我希望能够在numpy数组中绘制线条，以获取在线手写识别的离线特征。这意味着我不需要图像，但我需要在numpy数组中绘制一些位置，使其看起来像是给定大小的图像。
我希望能够指定图像大小，然后像这样绘制笔画：
import module
im = module.new_image(width=800, height=200)
im.add_stroke(from={'x': 123, 'y': 2}, to={'x': 42, 'y': 3})
im.add_stroke(from={'x': 4, 'y': 3}, to={'x': 2, 'y': 1})
features = im.get(x_min=12, x_max=15, y_min=0, y_max=111)
有没有简单的方法可以实现这个（最好是使用numpy / scipy直接实现）？
请注意，我想要灰度插值。因此，features 应该是一个取值范围在[0, 255]之间的矩阵。
感谢Joe Kington提供的答案！我正在寻找skimage.draw.line_aa。
import scipy.misc
import numpy as np
from skimage.draw import line_aa
img = np.zeros((10, 10), dtype=np.uint8)
rr, cc, val = line_aa(1, 1, 8, 4)
img[rr, cc] = val * 255
scipy.misc.imsave("out.png", img)
pip install scikit-image。 - Ben它有两种风味，带和不带linewidth。实际上前者不是后者的泛化，也与line_aa不完全相同，但对于我的目的来说它们就足够好，在图表上看起来也不错。
def naive_line(r0, c0, r1, c1):
# The algorithm below works fine if c1 >= c0 and c1-c0 >= abs(r1-r0).
# If either of these cases are violated, do some switches.
if abs(c1-c0) < abs(r1-r0):
# Switch x and y, and switch again when returning.
xx, yy, val = naive_line(c0, r0, c1, r1)
return (yy, xx, val)
# At this point we know that the distance in columns (x) is greater
# than that in rows (y). Possibly one more switch if c0 > c1.
if c0 > c1:
return naive_line(r1, c1, r0, c0)
# We write y as a function of x, because the slope is always <= 1
# (in absolute value)
x = np.arange(c0, c1+1, dtype=float)
y = x * (r1-r0) / (c1-c0) + (c1*r0-c0*r1) / (c1-c0)
valbot = np.floor(y)-y+1
valtop = y-np.floor(y)
return (np.concatenate((np.floor(y), np.floor(y)+1)).astype(int), np.concatenate((x,x)).astype(int),
np.concatenate((valbot, valtop)))
def trapez(y,y0,w):
return np.clip(np.minimum(y+1+w/2-y0, -y+1+w/2+y0),0,1)
def weighted_line(r0, c0, r1, c1, w, rmin=0, rmax=np.inf):
# The algorithm below works fine if c1 >= c0 and c1-c0 >= abs(r1-r0).
# If either of these cases are violated, do some switches.
if abs(c1-c0) < abs(r1-r0):
# Switch x and y, and switch again when returning.
xx, yy, val = weighted_line(c0, r0, c1, r1, w, rmin=rmin, rmax=rmax)
return (yy, xx, val)
# At this point we know that the distance in columns (x) is greater
# than that in rows (y). Possibly one more switch if c0 > c1.
if c0 > c1:
return weighted_line(r1, c1, r0, c0, w, rmin=rmin, rmax=rmax)
# The following is now always < 1 in abs
slope = (r1-r0) / (c1-c0)
# Adjust weight by the slope
w *= np.sqrt(1+np.abs(slope)) / 2
# We write y as a function of x, because the slope is always <= 1
# (in absolute value)
x = np.arange(c0, c1+1, dtype=float)
y = x * slope + (c1*r0-c0*r1) / (c1-c0)
# Now instead of 2 values for y, we have 2*np.ceil(w/2).
# All values are 1 except the upmost and bottommost.
thickness = np.ceil(w/2)
yy = (np.floor(y).reshape(-1,1) + np.arange(-thickness-1,thickness+2).reshape(1,-1))
xx = np.repeat(x, yy.shape[1])
vals = trapez(yy, y.reshape(-1,1), w).flatten()
yy = yy.flatten()
# Exclude useless parts and those outside of the interval
# to avoid parts outside of the picture
mask = np.logical_and.reduce((yy >= rmin, yy < rmax, vals > 0))
return (yy[mask].astype(int), xx[mask].astype(int), vals[mask])
权重调整的确相当任意，因此任何人都可以根据自己的口味进行调整。现在需要rmin和rmax以避免图片外的像素。比较如下：
如您所见，即使w=1，weighted_line也稍微粗一些，但以某种均匀的方式；同样，naive_line也略微均匀地变细。
有关基准测试的最后说明：在我的机器上，运行%timeit f(1,1,100,240)来测试各种函数（对于weighted_line，w=1）的结果是，line_aa需要90微秒，weighted_line需要84微秒（当然随着权重的增加时间会增加），而naive_line只需要18微秒。再做比较，将line_aa重新实现为纯Python代码（而不是包中的Cython代码）需要350微秒。
val * 255方法不太理想，因为它似乎只在黑色背景上正确工作。如果背景包含较暗和较亮的区域，则这似乎并不完全正确：
要使其在所有背景上正确工作，必须考虑到被抗锯齿线覆盖的像素的颜色。
这里是一个基于原始答案的小演示：
from scipy import ndimage
from scipy import misc
from skimage.draw import line_aa
import numpy as np
img = np.zeros((100, 100, 4), dtype = np.uint8) # create image
img[:,:,3] = 255 # set alpha to full
img[30:70, 40:90, 0:3] = 255 # paint white rectangle
rows, cols, weights = line_aa(10, 10, 90, 90) # antialias line
w = weights.reshape([-1, 1]) # reshape anti-alias weights
lineColorRgb = [255, 120, 50] # color of line, orange here
img[rows, cols, 0:3] = (
np.multiply((1 - w) * np.ones([1, 3]),img[rows, cols, 0:3]) +
w * np.array([lineColorRgb])
)
misc.imsave('test.png', img)
np.multiply((1 - w) * np.ones([1, 3]),img[rows, cols, 0:3]) +
w * np.array([lineColorRgb])
新颜色由图像原始颜色和线的颜色通过使用抗锯齿weights值进行线性插值计算而得。以下是一个结果，橙色线在两种不同背景上运行：
现在，在上半部分环绕线的像素变得更暗，而在下半部分像素变得更亮。
我想要绘制平滑（抗锯齿）的线条，并且不想安装另一个软件包来完成此任务。最终，我使用了Matplotlib的内部方法，该方法可以在我的计算机上以10us/线的速度将1000条线绘制到100x100的数组中。
def rasterize(lines, shape, **kwargs):
"""Rasterizes an array of lines onto an array of a specific shape using
Matplotlib. The output lines are antialiased.
Be wary that the line coordinates are in terms of (i, j), _not_ (x, y).
Args:
lines: (line x end x coords)-shaped array of floats
shape: (rows, columns) tuple-like
Returns:
arr: (rows x columns)-shaped array of floats, with line centres being
1. and empty space being 0.
"""
lines, shape = np.array(lines), np.array(shape)
# Flip from (i, j) to (x, y), as Matplotlib expects
lines = lines[:, :, ::-1]
# Create our canvas
fig = plt.figure()
fig.set_size_inches(shape[::-1]/fig.get_dpi())
# Here we're creating axes that cover the entire figure
ax = fig.add_axes([0, 0, 1, 1])
ax.axis('off')
# And now we're setting the boundaries of the axes to match the shape
ax.set_xlim(0, shape[1])
ax.set_ylim(0, shape[0])
ax.invert_yaxis()
# Add the lines
lines = mpl.collections.LineCollection(lines, color='k', **kwargs)
ax.add_collection(lines)
# Then draw and grab the buffer
fig.canvas.draw_idle()
arr = (np.frombuffer(fig.canvas.get_renderer().buffer_rgba(), np.uint8)
.reshape((*shape, 4))
[:, :, :3]
.mean(-1))
# And close the figure for all the IPython folk out there
plt.close()
# Finally, flip and reverse the array so empty space is 0.
return 1 - arr/255.
plt.imshow(rasterize([[[5, 10], [15, 20]]], [25, 25]), cmap='Greys')
plt.grid()
我想在我的项目中只使用numpy来绘制抗锯齿线条，因此我从scikit-image库中获取了该函数，这里是仅使用numpy的line_aa函数：
def line_aa(r0:int, c0:int, r1:int, c1:int):
"""Generate anti-aliased line pixel coordinates.
Parameters
----------
r0, c0 : int
Starting position (row, column).
r1, c1 : int
End position (row, column).
Returns
-------
rr, cc, val : (N,) ndarray (int, )
Indices of pixels (`rr`, `cc`) and intensity values (`val`).
``img[rr, cc] = val``.
References
----------
.. [1] A Rasterizing Algorithm for Drawing Curves, A. Zingl, 2012
http://members.chello.at/easyfilter/Bresenham.pdf
"""
rr = list()
cc = list()
val = list()
dc = abs(c0 - c1)
dr = abs(r0 - r1)
err = dc - dr
err_prime = 0
c, r, sign_c, sign_r = 0, 0, 0, 0
ed = 0
if c0 < c1:
sign_c = 1
else:
sign_c = -1
if r0 < r1:
sign_r = 1
else:
sign_r = -1
if dc + dr == 0:
ed = 1
else:
ed = np.sqrt(dc*dc + dr*dr)
c, r = c0, r0
while True:
cc.append(c)
rr.append(r)
val.append(np.fabs(err - dc + dr) / ed)
err_prime = err
c_prime = c
if (2 * err_prime) >= -dc:
if c == c1:
break
if (err_prime + dr) < ed:
cc.append(c)
rr.append(r + sign_r)
val.append(np.fabs(err_prime + dr) / ed)
err -= dr
c += sign_c
if 2 * err_prime <= dr:
if r == r1:
break
if (dc - err_prime) < ed:
cc.append(c_prime + sign_c)
rr.append(r)
val.append(np.fabs(dc - err_prime) / ed)
err += dc
r += sign_r
return (np.array(rr, dtype=np.intp),
np.array(cc, dtype=np.intp),
1. - np.array(val))
ImageDraw模块具有与您描述的相似的API。此外，请查看skimage.draw：http://scikit-image.org/docs/dev/api/skimage.draw.html。就此而言，如果需要抗锯齿和/或更高级的绘图方法，甚至可以使用matplotlib进行绘制。 - Joe Kingtonline_aa。谢谢！你想要发布答案还是我创建一个社区wiki？ - Martin Thoma