I am currently trying to make a group of arcs, with text above them. This seems to be working, but for only half the cases. The other half is ending up below the arc node and is invisible.
I have tried using node.tofront() , toback() etc but it is still not changing anything.
class pieslice extends Group{
pieslice(double centerx,double centery,double segpart,double totalseg){
Text value= new Text(String.format("%.2f",worth));
segment = totalseg;
Arc innerArc=new Arc();
Arc outerArc=new Arc();
outerArc.setType(ArcType.ROUND);
innerArc.setType(ArcType.ROUND);
value.setFill(Color.WHITE);
innerArc.setStrokeWidth(1);
innerArc.setRadiusX(150.0f);
innerArc.setRadiusY(150.0f);
outerArc.setRadiusX(innerArc.getRadiusX()+10);
outerArc.setRadiusY(innerArc.getRadiusY()+10);
outerArc.setFill(Color.WHITE);
innerArc.setStartAngle((360/segment)*segpart);
outerArc.setStartAngle((360/segment)*segpart);
innerArc.setCenterX(centerx);
innerArc.setCenterY(centery);
outerArc.setCenterX(centerx);
outerArc.setCenterX(centery);
innerArc.setLength(360/segment);
outerArc.setLength(360/segment);
innerArc.setStrokeWidth(1);
innerArc.setStroke(Color.BLACK);
innerArc.setFill(Color.color(Math.random(),Math.random(),Math.random()));
value.setX(150);
value.setFill(Color.BLACK);
value.getTransforms().add(new Rotate((360/segment)*segpart+((360/segment)/2),0,0));
System.out.println((360/segment)*segpart+((360/segment)/2));
this.getChildren().add(outerArc);
this.getChildren().get(0).setViewOrder(2);
this.getChildren().add(innerArc);
this.getChildren().add(value);}
I would expect that since I am adding the two arcs (Inner and outer for aesthetic effect only) and then the text, that the text would be rendered above the shapes, but that is not the case. Any ideas?
Related
TLDR: How to find boxes that are lined up horizontally
Given I have the data from an image like this:
We can visually see that we have two lines:
Tare: 11700 kg 10:40:58 am 16-May
Gross: 21300 kg 12:49:34 pm 9-Aug
The data I have for each blue box shown in the image is:
Top
Left
Width
Height
Coordinates for each corner of the box (X, Y)
My main thought is to start from the top of my "grid" and loop through each value of y, and then group boxes where they share the largest amount of matching "y" values, but it seems very over the top for something that seems simple.
Unsure really where to go from here
Example data set
I was able to get boxes lined up using this bit of code (in JavaScript), it essentially finds the first "most top left" box, and then finds any boxes that "intersect" with a line that starts from the middle of that first box
We don't care what order we get the boxes in, so as long as we start with the most left on any line we are golden.
function getMostTopLeftBox(boxes) {
const sorted = boxes.slice()
.sort(
(a, b) => {
if (a.Left === b.Left) {
return a.Top < b.Top ? -1 : 1;
}
return a.Left < b.Left ? -1 : 1;
}
);
return sorted[0];
}
function getAlignedBoxesFromSet(boxes) {
const mostTopLeftBox = getMostTopLeftBox(boxes);
const line = mostTopLeftBox.Top + (mostTopLeftBox.Height / 2);
return boxes
.filter(({ Top, Height }) => Top < line && (Top + Height) > line)
.sort(({ Left: a }, { Left: b }) => a < b ? -1 : 1)
}
function getAlignedBoxes(boxes) {
let remaining = boxes;
const lines = [];
const next = () => {
const line = getAlignedBoxesFromSet(remaining);
lines.push(line);
remaining = remaining.filter(box => line.indexOf(box) === -1);
if (!remaining.length) {
return;
}
return next();
};
next();
return lines;
}
The above code with the data set provided above gives us this result
However, it doesn't account for slight angles on the boxes, for example this image:
Another example of different boxes, with senstive information removed:
You can see from the above that the values below should be considered to be on the same line:
Product: [type]
Num Of [type]: 0
[value]: [value]
I may make a new question for this, but part of the answer to this is to figure out the actual curve of a line, and not just assume that the median angle of all lines is the actual "curve" of the line, so if I was to start with the most left box, then progress to the second box, now I have two distinct lines that I would want to find the smoothed curve for, which I would then use to find the next box, as I find each box I would want to adjust this curve to find the complete line, I will investigate this one further, if anyone has any hints, please do mention it.
I've managed to solve this, with a variant of the code posted in the question.
Here is a code sandbox of the solution, I will do a full write up of this, but here it is for now: https://codesandbox.io/s/102xnl7on3
Here is an example of grouped boxes based on angled lines calculated from the angle of all horizontal lines, if all the boxes were to be straight, then the result would be the lines being straight as well, so it should work in all scenarios.
Here is also an example where the lines are straight:
You can see the lines from the box before intersecting with the next box, it does this each time until it can find a complete line of boxes (till no more line up), this works out better than using an average angle from the entire data set.
I would like to be able to generate a mathematical curve for the already found boxes and apply that to find the next box, but for now, using the previous box as the anchor works pretty well.
I was using Visjs and displaying rectanglar nodes with text. Some of the nodes can have a couple of lines of text so I added a heuristic algorithm to work out roughly where the line breaks should go to avoid very wide, single line chunks of text in very wide but very short nodes.
The trouble is, even with physics turned on, I still get overlapping nodes.
Is it possible to tell the layout engine that, under no circumstances (or physics models), should any two nodes overlap?
Well, check out the physics configuration example: as you can see, barnesHut solver has avoidOverlap property which prevents overlapping even when springConstant is equal to zero. Try this:
var options = {
"physics": {
"barnesHut": {
"springConstant": 0,
"avoidOverlap": 0.2
}
}
}
and tweak the constants to fit your needs (the example linked above is useful for that).
From its documentation:
Accepted range: [0 .. 1]. When larger than 0, the size of the node is taken into account. The distance will be calculated from the radius of the encompassing circle of the node for both the gravity model. Value 1 is maximum overlap avoidance.
To be noted, though: I've seen a recommendation to start with low values of avoidOverlap (like 0.1) to simplify the task for the solver. I can't recall where exactly I've read this.
I used levelSeparation to adjust the horizontal node distance, and nodeDistance to adjust the vertical node distance:
const options = {
layout: {
hierarchical: {
direction: 'LR',
sortMethod: 'directed',
levelSeparation: 300,
},
},
physics: {
hierarchicalRepulsion: {
nodeDistance: 140,
},
},
...
}
While setting a QGraphicsItem rotation, I get different results upon the transformation origin point while using setRotation() and using:
transform = QTransform()
transform.rotate(myAngle)
myItem.setTransform(transform)
In both portion of code, I set setTransformOriginPoint() to the same point.
Results are:
While using setRotation() method, the item is rotated upon its transformation origin point.
While using the QTransform object, the item is rotated upon item's origin, that is, point (0,0).
My code is more complex than that, but I think It applies the same. The QGraphicsItem is in fact a QGraphicsItemGroup and I can check the issue adding just one item, and in my rotation procedure change the setRotation() method for the QTransform object. The latter, ignores the setTransformOriginPoint().
I'm having this issue for a while, and I dig a lot of resources. I browse the Qt C++ code, and I can see that the setRotation() method modifies a field calles rotation (a real value) in the TransformData structure within the QGraphicsItem. The origin point is also a two field real value in such a structure called xOrigin and yOrigin respectively. The transformation is stored in the tranform field. All this information is used in a variable called: transformData.
So, I don't get why the transformation set in the transformData->transform field is ignoring the values transformData->xOrigin and transformData->yOrigin at the time of being applied.
The code I used to test that issue is the following relevant part (I have an rotate item that receives mouse inputs and applies rotation to the item itself):
# This method using QTransform object....
def mouseMoveEvent(self, event):
if self.pressed:
parent = self.parentItem()
parentPos = parent.boundingRect().center()
newPoint = event.scenePos()
iNumber = (newPoint.x()-parentPos.x())-((newPoint.y()-parentPos.y()))*1j
angle = cmath.phase(iNumber)+1.5*math.pi
self.appliedRotation = (360-math.degrees(angle))%360 - self.angleOffset
transform = QTransform()
transform.rotate(self.appliedRotation)
self.parentItem().setTransform(transform)
# ...Against this one using setRotation()
def mouseMoveEvent(self, event):
if self.pressed:
parent = self.parentItem()
parentPos = parent.boundingRect().center()
newPoint = event.scenePos()
iNumber = (newPoint.x()-parentPos.x())-((newPoint.y()-parentPos.y()))*1j
angle = cmath.phase(iNumber)+1.5*math.pi
self.appliedRotation = (360-math.degrees(angle))%360 - self.angleOffset
self.parentItem().setRotation(self.appliedRotation)
On both, previously the setTransformOriginPoint() is set, but it's not a relevant part to show the code, but just to know that it is done.
I'm getting frustrated to not find a solution to it. As it seems so straightforward, why setting a rotation transformation matrix does not use the transformation origin point that I have set and while using setRotation() method works fine? That question took me to the source code, but now is more confusing as rotation is keeping separated from the transformation applied...
I was solving the same problem. I found out that QGraphicsItem::setTransformOriginPoint() is accepted only for QGraphicsItem::setRotation(). It is ignored for QGraphicsItem::setTransform().
I use this code to reach the same behavior for QTransform():
transform = QtGui.QTransform()
centerX = item.boundingRect().width()/2
centerY = item.boundingRect().height()/2
transform.translate( centerX , centerY )
transform.rotate( -rotation )
transform.translate( -centerX , -centerY )
item.setTransform( transform )
I have a chart with a DateTime axis as my horizontal and a Linear Axis for my vertical inside a Adobe Flex Line Chart. I want to use a Cartesian Data Canvas as a background element and draw custom set of background graphics mostly rectangles. When I have more than a single data point, the graphics work perfectly since they are supposed to span the width of the entire chart.
When I have only a single data point, however, I can't seem to get the rectangles to draw. Since I want my rectangles to span the entire width of the chart, I was thinking that I could get the x-coordinates from my axis, but this isn't working.
var canvasWidth:Number = chtCanvas.width;
var canvasHeight:Number = chtCanvas.height;
var minPt:Array;
var maxPt:Array;
var minPtDate:Date;
var maxPtDate:Date;
var minPtComplete:Point;
var maxPtComplete:Point;
// This works fine when there is more than 1 data point
minPt = chtCanvas.localToData(new Point(0, 0));
maxPt = chtCanvas.localToData(new Point(canvasWidth,canvasHeight));
//This does return a date object, but wont draw below
minPtDate = axisDate.minimum;
maxPtDate = axisDate.maximum;
//This returns NaN for the x
minPtComplete = chtCanvas.dataToLocal(minPtDate, axisSalary.minimum);
maxPtComplete = chtCanvas.dataToLocal(maxPtDate, axisSalary.maximum);
// Also tried this. Also returns NaN for the x value
//minPtComplete = chtCanvas.dataToLocal(axisDate.minimum, axisSalary.minumum);
//maxPtComplete = chtCanvas.dataToLocal(axisDate.maximum, axisSalary.maximum);
My actual drawing method is as follows:
// Tried this, works with points >2, does not draw with single data point
chtCanvas.drawRect(minPt[0], detail[i].MaxValue, maxPt[0], detail[i].MinValue);
//tried this, no effect with single point
//chtCanvas.drawRect(minPtDate, detail[i].MaxValue, maxPtDate, detail[i].MinValue);
// Tried this, no effect with single point
//chtCanvas.drawRect(minPtDate, minPt[1], maxPtDate, detail[i].MinValue);
// Tried this also
//chtCanvas.drawRect(minPtComplete.x, detail[i].MaxValue, maxPtComplete.x, detail[i].MinValue);
In this example, detail is an array collection of salary values and Im using the data value in the array to determine the vertical bounds of my rectangles.
I need to draw the rectangles the entire width of the chart (even when there is only a single data point). Thanks
Thanks to Heikki for his help. The following code works to use the axis values to draw on your Cartesian Data Canvas:
chtCanvas.drawRect(axisDate.minimum as Date, axisSalary.maximum, axisDate.maximum as Date, axisSalary.minimum);
Casting the values as Date really helped. The rest of the code used above is unecessary.
One thing to note, I was using a DateFormatter to format the date values from my data. What I didn't consider was that when using a DateTimeAxis, Flex will automatically add in extra dates to display on the axis. In my case, I was using a custom parse function to create MY points, but wasnt considering the points Flex was creating and also passing to my parse function (Therefore, they were not getting parsed correctly). Once I corrected this, the values laid out correctly in the case of multiple data points. I'm still having a bit of an issue with single data points and them not filling the chart entirely, but they are now drawing.
UPDATE:
Although there are signs of life, the minimum and maximum are still not drawing the entire width of the chart in some cases depending on the dataUnits and labelUnits combination.
UPDATE #2: SOLVED
Ok, so the axis does work as minimum/maximum values for the Cartesian Data Canvas but there is something important to remember. For a single point (and probably for multiple points as well, I just couldnt visually see the difference), when using a custom DateTimeAxis parse function such as what was in the Adobe Flex ASDoc tutorials:
private function axisDateParseFunction(item:String):Date
{
var inputDate:String = item;
inputDate = fmtDate.format(inputDate);
var newDate:Date = new Date();
if(inputDate)
{
var a:Array = inputDate.split('/');
newDate.fullYear = a[2];
newDate.month = a[0] - 1;
newDate.date = a[1];
newDate.hours = 0;
newDate.hoursUTC = 0;
newDate.minutes = 0;
newDate.minutesUTC = 0;
newDate.seconds = 0;
newDate.secondsUTC = 0;
newDate.milliseconds = 0;
newDate.millisecondsUTC = 0;
}
return newDate;
}
You MUST remember to set the UTC values as shown above also. Since the DateTimeAxis uses date AND time, when you create new Date objects, their time values also get set to the local system time. Remember to set those values to zero also or you will get points that dont exactly line up with your axis labels.
Using Flex 4 Builder
Is it possible to draw 2 rectangular shapes of "Box A" and "Box B" and place them apart, next, adding a magnetic line (black line) between them which will keep them connected without having to manually update the line xy position?
It depends on what you mean by manually, practically your black line should be drawn between two points defined by BoxA & BoxB coordinates, anytime you move either of the boxes, you should call a method that will refresh your line.
As long as your points are referenced to BoxA & BoxB positions, refreshing the line is only a matter of recalling the method you've used to draw it.
//Pseudo Code
define BoxA position
define BoxB position
define PointA PointA = new Point( BoxA.centerX , BoxA.centerY)
define PointB PointB = new Point( BoxB.centerX , BoxB.centerY)
define drawLine method // draw line between PointA & PointB
drawLine();
move( BoxB ); //will change the value of PointB
drawLine();