python计算最近xy点的距离
所以我有一个要点列表python计算最近xy点的距离,python,Python,所以我有一个要点列表 ["9.5 7.5", "10.2 19.1", "9.7 10.2", "2.5 3.6", "5.5 6.5", "7.8 9.8"] 以 ["2.2 4.6"] 现在我要做的是得到离起点最近的点,然后是离起点最近的点,依此类推 所以我开始计算距离 def dist(p1,p2): return math.sqrt((p2[0] - p1[0]) ** 2 + (p2[1] - p1[1]) ** 2) 不过,我还是想尽量接近我的起点,然后是最接近那个起点
["9.5 7.5", "10.2 19.1", "9.7 10.2", "2.5 3.6", "5.5 6.5", "7.8 9.8"]
以
["2.2 4.6"]
现在我要做的是得到离起点最近的点,然后是离起点最近的点,依此类推
所以我开始计算距离
def dist(p1,p2):
return math.sqrt((p2[0] - p1[0]) ** 2 + (p2[1] - p1[1]) ** 2)
不过,我还是想尽量接近我的起点,然后是最接近那个起点的点,以此类推
好吧,因为你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
所以我通过了我的起点
([2.2, 4.6], [2.5, 3.6])
所以现在我需要使用2.5,3.6作为我的起点,并找到下一个最接近的,以此类推
有人做过类似的事情吗?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"]))
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']
[[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])]
结果是使用
math.hypot
确定的最近点列表。您可以尝试以下代码。简单得多,简短得多。使用比较器根据与起点的距离对列表进行排序(2.2,4.6)
您可以简单地根据您的距离函数定义-f.e.:
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)
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']
[[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])]
欢迎来到StackOverflow。请按照您创建此帐户时的建议,阅读并遵循帮助文档中的发布指南。在这里申请。StackOverflow不是设计、编码、研究或教程服务。编辑得不错,你想得太多了。如果将sort()作为距离函数指定为
key=
,则可以通过在浮点列表上使用sort()
轻松解决此问题。你是邪恶的:)-我得到了[(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],(16.5604951679196,[10.2,19.1])
-但如果我发布我的6行程序,那将是一个复制、粘贴和理解solution@PatrickArtner啊,你在每一点上都包括了差异。