如何在numpy数组中绘制线条？
我希望能够将线条绘制到numpy数组中，以获得用于在线手写识别的离线特征。这意味着我根本不需要图像，但我需要在一个numpy数组中找到一些位置，该位置是给定大小的图像的外观。
我希望能够指定图像大小，然后像这样绘制笔划：
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)这样简单的事情是可能的吗(最好直接使用numpy / scipy)？
(请注意，我想要灰度插值。因此，features应该是一个0,255中的值矩阵。)
回答 3
Stack Overflow用户
发布于 2015-07-28 16:42:21
感谢Joe Kington的回答！我在找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)Stack Overflow用户
发布于 2017-11-20 03:25:51
我在寻找解决方案时偶然发现了这个问题，提供的答案很好地解决了它。然而，它并不真正适合我的目的，为此我需要一个“可伸缩的”解决方案(即在numpy中实现，没有显式循环)，并且可能还需要一个行宽选项。我最终实现了我自己的版本，因为最终它也比line_aa快得多，我想我可以分享它。
它有两种风格，有线宽和没有线宽。实际上，前者并不是后者的泛化，两者都不完全符合line_aa，但就我的目的而言，它们很好，在绘图中它们看起来也不错。
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)))我称其为" naive“，因为它非常类似于Wikipedia中的naive实现，但有一些抗锯齿，尽管不可否认不是完美的(例如，制作非常细的对角线)。
加权版本提供更粗的线条，更明显的抗锯齿效果。
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])不可否认，权重的调整是相当随意的，所以任何人都可以根据自己的喜好进行调整。现在需要rmin和rmax来避免图像之外的像素。比较：
正如您所看到的，即使使用w=1，weighted_line也稍微厚一些，但以一种同质的方式；类似地，naive_line的同质稍微薄一些。
关于基准测试的最后一点注意事项:在我的机器上，针对各种功能(针对weighted_line的w=1)运行%timeit f(1,1,100,240)时，line_aa的运行时间为90µs，weighted_line的运行时间为84µs (当然，时间会随着重量的增加而增加)，而naive_line的运行时间为18µs。同样为了进行比较，用纯Python (而不是包中的Cython )重新实现line_aa需要350微秒。
Stack Overflow用户
发布于 2019-08-01 23:32:52
我想绘制抗锯齿的线条，我想绘制数千条这样的线条，而不需要安装另一个包。我最终劫持了Matplotlib的内部结构，它以10us/line的速度在100x100数组上执行1000行代码，至少在我的机器上是这样。
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.下面是输出的样子：
plt.imshow(rasterize([[[5, 10], [15, 20]]], [25, 25]), cmap='Greys')
plt.grid()
https://stackoverflow.com/questions/31638651
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