The docs say that calling CTTypesetterCreateLine is the same as calling CTTypesetterCreateLineWithOffset with offset set to 0.0, but the description of what offset means is rather lacking: "The line position offset."
I've tried providing different values to it and it doesn't seem to have any impact on the typographic bounds or image bounds of the resulting CTLineRef, nor does it seem to affect the result of drawing the line using CTLineDraw. Can anyone clue me in as to the purpose of this extra parameter?
The offset is a tab offset. It doesn't apply to the line as a whole, but to the first tab-stop.
From http://lists.apple.com/archives/Coretext-dev/2011/Feb/msg00021.html
You create a string containing a tab like "A[tab]B" with Tab position at 200.
When you create a line with offset Zero and draw it at (x, y), it will appear like this.
A B
(x,y) (x+200,y)
When you create a line with offset 50 and draw it at (x + 50, y) (← you need to adjust X coordinate yourself), it will appear like this,
A B
(x+50,y) (x+200,y)
Note that "B" remains at the same position even though the line starts at a different position. If you were passing offset 0 and draw it at (x + 50, y), it would have been like the following.
A B
(x+50,y) (x+250,y)
Related
Write a function called connectTheDots that takes in a list of tuples as its input and an optional color input as well. The default color value should be black. Each tuple is a coordinate pair (x, y) for a turtle. The function will have the turtle trace out a picture by starting at the first coordinate and then moving to each coordinate in turn.
Your function should do the following:
a. Create a turtle, setting the turtle’s color and speed appropriately
b. Check if the input list is empty: if it is empty then nothing else should happen!
c. Without leaving a line behind, move the turtle to the first location given in the list. Then start leaving a line again. Note: recall how to pull values out of a list, and also know that the goto method can take a single (x, y) tuple as its input: myTurtle.goto( (25, 25) ) will move myTurtle to x = 25 and y = 25.
d. After the turtle is at the starting coordinate, move it to each coordinate in the list in turn.
This is what I have been able to do so far:
def connectTheDots(list1, color ="black"):
myTurtle = turtle.Turtle()
myTurtle.speed(1)
myTurtle.goto(list1[0])
for x,y in list1[1:]: #I'm unsure if this is correct
myTurtle.goto(x,y)
You have most of what you need but are probably making it more complicated than needed and are missing some small details.
For step "a" you need to explicitly set the color (you passed it in just fine). You are probably better off using a symbolic speed instead of a numeric one.
For step "b", if you have a proper for ... in loop, you don't need to explicitly check if the list is empty as the loop won't run if it is. Your splitting off the first item myTurtle.goto(list1[0]) works against you here as there may not be one, causing an IndexError.
For step "c" you need to add another command. Turtles start life in the center of the screen with their pens down. You need to raise the pen up after creating your turtle. But you don't need to explicitly move to the starting position, let your loop handle that.
The trick we'll use for step "c" and step "d" is to put the pen down after the goto() in the loop. The first time, this actually puts the pen down, after that, it's a harmless no-op:
import turtle
def connectTheDots(coordinates, color="black"):
myTurtle = turtle.Turtle()
myTurtle.speed("slowest")
myTurtle.color(color)
myTurtle.penup()
for coordinate in coordinates:
myTurtle.goto(coordinate)
myTurtle.pendown() # redundant after first iteration
dots = ((34, 56), (100, 240), (230, 105), (34, 56))
connectTheDots(dots, "green")
turtle.done()
If it bothers you that we're putting the pen down unnecessarily in the loop, then we can replace myTurtle.pendown() with:
if not myTurtle.isdown():
myTurtle.pendown()
This may have been asked before, but for my lack of correct English terms end up leaving me here. (I'm Finnish) This may be asked before, but what else could I have done?
But I have pygame code, which renders partion of bigger 'map'. I want to have behaviour to 'click' a squre and 'select' it.
The broblem is, how do I find the index of image I am currently overlapping with mouse?
Codelike close to what I have now
#...setup code...
map = [[0,0,0,0], [0,1,0,0], [0,0,0,0]]
while:
render()
#render completely fills the screen with images based on map's objects
mousepos=pyagem.mouse.get_pos()
selectedMapSquare=???
You just have to divide the absolute (screen) coordinates with the size of your squares. So, if the size of your squares is e.g. 32, you can use something like
x, y = pygame.mouse.get_pos()
# TODO: use a constant
w_x, w_y = x / 32, y /32
Now w_x is the index of the x axis, and w_y is the index of the y axis:
# TODO: bound/error checking
tile_under_mouse = map[w_y][w_x]
I have a weird issue in my bar graph realized using d3.js: the 1 px padding between each rectangle appears irregular. I gather either or both the width or x position are the culprit but i don't understand what i'm doing wrong: the width is a fraction of the svg area and the X position is obtained via a D3 scale.
I've put a demo here: http://jsfiddle.net/pixeline/j679N/4/
The code ( a scale) controling the x position:
var xScale = d3.time.scale().domain([minDate, maxDate]).rangeRound([padding, w - padding]);
The code controlling the width:
var barWidth = Math.floor((w/dataset.length))-barPadding;
Thank you for your insight.
It's irregular because you are rounding your output range (rangeRound). In some cases, the distance between two bars is 3 pixels and sometimes only 2. This is because the actual x position is a fractional value and ends up being rounded one way in some cases and the other way on other cases.
You can mitigate the effect but changing rangeRound to range, but that won't eliminate it entirely as you'll still get fractional pixel values for positions. The best thing to do is probably to simply increase the padding so that the differences aren't as obvious.
please , see following image, here you can see blue rectangle is custom shape bounds and custom shape is shoe , i want to find area of a portion written in image and i want that area in form of rectangle
do is there any path iterator concept ?
Note
custom shape i derived from image of the same size.
I would do it like this:
1.create table for all bounding box-rect perimeter lines
each value in it will represent the empty space length form border line to shape
something like this:
the values are found by simple image scanning until first non space color found
2.now bruteforce find the biggest rectangle area
x,y = top left corner
for xs = 1 to bounding box width
now scan the max valid height of rectangle from x to x + xs (x grows to the right)
// it should be the min y0[x..x+xs]
remember the biggest valid area/size combination
do this for all 4 combinations (star from the other corners)
I now Brute-force is slow but
you can divide perimeter lines not by pixels but with some step instead
also I am sure this can be optimized somehow
for example by derivation of perimeter find the extremes and check from them backwards
when the size will start shrinking then stop ...
of course take in mind that on complicated shapes this optimization will not work ...
EDIT - Thanks for all the answers everyone. I think I accidentally led you slightly wrong as the square in the picture below should be a rectangle (I see most of you are referencing squares which seems like it would make my life a lot easier). Also, the x/y lines could go in any direction, so the red dot won't always be at the top y boundary. I was originally going for a y = mx + b solution, but then I got stuck trying to figure out how I know whether to plug in the x or the y (one of them has to be known, obviously).
I have a very simple question (I think) that I'm currently struggling with for some reason. I'm trying to have a type of minimap in my game which shows symbols around the perimeter of the view, pointing towards objectives off-screen.
Anyway, I'm trying to find the value of the red point (while the black borders and everything in green is known):
It seems like simple trigonometry, but for some reason I can't wrap my head around it. I just need to find the "new" x value from the green point to the red point, then I can utilize basic math to get the red point, but how I go about finding that new x is puzzling me.
Thanks in advance!
scale = max(abs(x), abs(y))
x = x / scale
y = y / scale
This is the simple case, for a square from (-1, -1) to (1, 1). If you want a different sized square, multiply the coordinates by sidelen / 2.
If you want a rectangle instead of a square, use the following formula. (This is another solution to the arbitrarily-sized square version)
scale = max(abs(x) / (width / 2), abs(y) / (height / 2))
x = x / scale
y = y / scale
Let's call the length of one side of the square l. The slope of the line is -y/x. That means, if you move along the line and rise a distance y toward the top of the square, then you'll move a distance x to the left. But since the green point is at the center of the square, you can rise only l/2. You can express this as a ratio:
-y -l/2
——— = ———
x d
Where d is the distance you'll move to the left. Solving for d, we have
d = xl/2y
So if the green dot is at (0, 0), the red dot is at (-l/2, xl/2y).
All you need is the angle and the width of the square w.
If the green dot is at (0,0), then the angle is a = atan(y/x), the y-coordinate of the dot is w/2, and therefore the x-coordinate of the dot is tan(1/a) * (w/2). Note that tan(1/a) == pi/2 - tan(a), or in other words the angle you really want to plug into tan is the one outside the box.
Edit: yes, this can be done without trig, too. All you need is to interpolate the x-coordinate of the dot on the line. So you know the y-coordinate is w/2, then the x-coordinate is (w/2) * x/y. But, be careful which quadrant of the square you're working with. That formula is only valid for -y<x<y, otherwise you want to reverse x and y.