根据索引操作列表合并

本文介绍了根据索引操作列表合并的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

所以我有这2个列表列表:

So I have this 2 list of lists:

list1=[['user1', 186, 'Feb 2017, Mar 2017, Apr 2017', 550, 555], ['user2', 282, 'Mai 2017', 3579, 3579]] list2=[['user1', 186, 'Feb 2017, Mar 2017, Apr 2017, Mai 2017', 0, 740]]

^^这只是一个示例，我将在其中包含更多列表，但原理和格式将相同.

^^ That is just a example, I will have more lists inside but the principle and format will be the same.

我将说明如何格式化所需的输出:当list2 [0] [1] == list1 [0] [1](在我的示例中为186 == 186)替换list1上的所有list2 [0] [0]，但如果您没有匹配项list2 [0] [1] == list1 [0] [1](在我的示例中为user2-)，则仅保留list1 [0] [3](在我的示例中为550) 282不匹配)按原样使用该列表，仅将index [4]修改为0(将变为:['user2'，282，'Mai 2017'，3579，0])

I will explain how I would like to format my desired output: when list2[0][1] ==list1[0][1] (in my example 186==186) replace all list2[0] on list1[0] but only keep list1[0][3] (in my example 550), if you don't have a match list2[0][1] ==list1[0][1] (in my example user2 - 282 isn't matched) take that list as it is and modify only index[4] to 0 (will become like this: ['user2', 282, 'Mai 2017', 3579, 0])

我将输入所需的输出，以便您更好地理解它:

I will put my desired output so you will understand it better:

desiredlist = [['user2', 282, 'Mai 2017', 3579, 0], ['user1', 186, 'Feb 2017, Mar 2017, Apr 2017, Mai 2017', 550, 740]]

我想拥有一个可以做到这一点的功能，我开始研究它，最后我做到了:

I would like to have a function that can do that, I started to work on it, and I ended with this:

def mergesafirmacheta(safir,macheta): combined_list = macheta + safir final_dict = {tuple(i[:2]):tuple(i[2:]) for i in combined_list} merged_list = [list(k) + list (final_dict[k]) for k in final_dict] return merged_list

但是，如果我打印出所需列表= mergesafirmacheta(list2，list1)，我将会得到:

But if I print desiredlist = mergesafirmacheta(list2,list1) I will get:

[['user2', 282, 'Mai 2017', 3579, 3579], ['user1', 186, 'Feb 2017, Mar 2017, Apr 2017, Mai 2017', 0, 740]]

我如何获得所需的输出?我正在使用python 3！谢谢！

How I can get to my desired output?I'm using python 3! Thanks!

推荐答案

由于您似乎拥有结构化数据，因此我建议您创建类 User 并使用对象进行操作.

As you seem to have structured data, I would advise you to create class User and manipulate with objects.

class User: def __init__(self, nick, id_, months, a, b): self.nick = nick self.id_ = id_ self.months = months self.a = a self.b = b def __str__(self): return "{self.nick}, {self.id_}, {self.months}, {self.a}, {self.b}".format(**locals())

您将拥有更具可读性的代码.

You would have much more readable code.

list1 = [ User('user1', 186, 'Feb 2017, Mar 2017, Apr 2017', 550, 555), User('user2', 282, 'Mai 2017', 3579, 3579), ] list2 = [ User('user1', 186, 'Feb 2017, Mar 2017, Apr 2017, Mai 2017', 0, 740), ] for outdated in list1: updateds = [u for u in list2 if outdated.id_ == u.id_] if updateds: outdated.nick = updateds[0].nick outdated.months = updateds[0].months outdated.b = updateds[0].b else: outdated.b = 0 for user in list1: print(user)