我有以下字典:
points = {'Location1': (76, 81), 'Location2': (75, 105), 'Location3': (76, 130), 'Location4': (76, 152)}
如果我有一组坐标,我会尽力; coord =(x,y)查找与坐标最接近的值对的键.但我想检索对应最近的密钥.
我这样做了,但必须有一个更有效的方式.
points = {'Location1': (76, 81), 'Location2': (75, 105), 'Location3': (76, 130), 'Location4': (76, 152)}
array = [(76, 81), (75, 105), (76, 130), (76, 152)]
def find_nearest(array,coord):
dist = lambda s, d: (s[0] - d[0]) ** 2 + (s[1] - d[1]) ** 2
result = min(array, key=partial(dist, coord))
return result
found = find_nearest(array,coord)
print (list(points.keys())[list(points.values()).index(found)])
最佳答案 您根本不需要使用列表(数组),可以将字典(点)传递给min;字典的键将传递给关键功能:
>>> from functools import partial
>>>
>>> def find_nearest(points, coord):
... dist = lambda s, key: (s[0] - points[key][0]) ** 2 + \
... (s[1] - points[key][1]) ** 2
... return min(points, key=partial(dist, coord))
...
>>> points = {'Location1': (76, 81), 'Location2': (75, 105),
... 'Location3': (76, 130), 'Location4': (76, 152)}
>>> find_nearest(points, (0, 0))
'Location1'
>>> find_nearest(points, (100, 100))
'Location2'
>>> find_nearest(points, (100, 200))
'Location4'
通过直接访问lambda中的coord,您可以删除partial:
def find_nearest(points, coord):
dist = lambda key: (coord[0] - points[key][0]) ** 2 + \
(coord[1] - points[key][1]) ** 2
return min(points, key=dist)
要么
def find_nearest(points, coord):
x, y = coord
dist = lambda key: (x - points[key][0]) ** 2 + (y - points[key][1]) ** 2
return min(points, key=dist)