Re: depicting data in c# in realtime - C# Discussion Boards

depicting data in c# in realtime

30-Jan-14 9:06

yes. I' work around.
Richard Andrew x64,thanks for your attentions.

khomeyni29-Jan-14 2:38

29-Jan-14 2:38

hello

I want to depict a chart dynamically and every 3 millisecond a point is send to the chart .

when I run my program about 8sec it will be executed very more than 8sec( about 3 mins or more).(i used wpftoolkit charting, dynamic data display , ... but none of them works well).

also when I want to depict 6000 point in offline mode(storing data in an array and at last depicting it on a chart) ,yet there is a such problem.

Matlab depicts these data very rapidly .

is there any way to depict these data in c#?

Eddy Vluggen29-Jan-14 2:59

Eddy Vluggen

29-Jan-14 2:59

Matlab doesn't depict them in three milliseconds either. It may be quick, but it's not realtime. There's no realtime software under Windows, as Windows isn't a realtime OS. It's the OS that determines how many processortime you get.

khomeyni wrote:
is there any way to depict these data in c#?

Yes; and can be relatively quick too. Draw your graph once, and instead of redrawing when a new datapoint appears, scroll everything a single pixel and add the new point to the right.

Having a complete table in memory and charting the complete table every iteration will never be fast - not even if you use assembly.

Bastard Programmer from Hell Suspicious | :suss:

If you can't read my code, try converting it here[^]

khomeyni29-Jan-14 6:43

29-Jan-14 6:43

I've this simple code but it takes about 20sec to be depicted!

private void button1_Click(object sender, RoutedEventArgs e)
    {
        ObservableCollection<ChartItem> InputItems = new ObservableCollection<ChartItem>();
        for (int i = 0; i < 6000; i++)
        {
            InputItems.Add(new ChartItem(i, i));
        }
        InputLine.ItemsSource = InputItems;

    }
}
public class ChartItem : INotifyPropertyChanged
{
    public ChartItem(double t, double v)
    {
        time = t;
        myVar = v;
    }

    private double time;

    public double Time
    {
        get { return time; }
        set
        {
            time = value;
            OnPropertyChanged("Time");
        }
    }

    private double myVar;

    public double Value
    {
        get { return myVar; }
        set
        {
            myVar = value;
            OnPropertyChanged("Value");
        }
    }

    public event PropertyChangedEventHandler PropertyChanged;

    protected virtual void OnPropertyChanged(string propertyName)
    {
        if (this.PropertyChanged != null)
        {
            this.PropertyChanged(this, new PropertyChangedEventArgs(propertyName));
        }
    }
}

with using :

System.Windows.Controls.DataVisualization.Toolkit

and

System.Windows.Controls.DataVisualization.Toolkit

Eddy Vluggen29-Jan-14 8:01

Eddy Vluggen

29-Jan-14 8:01

If you're "looking" for faster code - you won't find it. It's something that needs to be written.

You plot the graph yourself, and update the graph with new info instead of redrawing it completely.

Human eyes aren't realtime either; you'd need 24 frames/second to get it to look convincingly realtime. Although Windows might delay your app (one a single-core machine, a virusscanner and some automatic updates would do the trick), it should be doable to update a picture in a reasonable time.

Bastard Programmer from Hell Suspicious | :suss:

If you can't read my code, try converting it here[^]

khomeyni30-Jan-14 4:26

Need help on chat between two computers

30-Jan-14 4:26

there is one another question: why can Matlab depict these charts so faster than c# ?

larsp77729-Jan-14 1:20

29-Jan-14 1:20

I have used this tutorial to make a chat:

http://www.youtube.com/watch?v=BDVfpPq3weo[^]

It works on the same computer but not between two.

The problem seem so be that I need more than one thread.

I get Cross-thread operation not valid.

OriginalGriff29-Jan-14 2:32

29-Jan-14 2:32

No, the problem is the reverse of that: "Cross-thread operation not valid" means that you are executing code on one thread that can only be executed on a different thread: normally, this occurs when you try to update a control from a different thread to that from which it was created (which must be the UI thread), either in a BackgoundWorker, a Thread instance or in the handler of a communications class that uses threading to handle it's events (the SerialPort does this for example).

Check your code: you may just have to start invoking the control instead of accessing it directly. For example:

private void AddNewTab(string tabName)
    {
    if (InvokeRequired)
        {
        Invoke(new MethodInvoker(delegate { AddNewTab(tabName); }));
        }
    else
        {
        TabPage tp = new TabPage(tabName);
        myTabControl.TabPages.Add(tp);
        }
    }

Those who fail to learn history are doomed to repeat it. --- George Santayana (December 16, 1863 – September 26, 1952)
Those who fail to clear history are doomed to explain it. --- OriginalGriff (February 24, 1959 – ∞)

larsp77729-Jan-14 2:43

29-Jan-14 2:43

Not sure I understand...Could you show how to change my code?

This is my code:

private void Form1_Load(object sender, EventArgs e)
       {
           //Set up socket
           sck = new Socket(AddressFamily.InterNetwork, SocketType.Dgram, ProtocolType.Udp);
           sck.SetSocketOption(SocketOptionLevel.Socket, SocketOptionName.ReuseAddress, true);

           //Get user IP
           txtLocalIP.Text = GetLocalIP();
           txtRemoteIP.Text = GetLocalIP();
       }


       private string GetLocalIP()
       {
           IPHostEntry host;
           host = Dns.GetHostEntry(Dns.GetHostName());
           foreach (IPAddress ip in host.AddressList)
           {
               if (ip.AddressFamily == AddressFamily.InterNetwork)
                   return ip.ToString();
           }
           return "127.0.0.1";
       }

       private void MessageCallBack(IAsyncResult aResult)
       {
           try
           {
               byte[] receivedData = new byte[1500];
               receivedData = (byte[])aResult.AsyncState;

               //Converting byte{] to string
               ASCIIEncoding aEncoding = new ASCIIEncoding();
               string receivedMessage = aEncoding.GetString(receivedData);

               //Adding this messegage into ListBox
               listMessage.Items.Add("Friend: " + receivedMessage);

               buffer = new byte[1500];
               sck.BeginReceiveFrom(buffer, 0, buffer.Length, SocketFlags.None, ref epRemote, new AsyncCallback(MessageCallBack), buffer);
           }
           catch (Exception ex)
           {
               MessageBox.Show(ex.ToString());
           }
       }


       private void btnConnect_Click(object sender, EventArgs e)
       {
           //Binding socket
           epLocal = new IPEndPoint(IPAddress.Parse(txtLocalIP.Text), Convert.ToInt32(txtLocalPort.Text));
           sck.Bind(epLocal);

           //Connecting to remote IP
           epRemote = new IPEndPoint(IPAddress.Parse(txtRemoteIP.Text), Convert.ToInt32(txtRemotePort.Text));
           sck.Connect(epRemote);

           //Listen to specific port
           buffer = new byte[1500];
           sck.BeginReceiveFrom(buffer, 0, buffer.Length, SocketFlags.None, ref epRemote, new AsyncCallback(MessageCallBack), buffer);
       }

       private void btnSend_Click(object sender, EventArgs e)
       {
           //Convert string message to byte[]
           ASCIIEncoding aEncoding = new ASCIIEncoding();
           byte[] sendingMessage = new byte[1500];
           sendingMessage = aEncoding.GetBytes(txtMessage.Text);

           //Sending the Encoded message
           sck.Send(sendingMessage);

           //Adding to the ListBox
           listMessage.Items.Add("Me: " + txtMessage.Text);
           txtMessage.Text = "";
       }

OriginalGriff29-Jan-14 2:54

29-Jan-14 2:54

Where do you get the error? Which line?

larsp77729-Jan-14 6:13

29-Jan-14 6:13

Ok... The other guy whose thread I can´t se now wrote:

I'm guessing that's the line that's causing the problem. The callback method will be called on a background thread; you will need to use
InvokeRequired
and
Invoke
to get back to the UI thread when you want to access the form's controls.
Invoke((Action<object>)listMessage.Items.Add, "Friend: " + receivedMessage);

I Think he was right about the line. Tried to do change the line accordingly but got an error:

int System.Windows.Forms.ListBox.ObjectCollection.Add(object)' has the wrong return type

Richard Deeming29-Jan-14 7:22

Richard Deeming

29-Jan-14 7:22

larsp777 wrote:
int System.Windows.Forms.ListBox.ObjectCollection.Add(object)' has the wrong return type

Whoops - didn't notice that the method returned an int. The delegate type will need to be a Func<object, int> instead of an Action<object>.

Invoke((Func<object, int>)listMessage.Items.Add, "Friend: " + receivedMessage);

"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer

OriginalGriff29-Jan-14 8:01

29-Jan-14 8:01

larsp77729-Jan-14 10:28

29-Jan-14 10:28

Thanks, Richard, it worked!

Well, thank you both for helping me out!

larsp77730-Jan-14 1:48

30-Jan-14 1:48

Do you know if there is a simple way for a third person can listen in what the others are Writing. (only has to listen to the sender)

I am making an application where you should be able to simulate someone eavesdropping the conversation.

Richard Deeming30-Jan-14 2:02

Richard Deeming

30-Jan-14 2:02

If all the computers are on the same LAN, you could use the broadcast address so that any computer can pick up the messages:
Broadcasting Using Socket-Oriented Approach[^]

If they're on different networks, or you don't want the overhead associated with broadcasting, then you'll need to use multicasting:
IP Multicasting in C#[^]

"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer

larsp77730-Jan-14 2:06

30-Jan-14 2:06

Thanks! It will probebly be on the same LAN.

larsp77729-Jan-14 6:15

29-Jan-14 6:15

The error occurs when I try to send a message. It occurs on both computers.

When I send from one computer the error shows on the other.

Richard Deeming29-Jan-14 3:19

Richard Deeming

29-Jan-14 3:19

larsp777 wrote:
listMessage.Items.Add("Friend: " + receivedMessage);

I'm guessing that's the line that's causing the problem. The callback method will be called on a background thread; you will need to use InvokeRequired and Invoke to get back to the UI thread when you want to access the form's controls.

Invoke((Func<object, int>)listMessage.Items.Add, "Friend: " + receivedMessage);

"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer

larsp77729-Jan-14 5:07

29-Jan-14 5:07

Thank you both!

I think that is the problem. Will check it out and get back.

Rahul VB29-Jan-14 20:39

Rahul VB

29-Jan-14 20:39

hey OG, please enlighten me more on this.

OriginalGriff29-Jan-14 22:40

29-Jan-14 22:40

Sorry Rahul, but it's a huge subject - I couldn't begin to do it justice in a small text box! Laugh | :laugh:

There are some good tutorials out there which explain threading and the UI pretty well:
http://stuff.seans.com/2009/05/21/net-basics-do-work-in-background-thread-to-keep-gui-responsive/[^] (Backgound and why to thread)

http://www.dreamincode.net/forums/topic/246911-c%23-multi-threading-in-a-gui-environment/[^] (Fairly advanced)

But basically when you start to use multiple threads you can't touch any controls, except from the thread that created them - which is called the UI thread (for User Interface) and is the original thread the form started on. If you try, you will get a "cross-threading" error telling you not to do that. The only way to get round it is to Invoke the control - which basically requests the UI thread to do the work for you.

Have a look at the BackgroundWorker thread - it provides a way to update the display without invoking via the ProgressChanged event, which is executed on the original thread.

larsp77729-Jan-14 10:35