pyspark randomForest特性重要性:如何从列号中获取列名
pyspark randomForest特性重要性:如何从列号中获取列名
提问于 2017-07-11 10:01:47
我在spark中使用标准的(字符串索引器+一个热编码器+ randomForest)管道，如下所示
labelIndexer = StringIndexer(inputCol = class_label_name, outputCol="indexedLabel").fit(data)
string_feature_indexers = [
StringIndexer(inputCol=x, outputCol="int_{0}".format(x)).fit(data)
for x in char_col_toUse_names
]
onehot_encoder = [
OneHotEncoder(inputCol="int_"+x, outputCol="onehot_{0}".format(x))
for x in char_col_toUse_names
]
all_columns = num_col_toUse_names + bool_col_toUse_names + ["onehot_"+x for x in char_col_toUse_names]
assembler = VectorAssembler(inputCols=[col for col in all_columns], outputCol="features")
rf = RandomForestClassifier(labelCol="indexedLabel", featuresCol="features", numTrees=100)
labelConverter = IndexToString(inputCol="prediction", outputCol="predictedLabel", labels=labelIndexer.labels)
pipeline = Pipeline(stages=[labelIndexer] + string_feature_indexers + onehot_encoder + [assembler, rf, labelConverter])
crossval = CrossValidator(estimator=pipeline,
estimatorParamMaps=paramGrid,
evaluator=evaluator,
numFolds=3)
cvModel = crossval.fit(trainingData)现在，在拟合之后，我可以使用cvModel.bestModel.stages[-2].featureImportances获得随机森林和特征重要性，但这并没有给出特征/列名，而只是给出了特征编号。
我得到的结果如下：
print(cvModel.bestModel.stages[-2].featureImportances)
(1446,[3,4,9,18,20,103,766,981,983,1098,1121,1134,1148,1227,1288,1345,1436,1444],[0.109898803421,0.0967396441648,4.24568235244e-05,0.0369705839109,0.0163489685127,3.2286694534e-06,0.0208192703688,0.0815822887175,0.0466903663708,0.0227619959989,0.0850922269211,0.000113388896956,0.0924779490403,0.163835022713,0.118987129392,0.107373548367,3.35577640585e-05,0.000229569946193])如何将其映射回某些列名或列名+值格式？
基本上是为了获得随机森林的特征重要性以及列名。
回答 2
Stack Overflow用户
发布于 2017-07-11 16:17:29
嘿，你为什么不通过列表扩展把它映射回原来的列呢？下面是一个示例：
# in your case: trainingData.columns
data_frame_columns = ["A", "B", "C", "D", "E", "F"]
# in your case: print(cvModel.bestModel.stages[-2].featureImportances)
feature_importance = (1, [1, 3, 5], [0.5, 0.5, 0.5])
rf_output = [(data_frame_columns[i], feature_importance[2][j]) for i, j in zip(feature_importance[1], range(len(feature_importance[2])))]
dict(rf_output)
{'B': 0.5, 'D': 0.5, 'F': 0.5}Stack Overflow用户
发布于 2018-01-30 15:28:17
在ml算法之后，我找不到任何方法来获得真正的列的初始列表，我正在使用这个作为当前的变通方法。
print(len(cols_now))
FEATURE_COLS=[]
for x in cols_now:
if(x[-6:]!="catVar"):
FEATURE_COLS+=[x]
else:
temp=trainingData.select([x[:-7],x[:-6]+"tmp"]).distinct().sort(x[:-6]+"tmp")
temp_list=temp.select(x[:-7]).collect()
FEATURE_COLS+=[list(x)[0] for x in temp_list]
print(len(FEATURE_COLS))
print(FEATURE_COLS)我在所有的索引器(_tmp)和编码器(_catVar)上保持了一致的后缀命名，比如：
column_vec_in = str_col
column_vec_out = [col+"_catVar" for col in str_col]
indexers = [StringIndexer(inputCol=x, outputCol=x+'_tmp')
for x in column_vec_in ]
encoders = [OneHotEncoder(dropLast=False, inputCol=x+"_tmp", outputCol=y)
for x,y in zip(column_vec_in, column_vec_out)]
tmp = [[i,j] for i,j in zip(indexers, encoders)]
tmp = [i for sublist in tmp for i in sublist]这可以进一步改进和推广，但目前这种繁琐的工作效果最好。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/45024192
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