查找由具有x，y，w，h像素坐標的矩形覆蓋的圖塊的坐標

Question

假設我有一個使用16x16像素的基於圖塊的系統。 您如何找出由浮點像素單位定義的矩形覆蓋的圖塊？

例如

rect(x=16.0,y=16.0, w=1.0, h=1.0) -> tile(x=1, y=1, w=1, h=1)
rect(x=16.0,y=16.0, w=16.0, h=16.0) -> tile(x=1, y=1, w=1, h=1) (still within same tile)
rect(x=24.0,y=24.0, w=8.0, y=8.0) -> (x=1,y=1,w=1,h=1) (still within same tile)
rect(x=24.0,y=24.0, w=8.1, y=8.1) -> (x=1,y=1,w=2,h=2)

我可靠地做到這一點的唯一方法是使用循環。 有沒有更好的辦法？ 除以16會給我關於邊緣情況的錯誤答案。 這是我在python中使用的一些示例代碼：

#!/usr/bin/env python

import math

TILE_W = 16
TILE_H = 16

def get_tile(x,y,w,h):
    t_x = int(x / TILE_W)
    t_x2 = t_x
    while t_x2*TILE_W < (x+w):
        t_x2 += 1
    t_w = t_x2-t_x

    t_y = int( y / TILE_H)
    t_y2 = t_y
    while t_y2*TILE_H < (y+h):
        t_y2 += 1
    t_h = t_y2-t_y

    return t_x,t_y,t_w,t_h

(x,y) = 16.0,16.0
(w,h) = 1.0, 1.0
assert get_tile(x,y,w,h) == (1,1,1,1)

(x,y) = 16.0,16.0
(w,h) = 15.0, 15.0
assert get_tile(x,y,w,h) == (1,1,1,1)

(x,y) = 16.0,16.0
(w,h) = 16.0, 16.0
assert get_tile(x,y,w,h) == (1,1,1,1)

(x,y) = 16.0,16.0
(w,h) = 16.1, 16.1
assert get_tile(x,y,w,h) == (1,1,2,2)

(x,y) = 24.0, 24.0
(w,h) = 1.0, 1.0
assert get_tile(x,y,w,h) == (1,1,1,1)

(x,y) = 24.0, 24.0
(w,h) = 8.0, 8.0
assert get_tile(x,y,w,h) == (1,1,1,1)

(x,y) = 24.0, 24.0
(w,h) = 8.1, 8.1
assert get_tile(x,y,w,h) == (1,1,2,2)

(x,y) = 24.0, 24.0
(w,h) = 9.0, 9.0
assert get_tile(x,y,w,h) == (1,1,2,2)

Answer 1

Matt的錯誤修復解決方案：

from __future__ import division
import math

TILE_W = TILE_H = 16

def get_tile(x,y,w,h):
    x1 = int(math.floor(x/TILE_W))
    x2 = int(math.ceil((x + w)/TILE_W))
    y1 = int(math.floor(y/TILE_H))
    y2 = int(math.ceil((y + h)/TILE_H))
    return x1, y1, x2-x1, y2-y1

Answer 2

這是通過您的測試用例的，告訴我是否有邊緣情況

TILE_W = TILE_H = 16

from math import floor, ceil

def get_tile2(x,y,w,h):
    x1 = int(x/TILE_W)
    y1 = int(y/TILE_H)
    x2 = int((x+w)/TILE_W)
    y2 = int((y+h)/TILE_H)
    if (x+w)%16 == 0: #edge case
        x2-=1
    if (y+h)%16 == 0: #edge case
        y2-=1
    tw = x2-x1 + 1
    th = y2-y1 + 1
    return x1, y1, tw, th

(x,y) = 16.0, 16.0
(w,h) = 1.0, 1.0
assert get_tile2(x,y,w,h) == (1,1,1,1)

(x,y) = 16.0, 16.0
(w,h) = 15.0, 15.0
assert get_tile2(x,y,w,h) == (1,1,1,1)

(x,y) = 16.0, 16.0
(w,h) = 16.0, 16.0
assert get_tile2(x,y,w,h) == (1,1,1,1)

(x,y) = 16.0, 16.0 
(w,h) = 16.1, 16.1
assert get_tile2(x,y,w,h) == (1,1,2,2)

我現在正在明確檢查邊緣情況，但是請注意，有時浮點比較似乎並不明顯，並且結果可能不符合預期。

Answer 3

您可以先將像素坐標ot整數對齊，然后再除以圖塊寬度：

xlower = int(floor(x))
xupper = int(ceil(x + w))

Answer 4

這可能會更濃縮一些，但是在這里。

def get_tile(x,y,w,h):

    x1 = int(x / TILE_W)
    x2 = (x + w) / TILE_W

    y1 = int(y / TILE_H)
    y2 = (x + w) / TILE_H

    if int(x2) == x2:
        x2 = int(x2 - 1)
    else:
        x2 = int(x2)

    if int(y2) == y2:
        y2 = int(y2 - 1)
    else:
        y2 = int(y2)

    tw = x2 - x1 + 1
    th = y2 - y1 + 1

    return x1, y1, tw, th