[英]python calculate distance closest xy points
so i have a list of points 所以我有一个要点清单
["9.5 7.5", "10.2 19.1", "9.7 10.2", "2.5 3.6", "5.5 6.5", "7.8 9.8"]
with a starting point of 起点为
["2.2 4.6"]
now what i am trying to do it is get the closest point to my starting point, then the closest point to that point and so on. 现在我正在尝试执行的操作是使最接近我的起点,然后最接近该点,依此类推。
So i get to calculate distance 所以我可以计算距离
def dist(p1,p2):
return math.sqrt((p2[0] - p1[0]) ** 2 + (p2[1] - p1[1]) ** 2)
but again, i'm trying to get the closest to my starting point, then the closest point to that one and so on. 但同样,我正在尝试最接近我的起点,然后最接近那个起点,依此类推。
ok, because you are complaing i didn't show enough code? 好的,因为您抱怨我没有显示足够的代码?
fList = ["2.5 3.6", "9.5 7.5", "10.2 19.1", "9.7 10.2", "5.5 6.5", "7.8 9.8"]
def distance(points):
p0, p1 = points
return math.sqrt((p0[0] - p1[0])**2 + (p0[1] - p1[1])**2)
min_pair = min(itertools.combinations(fList, 2), key=distance)
min_distance = distance(min_pair)
print min_pair
print min_distance
so i get passing my starting point I get 所以我通过了我的起点
([2.2, 4.6], [2.5, 3.6])
So now i need to use 2.5, 3.6 as my starting point and find the next closest and so on 所以现在我需要使用2.5、3.6作为起点,并找到下一个最接近的点,依此类推
Has anyone done anything similar? 有人做过类似的事情吗?
A possibility is to use a breadth-first search to scan all elements, and find the closest point for each element popped off the queue: 一种可能性是使用广度优先搜索来扫描所有元素,并为从队列中弹出的每个元素找到最接近的点:
import re, collections
import math
s = ["9.5 7.5", "10.2 19.1", "9.7 10.2", "2.5 3.6", "5.5 6.5", "7.8 9.8"]
def cast_data(f):
def wrapper(*args, **kwargs):
data, [start] = args
return list(map(lambda x:' '.join(map(str, x)), f(list(map(lambda x:list(map(float, re.findall('[\d\.]+', x))), data)), list(map(float, re.findall('[\d\.]+', start))))))
return wrapper
@cast_data
def bfs(data, start, results=[]):
queue = collections.deque([start])
while queue and data:
result = queue.popleft()
possible = min(data, key=lambda x:math.hypot(*[c-d for c, d in zip(result, x)]))
if possible not in results:
results.append(possible)
queue.append(possible)
data = list(filter(lambda x:x != possible, data))
return results
print(bfs(s, ["2.2 4.6"]))
Output: 输出:
['2.5 3.6', '5.5 6.5', '7.8 9.8', '9.7 10.2', '9.5 7.5', '10.2 19.1']
The result is the listing of closest points, as determined by using math.hypot
. 结果是通过使用math.hypot
确定的最接近点列表。
You can try the following code. 您可以尝试以下代码。 Much simpler and short. 简单多了。 Uses a comparator to sort the list depending on the distance from the starting point (2.2,4.6)
使用比较器根据到起点的距离对列表进行排序(2.2,4.6)
import math
data = ["9.5 7.5", "10.2 19.1", "9.7 10.2", "2.5 3.6", "5.5 6.5", "7.8 9.8"]
data.sort(key=lambda x: math.sqrt((float(x.split(" ")[0]) - 2.2)**2 +
(float(x.split(" ")[1]) -4.6)**2))
print(data)
# output ['2.5 3.6', '5.5 6.5', '7.8 9.8', '9.5 7.5', '9.7 10.2', '10.2 19.1']
You can simply sort a list by a key you define as you wish - fe by your distance function: 您可以简单地根据自己想要的键对列表进行排序 -fe通过距离函数进行排序 :
import math
def splitFloat(x):
"""Split each element of x on space and convert into float-sublists"""
return list(map(float,x.split()))
def dist(p1, p2):
# you could remove the sqrt for computation benefits, its a symetric func
# that does not change the relative ordering of distances
return math.sqrt((p2[0] - p1[0]) ** 2 + (p2[1] - p1[1]) ** 2)
p = ["9.5 7.5", "10.2 19.1", "9.7 10.2", "2.5 3.6", "5.5 6.5", "7.8 9.8"]
s = splitFloat("2.2 4.6") # your start point
p = [splitFloat(x) for x in p] # your list of points
# sort by distance between each individual x and s
p.sort(key = lambda x:dist(x,s))
d = [ (dist(x,s),x) for x in p] # create tuples with distance for funsies
print(p)
print(d)
Output: 输出:
[[2.5, 3.6], [5.5, 6.5], [7.8, 9.8], [9.5, 7.5], [9.7, 10.2], [10.2, 19.1]]
[(1.0440306508910546, [2.5, 3.6]), (3.8078865529319543, [5.5, 6.5]),
(7.64198926981712, [7.8, 9.8]), (7.854934754662192, [9.5, 7.5]),
(9.360021367496977, [9.7, 10.2]), (16.560495161679196, [10.2, 19.1])]
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