XL2002: XY Chart and Image Position...

T

Trevor Williams

Hi All,

I have a simple XY graph, plotting 1 point only. This graph is in front of
an image made up of several shapes.

What I would like to do is return the name of the shape that the plot sits
over.

Sounds simple, but where to start?

Any help appreciated.

Trevor
 
P

Peter T

Are you saying you have shapes on the worksheet, not on the chart, and
chart's Order is "on top".

Presumably ChartArea's & Plot's Fill are invisible so you can see through to
shapes and pictures, though that's not directly relevant.

Is speed important (say no!)

Regards,
Peter T
 
T

Trevor Williams

ha ha ha, thanks for the response Peter, and no, speed isn't important... (ish)

Yes, shapes exist on the worksheet, and yes the chart order is 'on top'

Regarding visibility, it will probably be the opposite - the chart will be
visible and the shapes won't.

Once the name of the shape which the plot lies within is returned, I will
use a lookup to show a specific screen.

I'll be happy to mail the image to you on a personal e-mail if you think it
will help.

Thanks again

Trevor
 
P

Peter T

Actually which object is on top and visibility is not relevant. You've got
me curious as to what you're doing with your lookup and how that relates to
a series point whose position is not easy to accurately control. If not
simple to explain I'd be pleased to look at what you're up to (my address is
disguised in the "reply to").

This macro is simplified in a number of respects but hopefully will work.
See comments about "ppp", for distribution to unknown users would require
API's. Uses brute force rather than calculating the point's position, hence
why I asked about speed. Speed could be improved by starting from top left
of the plot, though speed is better than I expected as should be fine to
"skip" pixels.

For testing I suggest make chartarea & Plot fill's invisible with the chart
"on top" so you can see what shape the point is over (you did say a single
series with single a point - right).

Sub SeriesPointOverShape()
Dim b As Boolean
Dim x As Long, y As Long, k As Long
Dim xx As Single, yy As Single
Dim elem As Long, arg1 As Long, arg2 As Long
Dim shp As Shape
Dim cht As Chart
Dim ppp As Single

' normally should get points per pixel with API's but
' but to simplify assume typical ppp for %90+ users at 0.75
ppp = 0.75

' oop pixels from top-leftt of chart until
' the series-1 is found with GetChartElement

Set cht = ActiveSheet.ChartObjects(1).Chart ' CHANGE to suit

xx = CLng(cht.Parent.Width / ppp) ' width in pixels
yy = CLng(cht.Parent.Height / ppp)

For k = 10 To 1 Step -2
'start by looping every 10th pixel to save time
For y = 1 To yy Step k
For x = 1 To xx Step k
Call cht.GetChartElement(x, y, elem, arg1, arg2)
If elem = xlSeries Then
'found the one & only series with single point, exit the loops
'(if need a particular series & point - check arg1 & arg2)
b = True
Exit For
End If
Next
If b Then Exit For
Next
If b Then Exit For
Next

If b Then
'convert chart pixel co-ord to worksheet point co-ord
xx = cht.Parent.Left + x * ppp
yy = cht.Parent.Top + y * ppp

b = False

'loop if/until our co-ord intersects a shape
For Each shp In ActiveSheet.Shapes
If shp.Name <> cht.Parent.Name Then
With shp
If xx >= .Left Then
If xx <= .Left + .Width Then
If yy >= .Top Then
If yy <= .Top + .Height Then
b = True ' got it
Exit For
End If
End If
End If
End If
End With
End If
Next

If b Then MsgBox shp.Name
End If

End Sub

Regards,
Peter T
 
T

Trevor Williams

Hi Peter - hope you had a good weekend.

The code you've sent works well - thank you...

However, the image that I'm using uses shapes whose thresholds overlap, and
the code seems to use the 'frame' of the shape, not the lines of the shape.
(hope that's clear.)

I couldn't see your disguised e-mail in your message - perhaps it's to well
disquised for me?! - I'd like to send it over to you if the offer's still
there.

Regards

Trevor
 
P

Peter T

Hi Trevor,

My address was in the "Reply to" field of my previous post, though not this
message. Here it is again, without the obvious punctuation -
pmbthornton gmail com

If by "shapes whose thresholds overlap" means overlapping shapes, then the
macro I posted would return the bottom-most of any shape that's over the
same screen position as the XY chart point. Could be adapted to return all
shapes under the point.

Not sure what this means -
" the code seems to use the 'frame' of the shape, not the lines of the
shape."

If you mean when the shape's border is close to the XY-point the macro may
be return incorrectly - indeed as posted that may well be the case, could be
a few pixels out. The reason for that is the macro assumes top-left pixel in
the chart to be same as the chart's object position (for use with
GetChartElement), but it isn't quite due to the chartobject's border.

After posting I tweaked a little and got it almost spot on, also made it a
little faster by only looping pixels in the PlotArea. FWIW I think it's a
really nice little macro that serves no useful purpose, at least that I can
think of ! So I would be interested to see your intentions with it.

Regards,
Peter T
 
J

Jon Peltier

I think "'frame' of the shape" means the rectangle circumscribing the shape,
like the square that's outlined by resizing handles when a circle is
selected.

It might be more robust to determine the coordinates of the mouse click, and
use algebraic algorithms to determine what mapped region the click occurred
within.

- Jon
 
P

Peter T

Trevor sent me his file. In effect he has a web-like grid of trapezoids and
wants to know which part of the web, ie trapezoid, his XY point is over. The
trapezoids are smaller than the rectangular shapes that contain them, it's
the parts of the rectangle that surround the trapezoids that overlap and
cause problems

Although the macro I posted can determine if the chart point is over
multiple shapes there's no way to determine which is the 'required' shape.
No way to set ZOrders in a consistent way.

Much better to dispense with the chart altogether, some trig and algorithms,
as you say, and do all with shapes. I've sent something to the OP along
those lines.

It was a great little macro (modified slightly from the original as posted).
But as it turns out it has no practical purpose, just as I thought! Or maybe
it might still have life for determining the position of any chart item
without tedious calculation from first principles.

Regards,
Peter T
 
P

Peter T

I had another look at this, based on what you suggest below and the file I
sent you off-line yesterday. The web like structure can be replicated with
series quite simply; in essence concentric octagons, 8 XY lines per series,
then XY lines intersecting equivalent 'corner' points of the hexagons. That
makes a large number of polygons, so far so good, except the polygons are
described by intersecting points in "different" series.

Would take quite a lot, I think, to work out which points in which series
describe the polygon that sits over a given XY position, with which to go on
and build a freeform over the points (or re-design an existing freeform as
required).

Regards,
Peter T
 
J

Jon Peltier

Peter -

Looking at the file you forwarded to me, there seems no need for different
colors, so you wouldn't need the freeforms. Even so, I don't think it would
be all that hard to determine the vertices for the freeforms, especially if
the shapes are regular and don't change from time to time. Or if the shapes
will be unchanging, draw them once in a reputable drawing program, and use
the image as the plot area fill of the chart.

- Jon
 
P

Peter T

Hi Jon,

I think there is a need for some means of visually indicating which "hole"
in the "web" contains the XY point.

Had another look (not sure why!) and it can all be done by highlighting the
XYLines between points in the various series that describe the 'hole'. No
shapes (freeforms) required (which would mask the main XY point if filled).

A bit fiddly to derive an algorithm to get the right points in the right
series but once done all works well.

Regards,
Peter T
 
G

gus valencia

Hi! good day! I am working in a very similar project, i have a chart (line chart, date and values of temperatures on that date) and i need to be able to put over the chart a FreeForm (the only free form in the working sheet), once the free form has being draw, and the user is happy with it's shape an position over the graph, the user will run a Macro (button) and it is expected that all points that were cover by the FreeForm (inside the Free Form perimeter) will be identified in a list.

i an relatively new on VBA, it is possible you to email me the example program taht you said it "works" here.
Thanks so much in advance!!!!
 
G

gus valencia

Hi! good day! I am working in a very similar project, i have a chart (line chart, date and values of temperatures on that date) and i need to be able to put over the chart a FreeForm (the only free form in the working sheet), once the free form has being draw, and the user is happy with it's shape an position over the graph, the user will run a Macro (button) and it is expected that all points that were cover by the FreeForm (inside the Free Form perimeter) will be identified in a list.

i an relatively new on VBA, it is possible you to email me the example program taht you said it "works" here.
Thanks so much in advance!!!!
 
P

Peter T

That's an old thread! As illustrated in the example I posted it’s possible
to work out if a point (or any/all points on the chart) is under a shape.
However the problem with respect to your objective is any Shape is simply a
rectangle, even though it may enclose some other shape such as a freeform.
If that’s something simple, same could be done but with a bit more work to
calculate the area (eg circle, triangle) within the rectangle. However with
a Freeform that’s going to be a lot more complicated, particularly if it
includes “indents”, even more so if curved lines are allowed.

A different approach would be to get the screen pixel colour over the point.
If it’s same as the Freeform’s fill colour that would indicate the point is
enclosed by the Freeform.

I’m pretty sure that should be doable, however I haven’t actually done it,
and it will take a lot more than five minutes!

Regards,
Peter T

PS similar sent in reply to the email you sent to me off-line
 

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