I want to get the angle of two points on a Line chart.
I know how to calculate an angle, the problem is that I need the x and y of the seriescollection.point and I have no idea how to get it.
Can someone help me with it?
EDIT:
Jean-François Corbett showed me how to get the points, I meant from top and left, and not point on the graph (on X scale and Y scale) though it can work.
I calculate it wrong. how can I calculate the angles in the picture below?
You ask how to get the (x,y) coordinates of points in a chart series. Here is how:
Dim c As Chart
Dim s As Series
Dim x As Variant
Dim y As Variant
Set c = ActiveChart
Set s = c.SeriesCollection.Item(1)
x = s.XValues
y = s.Values
EDIT As far as I can tell from the edited question, OP now wants the pixel coordinates of each point, with origin at the top left of the plot. To do so, you just need to scale by the axis width and span. The x axis is a bit tricky in the case of line plots (which I hate), because there is no min or max scale property; have to use the number of "categories" instead. The following code does this scaling:
Dim c As Chart
Dim s As Series
Dim xa As Axis
Dim ya As Axis
Dim x As Variant
Dim y As Variant
Dim i As Long
Set c = ActiveChart
Set s = c.SeriesCollection.Item(1)
Set xa = c.Axes(xlCategory)
Set ya = c.Axes(xlValue)
x = s.XValues
y = s.Values
For i = LBound(x) To UBound(x)
' Scale x by number of categories, equal to UBound(x) - LBound(x) + 1
x(i) = (i - LBound(x) + 0.5) / (UBound(x) - LBound(x) + 1) * xa.Width
' Scale y by axis span
y(i) = ya.Height - y(i) / (ya.MaximumScale - ya.MinimumScale) * ya.Height
Next i
Note that y increases along the negative y direction on the plot, since you want the origin to be at the top left.
Using this x and y, you can calculate your angle as seen on the screen.
The X and Y values are not directly accessible from the Point object, (as best as I can tell), but they represent actual values passed to the graph. Try accessing them from the worksheet where they are stored.
If that is unavailable, try Series.values, which returns an array of Y-values, and Series.XValues, which returns an array of X-values. (See MSDN Reference)
Related
Is there a way to change orientation (direction) of Hole Table axes with SolidWorks API?
I can do it manually by dragging the handles but recorded VBA macro does not contain actual changes.
This is what I would like to achieve:
Before
After
I don't have Visual Studio Tools on this PC so I cannot record a C# or VB macro and see if it contains more code. If somebody could check that on their PC I would be grateful.
I have figured it out. This time digging through SolidWorks API Help was useful.
By using HoleTable.DatumOrigin.SetAxisPoints() method it is possible to change points that define the Hole Table axes.
Important to notice is that SetAxisPoints() changes only the end points of the axis arrows (tips of the arrowheads). Start points get updated automatically.
You can get current point values with HoleTable.DatumOrigin.GetAxisPoints2() method.
Another thing to notice is that values in the hole table do not get updated automatically. They did update after I manually dragged a an axis point.
To get them update by the code set HoleTable.EnableUpdate property to False before and back to True after the call to SetAxisPoints().
Here is the code excerpt that does what I needed:
Dim ht As SldWorks.HoleTable
Dim htdo As SldWorks.DatumOrigin
Dim htdaxpts() As Double
Dim htdaxptsnew(0 To 3) As Double
Dim ystarty As Double
Dim yendx As Double
Dim yendy As Double
Dim xstartx As Double
Dim xendx As Double
Dim xendy As Double
'...
'here comes code to prepare for Hole Table insertion
'...
'insert the Hole Table
Set htann = theView.InsertHoleTable2(False, anchorx, anchory, swBOMConfigurationAnchor_BottomLeft, "A", holetemplatepath)
If Not htann Is Nothing Then
Set ht = htann.HoleTable
Set htdo = ht.DatumOrigin
'disable hole table update to get it refresh when done
ht.EnableUpdate = False
'get coordinates of the axis arrows (4 pairs of (x,y) doubles: X start(0,1), X end(2,3), Y start(4,5), Y end(6,7))
htdaxpts = htdo.GetAxisPoints2()
'take the values we use
xstartx = htdaxpts(0)
xendx = htdaxpts(2)
xendy = htdaxpts(3)
ystarty = htdaxpts(5)
yendx = htdaxpts(6)
yendy = htdaxpts(7)
'change direction only if Y arrow points up
If ystarty < yendy Then
yendy = ystarty - (yendy - ystarty)
End If
'change direction only if X arrow points left
If xstartx > xendx Then
xendx = xstartx - (xendx - xstartx)
End If
'change position only if X arrow is below Y arrow
If xendy < ystarty Then
'we can change end point only so change X end y only
xendy = xendy + (ystarty - xendy) * 2
End If
'prepare new axis points (2 pairs of (x,y) doubles: X end(0,1), Y end(2,3))
htdaxptsnew(0) = xendx
htdaxptsnew(1) = xendy
htdaxptsnew(2) = yendx
htdaxptsnew(3) = yendy
'set new axis end points
htdo.SetAxisPoints htdaxptsnew
'enable hole table update to refresh the values
ht.EnableUpdate = True
End If
I'm producing a line graph and I'm plotting points according to x and y values using points.addxy(X, Y). However the xvalues are plotted through regular intervals which I want to have labeled below the axis. My code is below:
Dim AllRaceArray() As String = GetAllRaceArray() 'Array of The race names in date order
Dim RaceIDs() As Integer = GetAllRaceID() 'Array Of The Race IDs Paralell to the above array
Dim Xnum As Integer = AllRaceArray.Length
Dim Interval As Integer = Math.Floor(Chart1.Size.Width / Xnum)
Chart1.Series.Clear()
For i = 0 To 3
If CompareSlotEmpty(i) = False Then
Chart1.Series.Add(i & CompPaddler(i).Name)
Chart1.Series(i).ChartType = SeriesChartType.Line
Chart1.Series(i).XValueType = ChartValueType.String
Chart1.Series(i).BorderWidth = 2
Chart1.Series(i).MarkerStyle = DataVisualization.Charting.MarkerStyle.Circle
Chart1.Series(i).MarkerSize = 8
For p = 0 To Xnum - 1
For q = 0 To CompPaddler(i).RacePoints.Length - 1
If CompPaddler(i).RacePoints(q).RaceID = RaceIDs(p) Then
Chart1.Series(i).Points.AddXY(Interval * p, CompPaddler(i).RacePoints(q).Points) ' this point plot creates the correct ordering
Chart1.Series(i).Points.AddXY(AllRaceArray(p), CompPaddler(i).RacePoints(q).Points) ' this point plot creates the correct labels
End If
Next
Next
End If
' EDIT
' This below Section has now been added but nothing appears below the axis at all now
For t = 0 To Chart1.Series(i).Points.Count - 1
Chart1.Series(i).Points(t).AxisLabel = AllRaceArray(t)
Next
'EDIT
Next
As annotated above One chart series creates the correct looking graph (http://tinypic.com/r/sobwbc/9) but x axis labels. Then the second plotter plots the correct labels but the ordering and scale gets messed up (http://tinypic.com/r/2w2g5s4/9).
TLDR: How to I change the xAxis Labels to strings that line up with regularly interval points?
Edit: I have added a loop to change the .axislabel but now nothing shows below the axis
I am trying to create a Sankey-diagram in Excel, and as a start to this, I am trying to create some "entry arrows" for the left part of the diagram, which will look roughly like this:
I created it by making a chevron arrow, and dragging the rightmost points of it to line up with the tip of the arrow.
Now, to do this for all the arrows I need, I want to do this programmatically, but I can't figure out if there is any way to do much with the nodes (?) of the shape. Trying to record a macro gave me nothing.
This is what I have so far, the macro aborts on the Debug.Print line, probably because the node object doesn't have a Left property :P
Sub energiInn()
Dim r As Range, c As Range
Dim lo As ListObject
Dim topp As Double, høgde As Double
Dim i As Long, farge As Long
Dim nd As Object
Set lo = Tabell.ListObjects("Energi_inn_elektrolyse")
Set r = lo.DataBodyRange
topp = 50
With SankeyDiagram.Shapes
For i = 1 To r.Rows.Count
høgde = Application.WorksheetFunction.Max(10, r.Cells(i, 2) / 50#)
With .AddShape(Type:=msoShapeChevron, Left:=50, top:=topp, Width:=200, Height:=høgde)
.Name = r.Cells(i, 1)
farge = fargekart((i - 1) Mod UBound(fargekart))
.Fill.ForeColor.RGB = RGB(farge Mod 256, (farge \ 256) Mod 256, farge \ 65536)
For Each nd In .Nodes
Debug.Print nd.Left
Next nd
End With
topp = topp + høgde
Next i
End With
Debug.Print r.Address
End Sub
Honestly, I am unsure if this can be done at all, but even if it is impossible, it would be nice to get it confirmed :)
What you're looking for is .Nodes.SetPosition. Because it's relative positioning, this can be a challenge. You need to use the objects position elements to make sure the points are moving in relation to the shape.
With .AddShape(Type:=msoShapeChevron, Left:=50, Top:=topp, Width:=200, Height:=høgde)
.Name = r.Cells(i, 1)
.Nodes.SetPosition 2, .Left + .Width, .Top
.Nodes.SetPosition 4, .Left + .Width, .Top + .Height
First argument is the node index. Next is the x position, which we want all the way to the right of the graphic, so we add the shapes position left to the width of the shape. Last is the y position, first point we want in the topmost corner, so we use the shapes top. Last point, we add the height to the top position to bring to the bottom corner.
I believe it would be more simple drawing this as free form using Shapes.BuildFreeform Method and then converting to shape using FreeformBuilder.ConvertToShape Method.
Example:
Sub drawEntryArrow()
Dim x1 As Single, y1 As Single, w As Single, h As Single
Dim oShape As Shape
x1 = 10
y1 = 10
w = 200
h = 200
With ActiveSheet.Shapes.BuildFreeform(msoEditingAuto, x1, y1)
.AddNodes msoSegmentLine, msoEditingAuto, x1 + w, y1
.AddNodes msoSegmentLine, msoEditingAuto, x1 + w, y1 + h
.AddNodes msoSegmentLine, msoEditingAuto, x1, y1 + h
.AddNodes msoSegmentLine, msoEditingAuto, x1 + w / 2, y1 + h / 2
.AddNodes msoSegmentLine, msoEditingAuto, x1, y1
Set oShape = .ConvertToShape
End With
End Sub
If you just want to get rid of the point at the right, you can simply delete the node (nodes of a chevron are counted clockwise starting at the top left):
.Nodes.Delete 3
To get access to all nodes with the nodes-property of a shape, however, as long as you deal with a standard shape type, you can't access the coordinates.
When you use the "edit points", a shape changes its type to msoShapeNotPrimitive - but I couldn't figure out how to do this using VBA.
UPDATE
Played around a bit (because I'm curious) - just as an example if someone wants to change a shape manually:
' First change Shape Type:
' WILL NOT WORK: sh.AutoShapeType = msoShapeNotPrimitive
' Instead, add a node and remove it immediately. This changes the shape type.
.Nodes.Insert c, msoSegmentLine, msoEditingCorner, 100, 100
.Nodes.Delete c + 1
' Now access the x-coordinate of node 2 and the y-coordinate of node 3
' (note that we cannot access the coordinates directly)
Dim pointsArray() As Single, x As Single, y As Single
pointsArray = .Nodes(2).Points
x = pointsArray(1, 1)
pointsArray = .Nodes(3).Points
y = pointsArray(1, 2)
' Now change the x-value of node 3
sh.Nodes.SetPosition 3, x, y
I'm trying to figure out how to code a straight line to the straight part of a curve, the curve should look something like the exponential, click the link to open the image:
Straight line to Curve and determining the x-intercept
Here is the code, I'm only using the exponential as an example
`
Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click
Dim s As String
Dim xl As New Excel.Application
Dim wb As Excel.Workbook
Dim ws As Excel.Worksheet
wb = xl.Workbooks.Add 'create new workbook
ws = wb.Worksheets(1) 'select sheet 1
ws.Activate()
Dim Currents() As Double
Dim PhotodiodeValues() As Double
Dim AppliedCurrent As Double
'AppliedCurrent = SerialPort1.ReadLine
AppliedCurrent = 0
If AppliedCurrent >= 0 And AppliedCurrent < 0.1 Then
Dim j As Integer = 1
For i As Double = 0 To 5 Step 0.5
ReDim Preserve Currents(j)
ReDim Preserve PhotodiodeValues(j)
MsgBox(i)
MsgBox("LDI " & CType(i, String))
s = ("LDI " & CType(i, String))
AppliedCurrent = i
If AppliedCurrent >= i And AppliedCurrent < (i + 0.1) Then
Currents(j) = CType(i, Double)
Label1.Text = Currents(j)
PhotodiodeValues(j) = CType(Math.E ^ (i), Double)
ws.Cells(j, 1) = Currents(j)
ws.Cells(j, 2) = PhotodiodeValues(j)
Else
System.Threading.Thread.Sleep(1000)
End If
j = j + 1
Next
Else
System.Threading.Thread.Sleep(1000)
End If
sfd1.ShowDialog() 'get file name
wb.SaveAs(sfd1.FileName) 'save data to file
wb.Close()
xl = Nothing 'dispose of excel
ScatterGraph1.PlotXY(Currents, PhotodiodeValues)
'SerialPort1.Close()
End Sub
End Class`
First off, I'll explain my thought process. If I have misunderstood, please let me know and I will update my answer. The slope dy/dx of the curve y = e^x is dy/dx = e^x, a monotonically increasing function of x for all real x. There is never a point at which the function becomes linear and, while it has a horizontal asymptote (y = 0) it has no vertical asymptote.
I take it that what you want is the equation of a tangent line taken at a point where the slope first becomes greater than some cutoff value m*. After that point, the graph of y = e^x "might as well" be a straight line for your intents and purposes.
So, we must first solve the equation m* = dy/dx = e^x for the x at which m* occurs. The range of e^x is all positive real numbers and e^x is monotonically increasing, so any positive real number m* will have a unique solution x*. indeed, x* = ln(m*). Our tangent line will pass through the point (x*, e^x*) and have slope m*. Recall that m* = e^x*, so the point is (ln(m*), m*) and the slope is m*.
With a point and the slope, we can figure out the equation of a line. We have that the slope from the given point to any other point must be m*; so, (y - y*)/(x - x*) = m*. Rearranging, (y - y*) = m*(x - x*), y = mx - mx* + y*, and finally y = mx + (y - mx) = mx + (m - mln(m)). The Y-intercept is therefore (m* - mln(m)). We can get the X-intercept by setting y = 0 and solving for x: 0 = mx + (m - mln(m)), mx = mln(m*) - m*, x = ln(m*) - 1.
In summary:
the equation of the line tangent to y = e^x with slope m* is y = mx + (m - mln(m)).
the Y-intercept of this line is (m* - mln(m)).
the X-intercept of this line is ln(m*) - 1
If the curve is known at compile time, I recommend hard-coding a closed form analytical solution for the derivative and any asymptotes. If the function is not known until runtime, the derivative at a given point can be approximated numerically using a variety of methods. Intuitively, the definition of the derivative as the limit of (f(x+d) - f(x)) / d as d approaches zero can be used to find approximations of the derivative where the derivative (likely) exists. For well-behaved analytic functions, you will typically be safe except in special cases.
If the function's derivative is monotonically non-decreasing, as in this example, you can find the point (if any) at which the function's slope meets or exceeds a certain cutoff using approximation (as above) in conjunction with something akin to binary search. Start at a value such as x = 0, and increase or decrease x by some multiplicatively increasing factor until you have passed your target. Now, with bounds on the values of x between which your target can be found, check the middle of the range, and then either the left or right half recursively until a suitably good x* is found.
I'm working with a MSchart and I'm trying to figure out how to show the x-axis value based on the mouse's position in the graph. It only shows when the mouse is on a datapoint. The y-axis is scaled linearly and works as it should.
The x-axis however doesn't since it is logarithmic and gives me results this:
(0 = 0, 10 = 1, 100 = 2, 1000 = 3)
Where x = 10, it gives me a returned value of 1, where x = 100, it gives me a returned value of 2, and etc.
dim position = e.Location
dim result = myChart.hitTest(position.x, position.y)
etoolTip.active = true
if(result.ChartElementType = DataVisualization.Charting.ChartElementType.datapoint)
dim x = result.ChartArea.AxisX.PixelPositionToValue(position.x)
dim y = result.chartarea.axisy.pixelPositionToValue(position.y)
etoolTip.SetToolTip(phaseNoiseChart, "X:"&x &"Y:"&y)
else
etoolTip.active = false
End If
How do I get the x-axis's true position based on the mouse's position for an axis that is logarithmic?
since the code belongs to mouse move event
e.X will give you the x co-ordinate
e.Y will give you the y co-ordinate
I clearly didn't understand what those numbers meant until I realized what the answer was telling me. The way I turn it back into actual coordinates is by 10^x.
dim x = result.ChartArea.AxisX.PixelPositionToValue(position.x)
x = 10^x
dim y = result.chartarea.axisy.pixelPositionToValue(position.y)
etoolTip.SetToolTip(phaseNoiseChart, "X:"&x & "Y:"&y)