找到点与线之间的距离，并获得该点在一条线中的投影距离 [英] Find the distance between points and line and achieve the projection distance that the point takes in a line
问题描述
我是python的新手.但是我有一个艰巨的任务: 我有带GPS坐标的CSV文件和1个带参考轨迹(坐标点)的文件，它们代表了车辆的轨迹.
I am new in python. But I have a challenging task for me: I have CSV files with GPS coordinates and 1 file with reference trajectory (coordinates points) which represent the track of the vehicle.
所以任务是对所有点计算到参考轨迹的垂直距离，并定义该点在参考轨迹的哪一公里
So the task is to all points calculate the perpendicular distance to reference trajectory and define in which kilometer this point is in the reference trajectory
我试图用QGIS解决它，但问题是数据太多，并且QGIS一直崩溃. 我正在使用距离矩阵.
I was trying to solve it with QGIS but the problem is that it is too much data and QGIS all the time crashed. I was using the distance matrix.
因此参考轨迹数据如下:
So the reference trajectory data looks like this:
datetime lon_deg lat_deg ki1ometers
0 27.03.2018 15:07 14.34559621 48.28282695 0
1 27.03.2018 15:07 14.34539589 48.283579 0.08492765648897423
2 27.03.2018 15:08 14.34509878 48.28437137 0.17573647034625345
3 27.03.2018 15:08 14.34476681 48.28520735 0.2718837851891085
4 27.03.2018 15:09 14.34440297 48.28607467 0.372012272777317
5 27.03.2018 15:10 14.34355387 48.28785601 0.5798125761498747
6 27.03.2018 15:10 14.34312139 48.28876254 0.6855708866782635
7 27.03.2018 15:11 14.34267986 48.28966368 0.7909635418697577
8 27.03.2018 15:11 14.34235909 48.29057934 0.895509507334529
9 27.03.2018 15:12 14.34193015 48.29147634 1.000178064181187
10 27.03.2018 15:12 14.34158939 48.2923968 1.1055875957864745
11 27.03.2018 15:13 14.34125444 48.29332421 1.2116463089787737
12 27.03.2018 15:13 14.34084938 48.29424082 1.31788253222638
13 27.03.2018 15:14 14.34041673 48.29515665 1.4246295164890292
14 27.03.2018 15:14 14.34001362 48.29608703 1.532295241219843
15 27.03.2018 15:15 14.33959522 48.29702238 1.6408091272201002
16 27.03.2018 15:15 14.33917898 48.29796904 1.7504838454702525
17 27.03.2018 15:16 14.33875624 48.29892358 1.8611345768980705
18 27.03.2018 15:16 14.33832484 48.29988211 1.9723928345544686
19 27.03.2018 15:17 14.337844699999998 48.30083163 2.083788039109954
20 27.03.2018 15:17 14.33733187 48.30177414 2.1952441083077696
21 27.03.2018 15:18 14.33680756 48.30271439 2.3067561380904458
22 27.03.2018 15:18 14.33637327 48.30366977 2.4177398933361665
23 27.03.2018 15:19 14.33579109 48.30456609 2.5263104564169723
计算参考轨迹上的距离和位置所需的数据:
And the data that I need to calculate distance and position on my reference trajectory:
datetime lon_deg lat_deg
2018-01-29 00:00:00.000 13.535165989333333 48.58077572716667
29.01.2018 0:00 13.535166009 48.580775726166664
2018-01-29 00:00:01.000 13.535165977166667 48.580775749
29.01.2018 0:00 13.5351658175 48.58077575
2018-01-29 00:00:02.000 13.535165976833333 48.58077567466667
29.01.2018 0:00 13.535165988166666 48.58077563316667
2018-01-29 00:00:03.000 13.535165978333334 48.580775599
29.01.2018 0:00 13.535166127833334 48.5807756575
2018-01-29 00:00:04.000 13.535166430833334 48.5807757935
29.01.2018 0:00 13.535166510166666 48.580775819
2018-01-29 00:00:05.000 13.5351665845 48.5807758835
29.01.2018 0:00 13.5351665215 48.580775906
2018-01-29 00:00:06.000 13.535166549166666 48.58077594583333
29.01.2018 0:00 13.535166521333334 48.58077594466667
2018-01-29 00:00:07.000 13.535166487 48.580775927666664
29.01.2018 0:00 13.5351670905 48.58077611433333
2018-01-29 00:00:08.000 13.5351669075 48.5807760195
29.01.2018 0:00 13.535166444166666 48.580775919
因此，我期望的输出是计算到直线轨迹的距离，以及该点在轨迹的特定公里数内.
So the output I expect is calculated distance to the line trajectory and in which specific kilometer of trajectory this point is.
我很乐意听到任何想法，因为我确实陷在这个问题中
I would be happy to hear any ideas because I really stuck in this problem
推荐答案
在如此小的距离(例如2.5公里的路程)中，您可能可以采用平面近似法并使用非常简单的方法:对于距离可以使用公式来计算三角形的高度，例如使用半周长.然后，您仍然必须计算该点是否在给定的线段(位于其上的垂直带的一部分)中，您可以在其中使用
In case of such small distances (like a trip of 2.5 km), you can probably live with a planar approximation and use very simple stuff: for the distance you can use formulas for calculating the height of a triangle, like the one working with semiperimeter. Then you will still have to calculate if the point is "in" the given line segment (part of the perpendicular strip built on it), where you can use the projection property of scalar (dot) product of vectors.
因此对于线段AB和点C(所有点都是元组/列表)，您将需要以下内容:
So for a line-segment AB, and a point C (all points are tuples/lists) you would need something like this:
pdist=lambda A,B:((A[0]-B[0])**2+(A[1]-B[1])**2)**(1/2)
def dist(A,B,C):
c=pdist(A,B)
rat=((C[0]-A[0])*(B[0]-A[0])+(C[1]-A[1])*(B[1]-A[1]))/c/c
if rat<0 or rat>1:
return None,None
a=pdist(B,C)
b=pdist(A,C)
s=(a+b+c)/2
alt=2*(s*(s-a)*(s-b)*(s-c))**(1/2)/c
return alt,rat
如果C的垂直基点在AB之外，或者(C到AB段的垂直距离)元组之间的比率为0 ... 1，则可以返回None,None，您可以将其用作计算两个相邻点之间的游览位置".
It returns None,None if the perpendicular base point for C is outside AB, or a tuple of perpendicular distance (of C from AB segment) and a ratio between 0...1 which you can use as weight for calculating "tour-position" between the two neighboring points.
然后对示例数据进行一些解析:
Then some parsing on your example data:
import re
rawtour='''0 27.03.2018 15:07 14.34559621 48.28282695 0
1 27.03.2018 15:07 14.34539589 48.283579 0.08492765648897423
2 27.03.2018 15:08 14.34509878 48.28437137 0.17573647034625345
3 27.03.2018 15:08 14.34476681 48.28520735 0.2718837851891085
4 27.03.2018 15:09 14.34440297 48.28607467 0.372012272777317
5 27.03.2018 15:10 14.34355387 48.28785601 0.5798125761498747
6 27.03.2018 15:10 14.34312139 48.28876254 0.6855708866782635
7 27.03.2018 15:11 14.34267986 48.28966368 0.7909635418697577
8 27.03.2018 15:11 14.34235909 48.29057934 0.895509507334529
9 27.03.2018 15:12 14.34193015 48.29147634 1.000178064181187
10 27.03.2018 15:12 14.34158939 48.2923968 1.1055875957864745
11 27.03.2018 15:13 14.34125444 48.29332421 1.2116463089787737
12 27.03.2018 15:13 14.34084938 48.29424082 1.31788253222638
13 27.03.2018 15:14 14.34041673 48.29515665 1.4246295164890292
14 27.03.2018 15:14 14.34001362 48.29608703 1.532295241219843
15 27.03.2018 15:15 14.33959522 48.29702238 1.6408091272201002
16 27.03.2018 15:15 14.33917898 48.29796904 1.7504838454702525
17 27.03.2018 15:16 14.33875624 48.29892358 1.8611345768980705
18 27.03.2018 15:16 14.33832484 48.29988211 1.9723928345544686
19 27.03.2018 15:17 14.337844699999998 48.30083163 2.083788039109954
20 27.03.2018 15:17 14.33733187 48.30177414 2.1952441083077696
21 27.03.2018 15:18 14.33680756 48.30271439 2.3067561380904458
22 27.03.2018 15:18 14.33637327 48.30366977 2.4177398933361665
23 27.03.2018 15:19 14.33579109 48.30456609 2.5263104564169723 '''
tour=list(map(lambda line:list(map(lambda x:float(x),re.match(r'[^\s]+\s+[^\s]+\s+[^\s]+\s+([\d\.]+)\s+([\d\.]+)\s+([\d\.]+)\s*',line).groups())),rawtour.split('\n')))
rawmarks='''2018-01-29 00:00:00.000 13.535165989333333 48.58077572716667
29.01.2018 0:00 13.535166009 48.580775726166664
2018-01-29 00:00:01.000 13.535165977166667 48.580775749
29.01.2018 0:00 13.5351658175 48.58077575
2018-01-29 00:00:02.000 13.535165976833333 48.58077567466667
29.01.2018 0:00 13.535165988166666 48.58077563316667
2018-01-29 00:00:03.000 13.535165978333334 48.580775599
29.01.2018 0:00 13.535166127833334 48.5807756575
2018-01-29 00:00:04.000 13.535166430833334 48.5807757935
29.01.2018 0:00 13.535166510166666 48.580775819
2018-01-29 00:00:05.000 13.5351665845 48.5807758835
29.01.2018 0:00 13.5351665215 48.580775906
2018-01-29 00:00:06.000 13.535166549166666 48.58077594583333
29.01.2018 0:00 13.535166521333334 48.58077594466667
2018-01-29 00:00:07.000 13.535166487 48.580775927666664
29.01.2018 0:00 13.5351670905 48.58077611433333
2018-01-29 00:00:08.000 13.5351669075 48.5807760195
29.01.2018 0:00 13.535166444166666 48.580775919'''
marks=list(map(lambda line:list(map(lambda x:float(x),re.match(r'[^\s]+\s+[^\s]+\s+([\d\.]+)\s+([\d\.]+)\s*',line).groups())),rawmarks.split('\n')))
和[[dist(A,B,C) for A,B in zip(tour,tour[1:])] for C in marks]生成距离矩阵，在这种情况下，该距离矩阵完全为空，因为两个坐标集相距太远，与建立在线段上的狭窄垂直条纹有关.经度约为48度时，一度经度约为111公里/秒.虽然整个示例游览仅2.5公里长.
And [[dist(A,B,C) for A,B in zip(tour,tour[1:])] for C in marks] produces the distance matrix, which is totally empty in this case as the two coordinate sets are too far away related to the narrow perpendicular strips built on the line segments. One degree of longitude is around 111 km-s at latitude ~48 degrees. While the entire example tour is only 2.5 km long.
为加快处理速度，请 https://docs.python .org/3/library/functools.html#functools.lru_cache 可以在pdist上使用，或者c -s(游览段的长度)可以明确地预先计算并存储在某个位置， /c/c除法可以被推迟，而temprat<0 or temprat>c**2可以用于if，其中c**2也可以被预先计算.当我注意到距离问题时，我只是失去了动力.
我认为您也必须检查点到点的距离.参见图的上部:蓝点落在灰色矩形的外面(当然，它们是无限的，并且向两个方向延伸，我只是将它们加盖以获得更好的视觉效果)，但是恰好在红色矩形的内部，因此是正交的-distance-only方法将表示最右边的线段是最接近所讨论点的线段，而查看点到点的距离可能会发现绿点是最近的位置.
For speeding up the thing, https://docs.python.org/3/library/functools.html#functools.lru_cache could be used on pdist, or the c-s (length of tour-segments) could be explicitly pre-calculated and stored somewhere, the /c/c division could be postponed and temprat<0 or temprat>c**2 could be used in the if, where c**2 could be pre-calculated too. Just I lost motivation when noticed the distance issue.
I think you will have to check point-point distances too. See the upper part of figure: the blue point falls outside of the gray rectangles (of course they are infinite, and extend towards both directions, I just capped them for better visuals), but happens to be inside the red one, so an orthogonal-distance-only approach would say that the rightmost segment was the closest one to the point in question, while looking at point-point distances could find the green point as closest location.
但是图像的下部是用来说明点对点的距离不能用作替代，它甚至也不能真正地驱动"正交计算，因为红色点是最接近的角点到蓝色的那一个并不意味着一个片段不可能更近，而红色的那个甚至都不是那个片段的终点.
However the lower part of the image is meant to illustrate that point-point distances can not be used as replacement, and it can not even really "drive" the orthogonal calculations either, as the red point being the closest corner-point to the blue one does not mean that a segment could not be closer, and the red one is not even an endpoint of that segment.
为了使它与经纬度坐标配合使用，存在许多令人毛骨悚然的公式，我现在不敢为您选择一个. https://en.wikipedia.org/wiki/Geographical_distance 可能是一个很好的起点.您可以依靠的一个捷径是您的轨迹包含以km-s为单位的距离，因此您可以将线段中某个点的行程距离作为其端点处行程距离的加权和来计算(而不是直接从经纬度计算距离) -lon坐标.)
For making it work with lat-lon coordinates lots of creepy formulaes exist, I would not dare to choose one for you at the moment. https://en.wikipedia.org/wiki/Geographical_distance may be a good starting point. One shortcut you can rely on is that your trajectories contain distances in km-s, so you can calculate trip-distance of a point in a segment as a weighted sum of the trip-distances at its endpoints (instead of calculating distances directly from lat-lon coordinates).
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