Question

我有一长串这样的令牌：

 a_lis = [['hi how are you'],
['im fine thanks'],
['cool lets go to the restaurant']
,...,
['I can't now']]

我想用这样的逗号分隔每个标记或术语（*）：

 a_lis = [['hi', 'how', 'are', 'you'],
['im', 'fine', 'thanks'],
['cool', 'lets', 'go', 'to', 'the', 'restaurant']
,...,
['I', 'can't', 'now']]

我试图：

[x.replace(' ', ', ') for x in a_lis]

但是，不起作用。艾米关于如何获得（*）的想法？

Answer 1

如果a_lis是

a_lis = [["hi how are you"],
["im fine thanks"],
["cool lets go to the restaurant"],
["I can't now"]]

然后我们可以将其展平为一个字符串列表，并split每个字符串

[s.split() for l in a_lis for s in l]

给我们

[['hi', 'how', 'are', 'you'], ['im', 'fine', 'thanks'], ['cool', 'lets', 'go', 'to', 'the', 'restaurant'], ['I', "can't", 'now']]

Answer 2

使用map的另一种方法

a_lis = [["hi how are you"],["good and you"]]
new_list = list(map(lambda x: x[0].split(' '), a_lis))
# new_list is [['hi', 'how', 'are', 'you'], ['good', 'and', 'you']]

Answer 3

a_list = ['hmm this is a thing', 'and another']
new_list = []
for i in a_list:
    new_list.append(i.split(' '))
print(new_list)

基本上，对带有空格作为参数的字符串使用.split()方法。另外，请确保在这些字符串周围加上引号！

如果您有一个列表列表，只需添加另一个for循环：

a_list = [['hmm, this is a thing', 'and another'], ['more things!']]
new_list = []
for i in a_list:
    sub_list = []
    for k in i:
        sub_list.append(k.split(' '))
    new_list.append(sub_list)

Answer 4

这是一种实用的方法：

Angular 6 Routes    

<a routerLink="/signup">Register</a>
 const appRoutes: Routes = [
  { path: 'signup', component: SignupComponent }
 ];

如何用逗号分隔嵌套列表的所有元素？

4 个答案: