不规则格式的Python列表

我有数据（使用Gensim的LDA结果），如下所示：

[(1, 0.97456828373415116)]
[(0, 0.91883125256489728), (1, 0.020225186991467976), (2, 0.020314851937259213), (3, 0.020382294889184499), (4, 0.020246413617191008)]
[(0, 0.93783520386426555), (1, 0.015481826214088806), (2, 0.015545735781026492), (3, 0.015535246185968628), (4, 0.015601987954650424)]
[(2, 0.98493696818505228)]
[(3, 0.99067359305252778)]
[(0, 0.73578249201070511), (3, 0.25197028613750805)]

我想转换为以下格式：

[(0, 0), (1, 0.97456828373415116), (2, 0), (3, 0), (4, 0)]
[(0, 0.91883125256489728), (1, 0.020225186991467976), (2, 0.020314851937259213), (3, 0.020382294889184499), (4, 0.020246413617191008)]
[(0, 0.93783520386426555), (1, 0.015481826214088806), (2, 0.015545735781026492), (3, 0.015535246185968628), (4, 0.015601987954650424)]
[(0, 0), (1, 0), (2, 0.98493696818505228), (3, 0), (4, 0)]
[(0, 0), (1, 0), (2, 0), (3, 0.96747728928637211), (4, 0)]
[(0, 0), (1, 0), (2, 0), (3, 0.99067359305252778), (4, 0)]
[(0, 0.73578249201070511), (1, 0), (2, 0), (3, 0.25197028613750805), (4, 0)]

---

您可以使用

map（）

函数将每个子列表更改为

dict

：

data = [[(1, 0.97456828373415116)],
[(0, 0.91883125256489728), (1, 0.020225186991467976), (2, 0.020314851937259213), (3, 0.020382294889184499), (4, 0.020246413617191008)],
[(0, 0.93783520386426555), (1, 0.015481826214088806), (2, 0.015545735781026492), (3, 0.015535246185968628), (4, 0.015601987954650424)],
[(2, 0.98493696818505228)],
[(3, 0.99067359305252778)],
[(0, 0.73578249201070511), (3, 0.25197028613750805)]]

results = list(map(dict, data))

然后使用

dict.get

方法，为字典中不存在的键指定默认值

0

：

for i in range(5):
    print(results[0].get(i, 0))

上述结果：

0
0.9745682837341512
0
0
0

---

一种非常简单的方法是将构造的dict与默认值一起使用，然后对其进行更新：

>>> d = dict([(0,0),(1,0),(2,0),(3,0)])
>>> print(d)
{0: 0, 1: 0, 2: 0, 3: 0}
>>> d.update([(0, 0.73578249201070511), (3, 0.25197028613750805)])
>>> print(d)
{0: 0.7357824920107051, 1: 0, 2: 0, 3: 0.25197028613750805}

编辑

结合hgwell关于输出元组列表的建议，这里有一个完整的函数（可能在某种程度上做得更好，但无论如何这是可行的）：

在行动中

>>> z = listify([[(1, 0.97456828373415116)],
                 [(0, 0.91883125256489728), (1, 0.020225186991467976), (2, 0.020314851937259213), (3, 0.020382294889184499), (4, 0.020246413617191008)],
                 [(2, 0.98493696818505228)]])
>>> pprint(z)
[[(0, 0), (1, 0.9745682837341512), (2, 0), (3, 0), (4, 0)],
 [(0, 0.9188312525648973),
  (1, 0.020225186991467976),
  (2, 0.020314851937259213),
  (3, 0.0203822948891845),
  (4, 0.020246413617191008)],
 [(0, 0), (1, 0), (2, 0.9849369681850523), (3, 0), (4, 0)]]

和

d.items（）

或在

python3

list（d.items（））

中提供所需的列表格式。@hgwells哈哈。我只知道在python3中必须有一种方法来做d.items（）这样的事情！谢谢我将使用一个函数定义进行编辑，该函数定义包含了这一点（我以后不会忘记它…）。

>>> z = listify([[(1, 0.97456828373415116)],
                 [(0, 0.91883125256489728), (1, 0.020225186991467976), (2, 0.020314851937259213), (3, 0.020382294889184499), (4, 0.020246413617191008)],
                 [(2, 0.98493696818505228)]])
>>> pprint(z)
[[(0, 0), (1, 0.9745682837341512), (2, 0), (3, 0), (4, 0)],
 [(0, 0.9188312525648973),
  (1, 0.020225186991467976),
  (2, 0.020314851937259213),
  (3, 0.0203822948891845),
  (4, 0.020246413617191008)],
 [(0, 0), (1, 0), (2, 0.9849369681850523), (3, 0), (4, 0)]]