Question

我有一个列表"testlist"，其中包含4个子列表

[
[name1,ip1,mask1,group1],
[name2,ip2,mask2,group1],
[name3,ip3,mask3,group2],
[name4,ip4,mask4,group2]
]

我想从“测试列表”中获得以下字典

{group1:[name1,name2], group2:[name3,name4]}

我在这里有这小段代码，它从每个子列表中获取“组”元素，然后使用所获取的元素作为键来更新字典。我坚持的是如何填充这些键的值？

def test():
dic={}
testlist = [
            [name1,ip1,mask1,group1],
            [name2,ip2,mask2,group1], 
            [name3,ip3,mask3,group2],
            [name4,ip4,mask4,group2]
           ]
for each in testlist:
    dic.update{each[3]:[]}

Answer 1

考虑到// Create an explicit intent for an Activity in your app Intent intent = new Intent(this, SplashActivity.class); intent.setFlags(Intent.FLAG_ACTIVITY_NEW_TASK | Intent.FLAG_ACTIVITY_CLEAR_TASK); PendingIntent pendingIntent = PendingIntent.getActivity(this, 0, intent, 0); NotificationCompat.Builder builder = new NotificationCompat.Builder(this, CHANNEL_ID) .setSmallIcon(R.drawable.notification_icon) .setContentTitle("My notification") .setContentText("Hello World!") .setPriority(NotificationCompat.PRIORITY_DEFAULT) // Set the intent that will fire when the user taps the notification .setContentIntent(pendingIntent) .setAutoCancel(true);每个子列表中的项目都是字符串，请尝试以下方法：

testlist

以下是相同的代码：

dic = {i : [j[0] for j in testlist if i==j[3]] for i in set([k[3] for k in testlist])}

Answer 2

在列表上使用“传统”循环（假设在某处定义了name1，ip1等）：

def test():
    dic = {}
    testlist = [
        [name1, ip1, mask1, group1],
        [name2, ip2, mask2, group1],
        [name3, ip3, mask3, group2],
        [name4, ip4, mask4, group2]
    ]
    for each in testlist:
        if each[3] not in dic:
            dic[each[3]] = []

        dic[each[3]].append(each[0])

    return dic

从列表创建字典

2 个答案: