0.016451082731822664), ('on', 0.014278088125928066),
('qc', 0.009752817881775656), ('mn', 0.008332886637563352), ('nt', 0.008250535392602258),
('nsw', 0.006874273287824427), ('bar', 0.005878684829852004), ('tor', 0.005741627328513831),
('wds', 0.004119216502907735)],
[('nb', 0.03053649661493629), ('ns', 0.01925207174326825), ('ham', 0.016207228280183325),
('bra', 0.013390785663058102), ('nia', 0.00878166482558038), ('knxr', 0.004648856466085521),
('nwm', 0.004463444159552605), ('md', 0.004377821331080258), ('ut', 0.004165890522922745),
('va', 0.0037484060754341083)]
]
data_list = []
for x in input_data:
d_list = []
for y in x:
d_list.append(y[0])
data_list.append(d_list)
# Result...
[['ab', 'dc', 'on', 'qc', 'mn', 'nt', 'nsw', 'bar', 'tor', 'wds'],
['nb', 'ns', 'ham', 'bra', 'nia', 'knxr', 'nwm', 'md', 'ut', 'va']]
使用串列推導式：- 這是通過洗掉 append() 方法和初始空串列來撰寫上述 for 回圈的一種速記方式。
data_list = [ [y[0] for y in x] for x in input_data ]
# Result...
[['ab', 'dc', 'on', 'qc', 'mn', 'nt', 'nsw', 'bar', 'tor', 'wds'],
['nb', 'ns', 'ham', 'bra', 'nia', 'knxr', 'nwm', 'md', 'ut', 'va']]
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