Question

所以我有这两个清单列表：

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]]

^^这只是一个例子，我将在里面有更多的列表，但原理和格式将是相同的。

我将解释如何格式化我想要的输出：当list2 [0] [1] == list1 [0] [1]（在我的例子中186 == 186）替换list1上的所有list2 [0] [0]但只保留list1 [0] [3]（在我的例子550中），如果你没有匹配list2 [0] [1] == list1 [0] [1]（在我的例子中） user2 - 282不匹配）按原样获取该列表并仅将索引[4]修改为0（将变为如此：[＆＃39; user2＆＃39;，282，＆＃39; Mai 2017＆＃ 39;，3579,0]）

我会输出我想要的输出，以便您更好地理解它：

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

我想有一个可以做到这一点的功能，我开始研究它，我结束了这个：

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

但如果我打印desiredlist = mergesafirmacheta（list2，list1），我会得到：

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

我如何达到我想要的输出？我使用的是python 3！谢谢！

Answer 1

由于您似乎有结构化数据，我建议您创建类用户并使用对象进行操作。

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())

你会有更多可读的代码。

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for outdated in list1:
    updateds = [u for u in list2 if outdated.id_ == u.id_]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        outdated.nick = updateds[0].nick
        outdated.months = updateds[0].months
        outdated.b = updateds[0].b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        outdated.b = 0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)

Answer 2

从示例中，我猜您将使用namedtuples，但解决方案基于列表理解：

desiredlist=[list2[0][:3]+[n2,list2[0][4]] if n1==list2[0][1] 
            else [id,n1,dates,n2,0] for id,n1,dates,n2,n3 in list1]

使用namedtuples你会有类似的东西：

from collections import namedtuple
Record=namedtuple("Record","Id n1 dates n2 n3")
Example1=Record('user1', 186, 'Feb 2017, Mar 2017, Apr 2017', 550, 555)
Example2=Record('user2', 282, 'Mai 2017', 3579, 3579)
Example3=Record('user1', 186, 'Feb 2017, Mar 2017, Apr 2017, Mai 2017', 0, 740)
list1=[Example1,Example2]
# Instead of creating a list of 1 element ( list2) I would rather compare it against the namedtuple directly
list2=Example3
# now the merging would look like this
desiredlist=[Record(*list2[:3],n2=i.n2,n3=list2.n3) if i.n1==list2.n1 
        else Record(*i[:4],n3=0) for i in list1]
# now desiredlist would look like:
[Record(Id='user1', n1=186, dates='Feb 2017, Mar 2017, Apr 2017, Mai 2017', n2=550, n3=740), Record(Id='user2', n1=282, dates='Mai 2017', n2=3579, n3=0)]

如果您不需要附加或更改每条记录的内容，它将使用namedtuples节省空间，它将帮助您保持记录更具可读性。否则你将不得不使用一个可变容器，无论是dict还是list

根据索引操作列表合并

2 个答案: