使用for循環的最小距離

Question

對於許多點，我正在計算位於(x, y)中的參考點的距離。我怎樣才能找到距離的最小值？這些是我寫的代碼行：

for k in range(0, 10): 
    dist = math.sqrt((x - data.X[k])**2 + (y - data.Y[k])**2) 

Answer 1

你的意思是這樣的嗎？

min=math.sqrt((x - data.X[0])**2 + (y - data.Y[0])**2) 
for k in range(0, 10): 
    dist = math.sqrt((x - data.X[k])**2 + (y - data.Y[k])**2) 
    if dist<min: 
     min=dist 

或者：

for k in range(0, 10): 
    dist = math.sqrt((x - data.X[k])**2 + (y - data.Y[k])**2) 
    try: 
     if dist<min: 
      min=dist 
    except NameError: 
     min=dist 

Answer 2

類是你的朋友。這有點多，但它更好，而且是可擴展的。

class point: 
    def __init__(self, x, y): 
     self.x = x 
     self.y = y 

    def __str__(self): 
     return '{0}, {1}'.format(self.x, self.y) 

    def distanceto(self, other): 
     return math.sqrt((self.x - other.x)**2 + (self.y - other.y)**2) 

    def closestpoint(self, pointlist): 
     pointinfo = [{'point':x, 'dist':self.distanceto(x)} for x in pointlist] 
     pointinfo.sort(key=lambda p: p.dist) 
     return pointinfo[0] 

而不是從文件中讀取點和seperately保存X和Y組件，爲什麼不將它們保存爲點的列表？

# all points read from the file. 
listofpoints = [] 
for i in range(0, 10): 
    listofpoints.append(point(data.X[i], data.Y[i])) 

# the point you'd like to test against. 
mytestpoint = point(0,0) 

您現在可以使用point成員方法測試所有點的差異。

closest = mytestpoint.closestpoint(listofpoints) 
print 'Closest point is at {0} and is a distance of {1} from {2}'.format(
    closest, 
    mytestpoint.distanceto(closest), 
    mytestpoint)