我在下面有C＃课程，并希望根据该人居住的地址创建不同家庭的列表。如果他们有相同的地址，他们就是家人。所以它应该有一个家庭清单，每个家庭都有一份家庭成员名单。
public class Person
{
public string firstName;
public string lastName;
public Address address;
public Person(string f, string l, string addr, string city, string state)
{
firstName = f;
lastName = l;
address= new Address(addr, city, state);
}
}
public class Family
{
public string familyAddress;//maybe I don't need this
public List<Person> personList;
public Family()
{
personList = new List<Person>();
}
public void AddPerson(Person p)
{
personList.Add(p);
}
public string FamilyAddress //maybe I dont need this
{
get
{
return familyAddress;
}
set
{
familyAddress= value;
}
}
}
public partial class Form1 : Form
{
public List<Family> familyList;
public Form1()
{
InitializeComponent();
}
private void Form1_Load(object sender, EventArgs e)
{
familyList= new List<Family>();
ReadCustomerFile("..\\Customers.txt");
}
private void ReadCustomerFile(file)
{
var lines = File.ReadAllLines(file);
var reg = new Regex("(?:^|,)(\"(?:[^\"]+|\"\")*\"|[^,]*)");
for (int i = 0; i < lines.Length; i++)
{
var data = reg.Matches(lines[i]).Cast<Match>().Select(m => m.Value).ToArray();
//data[0] contains first name
//data[1] contains last name
//data[2] contains street
//data[3] contains city
//data[4] contains state
//In here as I read the customer text file, i would
//like to add the person that has the same address into
//the same family.
//for example:
//familyList address: "800 NE Oregon St. Portland, OR" has
// "Steve Jones"
// "Sarah Jones"
// "Alisa Jones"
//familyList address: "2525 Lake Park. Salt Lake City, UT" has
// "Joey William"
// "Becky William"
// "Sam William"
}
}
}
更新＃1-基于Tempx评论：谢谢！
public partial class Form1 : Form
{
public List<Family> familyList;
public List<Person> personList;
public Form1()
{
InitializeComponent();
}
private void Form1_Load(object sender, EventArgs e)
{
familyList= new List<Family>();
personList = new List<Person>();
ReadCustomerFile("..\\Customers.txt");
}
private void ReadCustomerFile(file)
{
var lines = File.ReadAllLines(file);
var reg = new Regex("(?:^|,)(\"(?:[^\"]+|\"\")*\"|[^,]*)");
for (int i = 0; i < lines.Length; i++)
{
var data = reg.Matches(lines[i]).Cast<Match>().Select(m => m.Value).ToArray();
//data[0] contains first name
//data[1] contains last name
//data[2] contains street
//data[3] contains city
//data[4] contains state
Person person = new Person(data[0].Trim(','),
data[1].Trim(','),
data[2].Trim(','),
data[3].Trim(','),
data[4].Trim(','));
personList.Add(person);
}
//In here as I read the customer text file, i would
//like to add the person that has the same address into
//the same family.
//for example:
//familyList address: "800 NE Oregon St. Portland, OR" has
// "Steve Jones"
// "Sarah Jones"
// "Alisa Jones"
//familyList address: "2525 Lake Park. Salt Lake City, UT" has
// "Joey William"
// "Becky William"
// "Sam William"
var familyAddress = new Dictionary<Address, Family>();
Family family;
foreach (Person p in personList)
{
// Returns true if the familyAddess contains the address of the person
if (familyAddress.TryGetValue(p.address, out family))
//if such family exists add the person the family
family.AddPerson(p);
else // no family is found with the person's address
{
//create a new family
family = new Family();
//add person to the family
family.AddPerson(p);
//add the family to the familyAddress dictionary
familyAddress.Add(p.address, family);
}
}
}
}
本声明
if (familyAddress.TryGetValue(p.address, out family))
始终返回系列不存在，因此转到else语句
更新＃2-现在可以添加到家庭的人。但是现在有时人的地址在地址中有逗号和句号，即使街道名称，城市和州都相同。因此它导致Dictionary键TryGetValue（）返回不需要的结果。 TryGetValue（）是否可以忽略逗号和句点？感谢
public partial class Form1 : Form
{
public List<Family> familyList;
public List<Person> personList;
public Form1()
{
InitializeComponent();
}
private void Form1_Load(object sender, EventArgs e)
{
familyList= new List<Family>();
personList = new List<Person>();
ReadCustomerFile("..\\Customers.txt");
}
private void ReadCustomerFile(file)
{
var lines = File.ReadAllLines(file);
var reg = new Regex("(?:^|,)(\"(?:[^\"]+|\"\")*\"|[^,]*)");
for (int i = 0; i < lines.Length; i++)
{
var data = reg.Matches(lines[i]).Cast<Match>().Select(m => m.Value).ToArray();
Person person = new Person(data[0].Trim(','),
data[1].Trim(','),
data[2].Trim(','),
data[3].Trim(','),
data[4].Trim(','));
personList.Add(person);
}
var familyAddress = new Dictionary<string, Family>();
Family family;
foreach (Person p in personList)
{
// Returns true if the familyAddess contains the address of the person
if (familyAddress.TryGetValue(p.address.GetAddress(), out family))
//if such family exists add the person the family
family.AddPerson(p);
else // no family is found with the person's address
{
//create a new family
family = new Family();
//add person to the family
family.AddPerson(p);
//add the family to the familyAddress dictionary
familyAddress.Add(p.address, family);
}
}
}
}
UPDATE＃3-我决定将地址字符串格式化为正确的格式，然后再添加为Dictionary in Key。在Tempx建议之后转换为如下所示的列表：
var valueList = familyAddress.Values.ToList();
如何遍历列表？因为我尝试以下：
foreach (var h in valueList)
{
foreach(var p in h)
{
//I cannot access the Person data?
}
}
由于
答案 0 :(得分：1)
我相信您希望将此人添加到正确的家庭，并相信您可以使用词典来完成工作。
public List<Person> personList;
//a specific address points to a specific family
var familyAddress = new Dictionary<Address, Family>();
一开始我假设personList不是空的，而是familyAddress。
然后，在您的代码中，人员列表的for语句可以写为
foreach (person p in personList)
{
// Returns true if the familyAddess contains the address of the person
if (familyAddress .TryGetValue(p.Address, out family))
//if such family exists add the person the family
family.AddPerson(p);
else // no family is found with the person's address
{
//create a new family
family = new Family();
//add person to the family
family.AddPerson(p);
//add the family to the familyAddress dictionary
familyAddress.add(p.Address, family);
}
}
您可以检索家庭列表
familyAddress.Values.ToList();
PS：正如其他评论者所说，我也相信Family类familyAddress应该是Address的类型。
答案 1 :(得分：1)
如果您想使用LINQ，可以通过LINQ将StructType(StructField(A1,ArrayType(StringType,true),true), StructField(A2,StringType,true), StructField(A3,IntegerType,true),StructField(A4,ArrayType(StringType,true),true)
转换为List<Person>。这是一个扩展方法，可以做到这一点：
List<Family>然后你可以在代码中的任何地方调用你的方法......
public static class FamilyExtensions
{
public static List<Family> GroupToFamilies(this List<Person> personList)
{
var retval = from p in personList
group p by new p.address into f
select new Family() { personList = f.ToList() }; //If familyAddress was of type Address, you could add: familyAddress = f.Key
return retval.ToList();
}
}
请注意，如果您想按List<Person> people = new List<Person>();
people.Add(new Person("Papa", "Bear", "123 Forest Rd", "Somewhere", "AK"));
people.Add(new Person("Mama", "Bear", "123 Forest Rd", "Somewhere", "AK"));
people.Add(new Person("Some", "Bear", "567 Woodland Ln", "Somewhere", "AK"));
people.Add(new Person("Other", "Bear", "890 Canopy Cr", "Somewhere", "AK"));
var families = people.GroupToFamilies();
进行分组，则需要覆盖p.address课程中的GetHashCode()和Equals(object obj)。如果您不想覆盖这两种方法，则可以使用组合键进行分组：
Address如果您确实希望按from p in personList
group p by new { p.address.addr, p.address.city, p.address.state } into f
select new Family() { personList = f.ToList() };
进行分组，请覆盖Address类上的GetHashCode()和Equals(object obj)，如此...
Address