如何在python中映射列表

Question

我下面有2個清單

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]]

我想這樣映射Token_Sentence（將文本轉換為索引號）

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

這是我的代碼

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]

但是問題是運行時間很長（我的真實數據列表很大）。

還有其他比我更快，更簡單的方法來實現我的目標嗎？

Answer 1

您可以從Mapping創建Mapping ：

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]

輸出：

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

Answer 2

上面的答案是很好的，只想展示是否不需要轉換為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 ])

輸出：

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