Question

我尝试使用PIL /枕头的ImageDraw模块在图像上绘制粗矩形。

我尝试使用draw.rectangle([x1, y1, x2, y2], outline='yellow', width=3)，但它似乎不喜欢width参数。

我可以模仿我想用一堆线做的事情，但我想知道是否有一种正确的方法。

'''
coordinates = [(x1, y1), (x2, y2)]

    (x1, y1)
        *--------------
        |             |
        |             |
        |             |
        |             |
        |             |
        |             |
        --------------*
                      (x2, y2)

'''

def draw_rectangle(drawing, xy, outline='yellow', width=10):
    top_left = xy[0]
    bottom_right = xy[1]
    top_right = (xy[1][0], xy[0][1])
    bottom_left= (xy[0][0], xy[1][1])

    drawing.line([top_left, top_right], fill=outline, width=width)
    drawing.line([top_right, bottom_right], fill=outline, width=width)
    drawing.line([bottom_right, bottom_left], fill=outline, width=width)
    drawing.line([bottom_left, top_left], fill=outline, width=width)

Answer 1

更新 - Pillow＆gt; = 5.3.0 rectangle现在支持width参数： PIL.ImageDraw.ImageDraw.rectangle(xy, fill=None, outline=None, width=0)

上一个回答：

这是一个绘制第一个初始矩形的方法，然后是向内的其他矩形 - 注意，沿边界居中的线宽不。

def draw_rectangle(draw, coordinates, color, width=1):
    for i in range(width):
        rect_start = (coordinates[0][0] - i, coordinates[0][1] - i)
        rect_end = (coordinates[1][0] + i, coordinates[1][1] + i)
        draw.rectangle((rect_start, rect_end), outline = color)

# example usage

im = Image.open(image_path)
drawing = ImageDraw.Draw(im)

top_left = (50, 50)
bottom_right = (100, 100)

outline_width = 10
outline_color = "black"

draw_rectangle(drawing, (top_left, bottom_right), color=outline_color, width=outline_width)

Answer 2

您也可以用四点画一条线，而不是四条线，制成一个矩形：

def drawrect(drawcontext, xy, outline=None, width=0):
    (x1, y1), (x2, y2) = xy
    points = (x1, y1), (x2, y1), (x2, y2), (x1, y2), (x1, y1)
    drawcontext.line(points, fill=outline, width=width)

# example
from PIL import Image, ImageDraw
im = Image.new("RGB", (150, 150), color="white")
draw = ImageDraw.Draw(im)

drawrect(draw, [(50, 50), (100, 100)], outline="red", width=5)

im.show()

Answer 3

PIL矩形现在支持width参数。

from PIL import Image, ImageDraw

height = width = 800
img = Image.new('RGB', (height, width), (255, 255, 255))
draw = ImageDraw.Draw(img)
draw.rectangle([100,100,500,400], width = 10, outline="#0000ff")
img.show()

Answer 4

此方法适用于Ubuntu 18.04上的PIL v1.1.7和pillow v 5.3.0。角不是正方形，但是至少每个角仅偏离1/2像素，这与用宽度参数绘制4条线以创建矩形（每个角至少偏离1个像素）的方法相反。我认为枕头/枕头中的线条绘制算法中仍然存在错误。

var schoolToUpdate = schoolDbContext.Schools.Where(schoolId = 10)
   .Select(school = new
   {
       ... // you get a copy of the values: fast, but not suitable for updates
   })
   .FirstOrDefault();

，然后在您的代码中将具有：

School schoolToUpdate = schoolDbContext.Schools.Where(schoolId = 10)
   .FirstOrDefault()

感谢gdwarf的解决方案，这使我走上了通往这一道路的道路。

Answer 5

您可以绘制视图，例如：

draw.rectangle([(x, y),(x+w,y+h) ], outline=(0,0,255,255))
draw.rectangle([(x+1, y+1),(x+w-1,y+h-1) ], outline=(0,0,255,255))
draw.rectangle([(x+2, y+2),(x+w-2,y+h-2) ], outline=(0,0,255,255))
...

循环和函数中的原因

。

有没有办法在PIL中指定矩形的宽度？

5 个答案: