我有两个spark DataFrame。
架构DataFrame A(存储群集质心):
public void updateSchool(String username, String password, String school) throws Exception {
User user = userDao.findByUsername(userName);
if (user != null && user.getPassword().equals(password)) {
user.setSchool(school);
userDao.save(user);
}
}
DataFrame B架构(数据点):
cluster_id, dim1_pos, dim2_pos, dim3_pos, ..., dimN_pos
DataFrame A中有大约100行,这意味着我有100个群集质心。我需要将DataFrame B中的每个实体映射到最接近的簇(就欧几里德距离而言)。
我该怎么做?
我想要一个带有架构的DataFrame: entity_id,cluster_id 作为我的最终结果。
答案 0 :(得分:2)
我最终使用VectorAssembler来放置所有dimX列'值为单个列(对于每个数据帧)。
一旦完成,我只是使用UDF的组合来回答。
import numpy as np
featureCols = [dim1_pos, dim2_pos, ..., dimN_pos]
vecAssembler = VectorAssembler(inputCols=featureCols, outputCol="features")
dfA = vecAssembler.transform(dfA)
dfB = vecAssembler.transform(dfB)
def distCalc(a, b):
return np.sum(np.square(a-b))
def closestPoint(point_x, centers):
udf_dist = udf(lambda x: distCalc(x, point_x), DoubleType())
centers = centers.withColumn('distance',udf_dist(centers.features))
centers.registerTempTable('t1')
bestIndex = #write a query to get minimum distance from centers df
return bestIndex
udf_closestPoint = udf(lambda x: closestPoint(x, dfA), IntegerType())
dfB = dfB.withColumn('cluster_id',udf_closestPoint(dfB.features))
答案 1 :(得分:1)
如果Spark数据帧不是很大,您可以使用toPandas()
将其变为pandas数据框并使用scipy.spatial.distance.cdist()
(阅读this了解更多信息)
示例代码:
import pandas as pd
from scipy.spatial.distance import cdist
cluster = DataFrame({'cluster_id': [1, 2, 3, 7],
'dim1_pos': [201, 204, 203, 204],
'dim2_pos':[55, 40, 84, 31]})
entity = DataFrame({'entity_id': ['A', 'B', 'C'],
'dim1_pos': [201, 204, 203],
'dim2_pos':[55, 40, 84]})
cluster.set_index('cluster_id',inplace=True)
entity.set_index('entity_id',inplace=True)
result_metric= cdist(cluster, entity, metric='euclidean')
result_df = pd.DataFrame(result_metric,index=cluster.index.values,columns=entity.index.values)
print result_df
A B C
1 0.000000 15.297059 29.068884
2 15.297059 0.000000 44.011362
3 29.068884 44.011362 0.000000
7 24.186773 9.000000 53.009433
然后,您可以使用idxmin()
指定轴,以查找指标每行的最小对,如下所示:
# get the min. pair
result = DataFrame(result_df.idxmin(axis=1,skipna=True))
# turn the index value into column
result.reset_index(level=0, inplace=True)
# rename and order the columns
result.columns = ['cluster_id','entity_id']
result = result.reindex(columns=['entity_id','cluster_id'])
print result
entity_id cluster_id
0 A 1
1 B 2
2 C 3
3 B 7