Question

我在下面有2个列表

Token_Sentence=[['This','is','a','book'],['This','is','a','cat'],['Those','are','two','books']]
Mapping=[['This',1],['is',2],['a',3],['book',4],['cat',5],['Those',6],['are',7],['two',8],['books',9]]

我想像这样

映射Token_Sentence（将文本转换为索引号）

[[1,2,3,4],[1,2,3,5],[6,7,8,9]]

这是我的代码

for a in range(len(Token_Sentence)):
    for b in range(len(Token_Sentence[a])):
        for c in range(len(Mapping)):
            if Token_Sentence[a][b]==Mapping[c][0]:
                Token_Sentence[a][b]=Mapping[c][1]

但问题是它需要很长时间才能运行（我的真实数据列表非常大）。

还有其他方法可以实现比我更快更简单的目标吗？

Answer 1

您可以从Mapping：

创建映射

Token_Sentence=[['This','is','a','book'],['This','is','a','cat'],['Those','are','two','books']]
Mapping=[['This',1],['is',2],['a',3],['book',4],['cat',5],['Those',6],['are',7],['two',8],['books',9]]
d = dict(Mapping)
new_sentence = [[d[b] for b in i] for i in Token_Sentence]

输出：

[[1, 2, 3, 4], [1, 2, 3, 5], [6, 7, 8, 9]

Answer 2

上面的答案很好，只是想表明你是否想要不转换为dict：

Token_Sentence=[['This','is','a','book'],['This','is','a','cat'],['Those','are','two','books']]
Mapping=[['This',1],['is',2],['a',3],['book',4],['cat',5],['Those',6],['are',7],['two',8],['books',9]]

print([[k[1] for j in i for k in Mapping if j==k[0]] for i in Token_Sentence ])

输出：

[[1, 2, 3, 4], [1, 2, 3, 5], [6, 7, 8, 9]]

如何在python中映射列表

2 个答案: