将列表列表与字典进行比较，并将输出作为元组列表的列表

Posted 2021-04-09

我有一个字符串列表和一个字典列表：

docsp = [['how', 'can', 'I', 'change', 'this', 'car'], ['I', 'can', 'delete', 'this', 'module']]

ent = {'change' : 'action', 'car' : 'item', 'delete' : 'action'}

我想将字典与列表列表进行比较，并将列表元素与字典键值对标记为新的元组列表，如下所示。

newlist = [[('how', 'O'), ('can', 'O'), ('I', 'O'), ('change', 'action'), ('this', 'O'), ('car', 'item')], [('I', 'O'), ('can', 'O'), ('delete', 'action'), ('this', 'O'), ('module', 'O')]]

我尝试了以下代码：

n = []
for k,v in ent.items():
    for i in docsp:
        for j in i:
            if j==k:
                n.append((j,v))
            n.append((j, 'O'))
n

并获得以下输出：

[('how', 'O'),
 ('can', 'O'),
 ('I', 'O'),
 ('change', 'action'),
 ('change', 'O'),
 ('this', 'O'),
 ('car', 'O'),
 ('I', 'O'),
 ('can', 'O'),
 ('delete', 'O'),
 ('this', 'O'),
 ('module', 'O'),
 ('how', 'O'),
 ('can', 'O'),
 ('I', 'O'),
 ('change', 'O'),
 ('this', 'O'),
 ('car', 'item'),
 ('car', 'O'),
 ('I', 'O'),
 ('can', 'O'),
 ('delete', 'O'),
 ('this', 'O'),
 ('module', 'O'),
 ('how', 'O'),
 ('can', 'O'),
 ('I', 'O'),
 ('change', 'O'),
 ('this', 'O'),
 ('car', 'O'),
 ('I', 'O'),
 ('can', 'O'),
 ('delete', 'action'),
 ('delete', 'O'),
 ('this', 'O'),
 ('module', 'O')]

我经历了这个link，但无法修改它以获得我的预期输出。

答案

这是一个利用列表理解的解决方案：

docsp = [['how', 'can', 'I', 'change', 'this', 'car'], ['I', 'can', 'delete', 'this', 'module']]

ent = {'change' : 'action', 'car' : 'item', 'delete' : 'action'}

result = [[(docsp[i][j], ent.get(docsp[i][j], 'O')) for j in range(len(docsp[i]))] 
                                                    for i in range(len(docsp))]

# [[('how', 'O'),
#   ('can', 'O'),
#   ('I', 'O'),
#   ('change', 'action'),
#   ('this', 'O'),
#   ('car', 'item')],
#  [('I', 'O'),
#   ('can', 'O'),
#   ('delete', 'action'),
#   ('this', 'O'),
#   ('module', 'O')]]