Question

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)]

Answer 1

您可以使用<html> <body> <p> This is page Two! </p> </body> </html>功能将每个子列表更改为dict：

map()

然后使用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))方法并为字典中不存在的键指定默认值dict.get：

上述结果：

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

Answer 2

一种非常简单的方法是使用带有默认值的构造dict，然后更新它：

>>> 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}

修改

结合hgwell的建议来输出元组列表，这里是一个完整的函数（可能会以某种方式更好地完成，但无论如何都可以）：

def listify(l): res = [] for j in l: d = dict([(0,0),(1,0),(2,0),(3,0),(4,0)]) d.update(j) res.append(list(d.items())) return res

并且在行动......

>>> 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)]]

Python列表的格式不规则

2 个答案: