Python最近的字典坐标

时间:2017-03-26 06:37:22

标签: python list python-3.x dictionary

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)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。

我是这样做的,但必须有一种更有效的方式。

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)])

1 个答案:

答案 0 :(得分:1)

您根本不需要使用列表(array),您可以将字典(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)

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)