k表示欧几里得到每个质心的距离,避免从df的其余部分分离特征。

我有一个测向仪:

    id      Type1   Type2   Type3   
0   10000   0.0     0.00    0.00    
1   10001   0.0     63.72   0.00    
2   10002   473.6   174.00  31.60   
3   10003   0.0     996.00  160.92  
4   10004   0.0     524.91  0.00

我将k-均值应用于这个df,并将结果集群添加到df:

kmeans = cluster.KMeans(n_clusters=5, random_state=0).fit(df.drop('id', axis=1))
df['cluster'] = kmeans.labels_

现在,我尝试在df中添加每个点(即df中的行)和每个质心之间的欧几里得距离的列:

def distance_to_centroid(row, centroid):
    row = row[['Type1',
               'Type2',
               'Type3']]
    return euclidean(row, centroid)

df['distance_to_center_0'] = df.apply(lambda r: distance_to_centroid(r, kmeans.cluster_centers_[0]),1)

这将导致此错误:

---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-34-56fa3ae3df54> in <module>()
----> 1 df['distance_to_center_0'] = df.apply(lambda r: distance_to_centroid(r, kmeans.cluster_centers_[0]),1)

~\_installed\anaconda\lib\site-packages\pandas\core\frame.py in apply(self, func, axis, broadcast, raw, reduce, result_type, args, **kwds)
   6002                          args=args,
   6003                          kwds=kwds)
-> 6004         return op.get_result()
   6005 
   6006     def applymap(self, func):

~\_installed\anaconda\lib\site-packages\pandas\core\apply.py in get_result(self)
    140             return self.apply_raw()
    141 
--> 142         return self.apply_standard()
    143 
    144     def apply_empty_result(self):

~\_installed\anaconda\lib\site-packages\pandas\core\apply.py in apply_standard(self)
    246 
    247         # compute the result using the series generator
--> 248         self.apply_series_generator()
    249 
    250         # wrap results

~\_installed\anaconda\lib\site-packages\pandas\core\apply.py in apply_series_generator(self)
    275             try:
    276                 for i, v in enumerate(series_gen):
--> 277                     results[i] = self.f(v)
    278                     keys.append(v.name)
    279             except Exception as e:

<ipython-input-34-56fa3ae3df54> in <lambda>(r)
----> 1 df['distance_to_center_0'] = df.apply(lambda r: distance_to_centroid(r, kmeans.cluster_centers_[0]),1)

<ipython-input-33-7b988ca2ad8c> in distance_to_centroid(row, centroid)
      7                 'atype',
      8                 'anothertype']]
----> 9     return euclidean(row, centroid)

~\_installed\anaconda\lib\site-packages\scipy\spatial\distance.py in euclidean(u, v, w)
    596 
    597     """
--> 598     return minkowski(u, v, p=2, w=w)
    599 
    600 

~\_installed\anaconda\lib\site-packages\scipy\spatial\distance.py in minkowski(u, v, p, w)
    488     if p < 1:
    489         raise ValueError("p must be at least 1")
--> 490     u_v = u - v
    491     if w is not None:
    492         w = _validate_weights(w)

ValueError: ('operands could not be broadcast together with shapes (7,) (8,) ', 'occurred at index 0')

出现此错误的原因似乎是 id 不包括在 row 函数中的变量 distance_to_centroid . 为了解决这个问题,我可以把df分成两部分( 身份证件 在df1和df2中的其余列)。但是,这是非常手动的,不允许轻松更改列。有没有一种方法可以在不拆分原始df的情况下将到每个质心的距离转换为原始df?同样,有没有更好的方法可以找到欧几里得距离,而不需要手动将列输入 行 变量,以及手动创建多少列作为集群?

预期结果:

    id      Type1   Type2   Type3   cluster    distanct_to_cluster_0
0   10000   0.0     0.00    0.00    1          2.3
1   10001   0.0     63.72   0.00    2          3.6 
2   10002   473.6   174.00  31.60   0          0.5 
3   10003   0.0     996.00  160.92  3          3.7 
4   10004   0.0     524.91  0.00    4          1.8