计算三均值的欧氏距离

Question

我在计算欧几里德距离时遇到了麻烦

后来我引用了函数，它给了我这个错误：

TypeError:输入类型不支持ufunc“按位\u和”，并且无法根据强制转换规则“safe”将输入安全强制为任何受支持的类型

它是硬编码K-均值算法所需要的

def euclideanDist(df,pointIDX,mean_1,mean_2,mean_3):

    point = df.iloc[pointIDX][['Shoe_Size','Height']].values
    mean_1 = mean_1[['Shoe_Size','Height']].values
    mean_2 = mean_2[['Shoe_Size','Height']].values
    mean_3 = mean_3[['Shoe_Size','Height']].values

    dist_Total_1 = sum([a-b for a,b in zip(point,mean_1)])**2
    dist_Total_2 = sum([a-b for a,b in zip(point,mean_2)])**2
    dist_Total_3 = sum([a-b for a,b in zip(point,mean_3)])**2

    if dist_Total_1 < dist_Total_2 & dist_Total_3: 
        df.loc[pointIDX,'class'] = 1

    elif dist_Total_2 < dist_Total_3 > dist_Total_1:
        df.loc[pointIDX, "class"] = 2

    else:
        df.loc[pointIDX,'class'] = 3

    return df

Answer 1

这里有一些语法问题

if dist_Total_1 < dist_Total_2 & dist_Total_3: 
    df.loc[pointIDX,'class'] = 1

elif dist_Total_2 < dist_Total_3 > dist_Total_1:
    df.loc[pointIDX, "class"] = 2

我相信你真正想要的是

if dist_Total_1 < dist_Total_2 and dist_Total_1 < dist_Total_3: 
    df.loc[pointIDX,'class'] = 1

elif dist_Total_2 < dist_Total_3 and dist_Total_2 < dist_Total_1:
    df.loc[pointIDX, "class"] = 2

你的距离计算似乎也不符合我对欧几里德距离的理解。也许是这个

dist_Total_1 = sum([(a-b)**2 for a,b in zip(point,mean_1)])**0.5

依此类推dist_Total_2和dist_Total_3