[英]Fastest way to get all the points between two (X,Y) coordinates in python
So I have a shapely
LineString
:所以我有一个shapely
的LineString
:
print np.round(shapely_intersecting_lines.coords).astype(np.int)
>>> array([[ 1520, -1140],
[ 1412, -973]])
This can be interpreted as a numpy
array as well as seen above.这可以解释为一个numpy
数组,如上所示。
I want to get all the points in between, that is I want to get the points of the line in between as integer values.我想获得介于两者之间的所有点,也就是说,我想获得介于两者之间的线的点作为整数值。 The output should be something like this:输出应该是这样的:
array([[ 1520, -1140],
[ 1519, -1139],
[ 1519, -1138],
...,
[ 1413, -975],
[ 1412, -974],
[ 1412, -973]], dtype=int32)
I posted this earlier in gis.stackexchange hoping there was a solution in shapely
that was efficient.我早些时候在gis.stackexchange上发布了这个,希望有一个shapely
的解决方案。 The solution was good at first, however, the solution is now too slow as I run this over 50000 times in my code.解决方案一开始很好,但是现在解决方案太慢了,因为我在代码中运行了超过 50000 次。 On my computer each loop takes about 0.03s resulting in over a day of running.在我的电脑上,每个循环大约需要 0.03 秒,导致运行一天以上。 It is too slow for what I need here and was hoping to see if anyone knows of a vectorized solution to this.对于我在这里需要的东西来说太慢了,希望看看是否有人知道对此的矢量化解决方案。
Bresenham may be smart but I'm pretty sure brute force vectorization is faster. Bresenham可能很聪明,但我很确定蛮力矢量化更快。 I've written two variants - the first is easier to read, the second is faster (80 us vs 50 us). 我写了两个变体 - 第一个更容易阅读,第二个更快(80 us vs 50 us)。
Update Fixed a bug (thanks @Varlor) and added an nd variant. 更新修复了一个错误(感谢@Varlor)并添加了一个nd变体。
import numpy as np
from timeit import timeit
def connect(ends):
d0, d1 = np.abs(np.diff(ends, axis=0))[0]
if d0 > d1:
return np.c_[np.linspace(ends[0, 0], ends[1, 0], d0+1, dtype=np.int32),
np.round(np.linspace(ends[0, 1], ends[1, 1], d0+1))
.astype(np.int32)]
else:
return np.c_[np.round(np.linspace(ends[0, 0], ends[1, 0], d1+1))
.astype(np.int32),
np.linspace(ends[0, 1], ends[1, 1], d1+1, dtype=np.int32)]
def connect2(ends):
d0, d1 = np.diff(ends, axis=0)[0]
if np.abs(d0) > np.abs(d1):
return np.c_[np.arange(ends[0, 0], ends[1,0] + np.sign(d0), np.sign(d0), dtype=np.int32),
np.arange(ends[0, 1] * np.abs(d0) + np.abs(d0)//2,
ends[0, 1] * np.abs(d0) + np.abs(d0)//2 + (np.abs(d0)+1) * d1, d1, dtype=np.int32) // np.abs(d0)]
else:
return np.c_[np.arange(ends[0, 0] * np.abs(d1) + np.abs(d1)//2,
ends[0, 0] * np.abs(d1) + np.abs(d1)//2 + (np.abs(d1)+1) * d0, d0, dtype=np.int32) // np.abs(d1),
np.arange(ends[0, 1], ends[1,1] + np.sign(d1), np.sign(d1), dtype=np.int32)]
def connect_nd(ends):
d = np.diff(ends, axis=0)[0]
j = np.argmax(np.abs(d))
D = d[j]
aD = np.abs(D)
return ends[0] + (np.outer(np.arange(aD + 1), d) + (aD>>1)) // aD
ends = np.array([[ 1520, -1140],
[ 1412, -73]])
ends_4d = np.array([[ 100, -302, 101, -49],
[ -100, -45, 112, 100]])
print(connect(ends))
print(connect_nd(ends_4d))
assert np.all(connect(ends)==connect2(ends))
assert np.all(connect(ends)==connect_nd(ends))
assert np.all(connect(ends)==connect(ends[:, ::-1])[:, ::-1])
assert np.all(connect(ends)==connect(ends[::-1])[::-1])
print(timeit('f(ends)', globals={'f': connect, 'ends': ends}, number=10000)*100, 'us')
print(timeit('f(ends)', globals={'f': connect2, 'ends': ends}, number=10000)*100, 'us')
print(timeit('f(ends)', globals={'f': connect_nd, 'ends': ends}, number=10000)*100, 'us')
Sample output: 样本输出:
[[ 1520 -1140]
[ 1520 -1139]
[ 1520 -1138]
...,
[ 1412 -75]
[ 1412 -74]
[ 1412 -73]]
[[ 100 -302 101 -49]
[ 99 -301 101 -48]
[ 98 -300 101 -48]
...,
[ -98 -47 112 99]
[ -99 -46 112 99]
[-100 -45 112 100]]
78.8237597000034 us
48.02509490000375 us
62.78072760001123 us
Good base code to start:好的基本代码开始:
def points_between(x1, y1, x2, y2):
d0 = x2 - x1
d1 = y2 - y1
count = max(abs(d1+1), abs(d0+1))
if d0 == 0:
return (
np.full(count, x1),
np.round(np.linspace(y1, y2, count)).astype(np.int32)
)
if d1 == 0:
return (
np.round(np.linspace(x1, x2, count)).astype(np.int32),
np.full(count, y1),
)
return (
np.round(np.linspace(x1, x2, count)).astype(np.int32),
np.round(np.linspace(y1, y2, count)).astype(np.int32)
Using generators so that way you will save memory 使用生成器,这样你将节省内存
import numpy as np
import math
d = np.array(
[
[1520,-1140],
[1412,-973]
],dtype=float);
rise = d[0,1]-d[1,1];
run = d[0,0]-d[1,0];
slope = rise/run;
print slope
fromPt = d[0];
sign = 1
if (slope<0):
sign = -1;
points = ([d[0,0]+sign*i,math.floor(d[0,1]+(sign*i*slope))] for i in xrange(1+int(math.ceil(abs(d[0,0]-d[1,0])))))
for pt in points:
print pt
Here's Bresenhams line algorithm as a generator. 这里是Bresenhams线算法作为生成器。 If you want a list just call list() on the output: 如果你想要一个列表,只需在输出上调用list():
def line(x0, y0, x1, y1):
deltax = x1-x0
dxsign = int(abs(deltax)/deltax)
deltay = y1-y0
dysign = int(abs(deltay)/deltay)
deltaerr = abs(deltay/deltax)
error = 0
y = y0
for x in range(x0, x1, dxsign):
yield x, y
error = error + deltaerr
while error >= 0.5:
y += dysign
error -= 1
yield x1, y1
here is a short solution using np library这是使用 np 库的简短解决方案
import numpy as np
x = [1,2,3]
y = [5,6,7]
ks,cs = np.meshgrid(x,y)
np.vstack((ks.flatten(),cs.flatten())).T
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