基于索引操作列表合并

Question

我有两张单子：

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，我会有更多的清单内，但原则和格式将是相同的。你知道吗

我将解释如何格式化我想要的输出：当list2[0][1]==list1[0][1]（在我的示例中186==186）替换list1[0]上的所有list2[0]，但只保留list1[0][3]（在我的示例550中），如果没有匹配的list2[0][1]==list1[0][1]（在我的示例中user2-282不匹配），则将该列表保持原样，只将索引[4]修改为0（将变成这样：['user2'，282，'Mai 2017'，3579，0]）

我会把我想要的输出，让你更好地理解它：

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

我想有一个函数可以做到这一点，我开始研究它，并以以下内容结束：

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desiredlist=mergesafirmacheta（list2，list1），我将得到：

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

我怎样才能得到我想要的结果？我在用Python3！谢谢！你知道吗

Answer 1

从这个示例中，我猜您将受益于使用namedtuples，但是解决方案是基于列表理解的：

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，您可以得到如下结果：

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个记录的内容，它将使用namedtuples节省空间，并帮助您保持记录的可读性。否则，您将不得不使用可变容器，无论是dict还是list

Answer 2

由于您似乎拥有结构化数据，我建议您创建classUser并使用对象进行操作。你知道吗

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())

你会有更可读的代码。你知道吗

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)