字典中的Python最近坐标

我有以下字典：

points = {'Location1': (76, 81), 'Location2': (75, 105), 'Location3': (76, 130), 'Location4': (76, 152)}

如果我有一组坐标的话，我会尝试去计算；coord=（x，y）查找与坐标值对最接近的键。但我要检索与最近的

我是这样做的，但必须有一个更有效的方法

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)])

---

您根本不需要使用列表（

array

），您可以将字典（

points

）传递到

min

；字典的键将传递给

键

函数：

>>> 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

，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)

或

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)