String concatenation C# with List of Class

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String concatenation C# with List of Class



this may be very fundamental C# question as I only study C# myself there are things that I do not have the logic to start.



I have a class CustomerSite with string Customer get;set and string Site get;set;


CustomerSite


string Customer get;set


string Site get;set;



I create a list List<CustomerSite> listCustomerSite= new List<CustomerSite>();


List<CustomerSite> listCustomerSite= new List<CustomerSite>();



Assume, I have a list with the following data


SAMSUNG CHINA
SAMSUNG AMERICA
SAMSUNG AFRICA
LG CHINA
APPLE AMERICA
APPLE CHINA



I would like to have 1 concatenated string



string Result = "APPLE (AMERICA, CHINA), LG (CHINA), SAMSUNG (AFRICA, AMERICA, CHINA)"


string Result = "APPLE (AMERICA, CHINA), LG (CHINA), SAMSUNG (AFRICA, AMERICA, CHINA)"



How could I do that?



My idea is to use a dictionary to keep a list of distinct Customers and adding the site to the string but I still have no clue how to deal with sorting (AFRICA --> AMERICA --> CHINA)


(AFRICA --> AMERICA --> CHINA)


Dictionary<string, int> dictCustomer = new Dictionary<string, int>();
foreach (var i in listCustomerSite)

if (!dictCustomer.ContainsKey(i.Customer))

dictCustomer.Add(i.Customer, 0);
Result = Result + "," + "i.Customer" + "( i.Site) ";

else

Result.Replace(")", "," + i.Site + ")");






maybe offtopic for your question but keep in mind that a dictionary doesn't guarantee the insertion order.
– gsharp
Aug 10 at 16:45




4 Answers
4



You can utilize LINQ and String.Join to group your collection by customer, convert each grouping to a formatted string with sorted comma-separated sites, and then combine them to a single string:


String.Join


var customersWithLocations = listCustomerSite
.GroupBy(cs => cs.Customer)
.Select(g => $"g.Key (String.Join(", ", g.Select(cs => cs.Site).OrderBy(s => s)))")
.ToArray();

string result = String.Join(", ", customersWithLocations);





Just to make it clear: .ToArray() is not actually required in the first statement, because String.Join can accept IEnumerable<string> as an argument, but I personally prefer not to store IEnumerable<T> as local variables to avoid possible multiple enumeration, while combining them into a single statement would make code less readable and messy. It shouldn't notably affect performance, but you may remove .ToArray() if customersWithLocations is not used in code further.
– Yeldar Kurmangaliyev
Aug 10 at 16:19



.ToArray()


String.Join


IEnumerable<string>


IEnumerable<T>


.ToArray()


customersWithLocations





OP states "I still have no clue how to deal with sorting (AFRICA --> AMERICA --> CHINA)". This answer doesn't address the sorting requirement.
– Jon
Aug 10 at 16:21





@Jon, my bad, didn't notice it actually. I have update my answer, thanks!
– Yeldar Kurmangaliyev
Aug 10 at 16:22





I would add a .OrderBy(g => g.Key) after the GroupBy to sort the customers.
– Olivier Jacot-Descombes
Aug 10 at 16:31


.OrderBy(g => g.Key)


GroupBy





Thank you very much - works perfectly
– Secret
Aug 10 at 17:11



A one line LINQ statment can do it all:


var list = listCustomerSite
.GroupBy(c => c.Customer)
.Select(g => $"g.Key (string.Join(", ", g.Select(c => c.Site)))")
.ToList();





This actually does not fit the requirements 100%. OP wanted a single string result: "I would like to have 1 concatenated string". This is a list...
– maccettura
Aug 10 at 16:09



"I would like to have 1 concatenated string"





... nor does it address the requirement to have the output sorted alphabetically.
– Jon
Aug 10 at 16:20



This should produce the desired output, sorted as required:


public class CustomerSite

public CustomerSite(string customer, string site)

Customer = customer;
Site = site;


public string Customer get; set;

public string Site get; set;


class Program

static void Main(string args)

var listCustomerSite = new List<CustomerSite>()

new CustomerSite("SAMSUNG", "CHINA"),
new CustomerSite("SAMSUNG", "AMERICA"),
new CustomerSite("SAMSUNG", "AFRICA"),
new CustomerSite("LG", "CHINA"),
new CustomerSite("APPLE", "AMERICA"),
new CustomerSite("APPLE", "CHINA")
;

var list = from cs in listCustomerSite
group cs by cs.Customer into g
orderby g.Key
select $"g.Key (string.Join(", ", g.OrderBy(c => c.Site).Select(c => c.Site)))";

Console.WriteLine(string.Join(", ", list));
Console.ReadKey();




Output:


APPLE (AMERICA, CHINA), LG (CHINA), SAMSUNG (AFRICA, AMERICA, CHINA)



Hope this helps!



The idea of using a dictionary is good; however, I would operate in two steps:



I'm not going to use LINQ, as it will be more instructive and show the fundamental way of doing things (since you are learning C#).



I am using a SortedDictionary<TKey, TValue> as it automatically sorts the entries by the key (the customers in our case).


SortedDictionary<TKey, TValue>


var dictCustomer = new SortedDictionary<string, SortedSet<string>>();



The idea is to use the customer name as key and a sorted set of sites as value. A SortedSet<T> not only keeps the entries sorted, but also eliminates duplicates (which is not required here).


SortedSet<T>


foreach (CustomerSite customerSite in listCustomerSite)
if (dictCustomer.TryGetValue(customerSite.Customer, out var siteSet))
siteSet.Add(customerSite.Site);
else
siteSet = new SortedSet<string> customerSite.Site ;
dictCustomer.Add(customerSite.Customer, siteSet);




Now, let us build the result string:


var sb = new StringBuilder();
foreach (var keyValuePair in dictCustomer)
if (sb.Length > 0)
sb.Append(", ");

sb.Append(keyValuePair.Key).Append(" ("); // Appends the customer and " (".
bool nextSite = false;
foreach (string site in keyValuePair.Value)
if (nextSite)
sb.Append(", ");

sb.Append(site);
nextSite = true;

sb.Append(")");

string result = sb.ToString();



Note that a StringBuilder is more efficient than concatenating strings directly, since a StringBuilder uses a buffer internally which grows only as needed, while string concatenation creates a new string object at each step and involves copying an ever-growing string over and over.


StringBuilder


StringBuilder






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