Equally spaced circle in UI

pavlinb

@cvp Is it possible on iPhone (split view)?

cvp

@pavlinb Sorry but I don't know. I have a very old iPhone 5s not allowed for iOS 13 nor iPadOS

cvp

@pavlinb it seems that does it exist on iPhone but hoped on iOS 14.
It is ok on my iPad mini 4 and it is not very bigger than the biggest iPhone

mikael

@cvp, you can open and unzip a zip in Pythonista.

cvp

@mikael sure. It was only to show that you can download a file (zip or not) from GitHub without needing a Pythonista script to import it, via iCloud then split view.
I know there are several ways.
We are not yet using a lot download to Files, then drag and drop to another app, like Pythonista

Personally, I use my script

AZOM

@mikael
How do I change the circle’s position (horizontally) with your code that you first gave me?

mikael

@AZOM, GridView has a pack_x argument:

g = GridView(
    count_x=1,
    pack_x=GridView.START)

Using END instead would move them to the right edge.

You can also control the placement down to the pixel by providing an additional gap argument, which in this case is essentially the distance to the edge.

See the end of this page for the documentation.

AZOM

@mikael
Yeah, thanks for all.

mikael

@adomanim, here you go. You need the vector.py from here.

import ui
import vector

chars = 'ABCDEFGH'
start_angle = 0 # First character on the right

circle_color = 'red'
char_color = 'white'
char_font = ('Apple SD Gothic Neo', 32)

diameter = min(ui.get_screen_size())/2

root = ui.View()
root.present()

pointer = vector.Vector()
pointer.magnitude = diameter/2
pointer.degrees = start_angle

for c in chars:
    label = ui.Label(
        text=c,
        text_color=char_color,
        alignment=ui.ALIGN_CENTER,
        font=char_font)
    label.center = root.bounds.center() + tuple(pointer)
    pointer.degrees += 360/len(chars)
    root.add_subview(label)
    
    
class CircleView(ui.View):
    
    def layout(self):
        self.corner_radius = self.width/2
        
circle = CircleView(
    width=diameter, height=diameter,
    border_width=1, border_color=circle_color,
    center = root.bounds.center()
)

root.add_subview(circle)
circle.send_to_back()

AZOM

This is for my friend: can someone make a 3 points generator on the circumference of a circle that has a diameter of 2 units. And with these three x, y coordinates, look if the point (0,5:0) is in the triangle made with the three points?

mikael

@AZOM, sounds like a school exercise. Wouldn’t it be more useful for your friend to spend the time on cracking it?

AZOM

@mikael

No, we are in high school, and this is not a homework or something like that, he just like maths and wants help to “verify” his answer to a question he gets somewhere on the web.

AZOM

@mikael

The only thing I am doing wrong is with rounding up numbers (I think it is my only problem) beau a se if 2 values that I compare are like at 0,000001 appart, they are not equal. Also, I am using cos and sin to try to find the three points. And my friend doesn’t program btw.

mikael

@AZOM, try

0.1 + 0.2 == 0.3

If you are not satisfied with the result, you might try to use the Decimal class, where you can also

getcontext().prec == 6

mikael

@AZOM, meanwhile, for your problem, I would use ui.Path with the three points, then hit_test with the given point. I do not know whether this would be any more accurate than the method you are using, though.

AZOM

@mikael

Right, now, I don’t have my program, but I’m going to send it at ~5h pm.

mikael

@AZOM, going to get my beauty sleep now, but here’s what I meant by the hit test:

import random
import ui
import vector

random_angle = lambda: random.random() * 360

p = ui.Path()
point = vector.Vector(0, 1)

point.degrees = random_angle()
p.move_to(*point)
print(point)
for _ in range(2):
    point.degrees = random_angle()
    p.line_to(*point)
    print(point)
p.close()
    
print('IN' if p.hit_test(0.5,0) else 'OUT')

JonB

So, are you trying to input 3 points (not all colinear) and find the circle (radius and x,y). Then want to determine if a fourth point is on the edge of the circle?

A simple method is to just subtract the center, then compute the length, then compare to the circle radius. But, as you found, you want to use abs(point_to_center_dist-circle_radius)< threshold, where you'll have to decide what is an acceptable threshold.