简体   繁体   English

元组列表中的列表理解

[英]list comprehension within list of tuple

my problem is something like this:我的问题是这样的:

a=[(0,0,'customer',["Hi, I'm user"]),
 (0,1,'agent',['Hi Welcome']),
 (0,2,'customer',["i would like to know"]),
 (0, 3, 'agent', ['Yes']),
 (0, 3, 'agent', ['Only credit']),
 (0, 4, 'customer', ['oic...']),
 (0, 4, 'customer', ['sub line?']),
 (0, 4, 'customer', ['is it?']),
 (0, 5, 'agent', ['no subline']),
 (0, 6, 'customer', ['oic...']),
 (0, 6, 'customer', ['by']),
 (0, 6, 'customer', ['bye'])]

Need to convert to

a=[(0,0,'customer',["Hi, I'm user"]),
 (0,1,'agent',['Hi Welcome']),
 (0, 2,'customer',["i would like to know"]),
 (0, 3, 'agent', ['Yes','Only credit']),
 (0, 4, 'customer', ['oic...','sub line?','is it?']),
 (0, 5, 'agent', ['no subline']),
 (0, 6, 'customer', ['oic...','by','bye'])]

I want to merge the list of responses based on the speaker (agent/customer).我想根据发言人(代理/客户)合并响应列表。 cant find quick logic... any help here?找不到快速逻辑...这里有什么帮助吗?

You can't change it in_place as it's a tuple, so make another copy and check for pre-existing values to append to the list.您不能在 in_place 更改它,因为它是一个元组,因此制作另一个副本并检查 append 到列表中的预先存在的值。

a=[(0,0,'customer',["Hi, I'm user"]),
 (0,1,'agent',['Hi Welcome']),
 (0,2,'customer',["i would like to know"]),
 (0, 3, 'agent', ['Yes']),
 (0, 3, 'agent', ['Only credit']),
 (0, 4, 'customer', ['oic...']),
 (0, 4, 'customer', ['sub line?']),
 (0, 4, 'customer', ['is it?']),
 (0, 5, 'agent', ['no subline']),
 (0, 6, 'customer', ['oic...']),
 (0, 6, 'customer', ['by']),
 (0, 6, 'customer', ['bye'])]

b = []
str2idx = []
idx = 0
for p in a:
  if p[:3] in str2idx:
    b[str2idx.index(p[:3])][3].append(p[3][0])
  else:
    b.append(p)
    str2idx.append(p[:3])
    idx += 1

print(b)

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM