python - 函数: Returning min value in dictionary

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我很难理解字典功能。我试图编写的函数需要找到当前位置的最近位置(在字典中给出)并返回它。我被告知涉及一个距离公式，但不确定如何将其实现到函数字典中。当没有找到任何内容时，它应该不返回任何内容。

def closest_location(d, place, now):
        close_lst = []# New list
        for d in closest.place():
            for d in closest.now():
                if now != place:
                    return None
                elif now <= place: #If location at now is less than place we want to go to...
                    close_val = now - place
                close_lst.append(close_val)
        return(min(d, key=close_lst.get))# returns closest value in list?

测试:

check that closest({(3,1):'gas', (1,4):'gas', (2,1):'food', (5,5):'food'},'food',(5,5)) == (5,5).
check that closest({(3,1):'gas', (1,4):'gas', (2,1):'food', (5,5):'food'},'hotel',(1,4)) == None.

Answer 1

假设元组是网格上的 x,y 坐标，距离公式由毕达哥拉斯给出:

(x1 - x2)^2 + (y1 - y2)^2 = distance^2

我们只是将其提取到它自己的函数中以提高可读性。

from math import sqrt
def find_distance(a,b):
    ax, ay = a
    bx, by = b
    return sqrt((ax-bx)**2 + (ay-by)**2)

def closest_location(d, place, now):
    locations = [loc for loc, attraction in d.items() if attraction==place]
    if not locations:
        return None
    else:
        return min(locations, key=lambda x: find_distance(x, now))