我们说我有这堂课:
class Person {
public int ID;
public string Name;
}
然后我有一份人员名单。
List<Person> persons = new List<Person>();
其中充满了很多随机的人。如何查询列表以获取ID最低的人?列表中的对象是随机顺序,因此ID最低的人可能不是第一个元素。我可以在没有先排序列表的情况下实现这一目标吗?
答案 0 :(得分:7)
这是没有对列表进行排序,只是迭代列表一次。
Person minIdPerson = persons[0];
foreach (var person in persons)
{
if (person.ID < minIdPerson.ID)
minIdPerson = person;
}
答案 1 :(得分:4)
您可以使用More Linq库中的MinBy
方法:
var person = persons.MinBy(x => x.ID);
如果您不能使用第三方库,您可以先获取最小ID,然后获取具有最小ID的人:
var minID = person.Min(x => x.ID);
var person = persons.First(x => x.ID == minID);
答案 2 :(得分:2)
使用LINQ的 Min 扩展方法:
persons.Min(p => p.ID)
修改强>
我的不好,前面的方法只返回最低的ID,所以如果你只想使用内置的LINQ方法,请转到:
persons.Aggregate(
(personWithMinID, currentPerson) =>
currentPerson.ID <= personWithMinID.ID ? currentPerson : personWithMinID)
答案 3 :(得分:0)
List<AnswerInfo> answerinfo;
Public void SampleFunction()
{
answerinfo = new List<AnswerInfo>();
//custom hash table storing elapsed time for all users
float LocalScoreTime = (float.Parse)((string)local.CustomProperties["elapsedTime"]);
AnswerInfo objc = new AnswerInfo();
if (CorrectAnswer)
{
foreach (PhotonPlayer _player in PhotonNetwork.otherPlayers)
{
objc.ID = _player.ID;
objc.AnsCorrect = (bool)_player.CustomProperties["RemoteAnswer"];
objc.AnsTime = (float.Parse)((string)_player.CustomProperties["elapsedTime"]);
answerinfo.Add(objc);
}
//This can work too and can be used in future
//var minID = answerinfo.Min(x => x.AnsTime);
//var person = answerinfo.First(x => x.AnsTime == minID);
AnswerInfo minTimePerson = answerinfo[0];
float minTime = 30;
foreach (AnswerInfo user in answerinfo)
{
if (user.AnsCorrect)
{
if (user.AnsTime < minTime)
{
minTime = user.AnsTime;
minTimePerson = user;
}
}
}
Debug.LogFormat("Remote User ID with correct Answer: {0} and lowest time {1}",minTimePerson.ID,minTimePerson.AnsTime);
if(LocalScoreTime < minTimePerson.AnsTime)
{
local.AddScore(1);
localplayerscore_textfield.color = Color.green;
}
}
}
[Serializable]
public class AnswerInfo
{
public bool AnsCorrect;
public float AnsTime;
public int ID;
}