Python 字典中最接近的坐标

Question

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，我正在尝试; coord = (x， y( 以查找与坐标最接近的值对的键。但是我想检索与最接近的密钥相对应的密钥。

我这样做了，但必须有一种更有效的方法。

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)])

Answer 1

你根本不需要使用列表(array(，你可以将字典(points(传递给min;字典的键将传递给key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，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)

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)