如何在 DataFrame 中将 Column 声明为分类特征以在 ml 中使用

How can I declare a Column as a categorical feature in a DataFrame for use in ml

如何声明我的 DataFrame 中的给定列包含分类信息?

我有一个从数据库加载的 Spark SQL DataFrame。此 DataFrame 中的许多列都有分类信息,但它们被编码为 Longs(出于隐私考虑)。

我希望能够告诉 spark-ml,尽管此列是数值列,但信息实际上是分类的。类别索引可能有一些漏洞,这是可以接受的。 (例如,一列可能具有值 [1, 0, 0 ,4])

我知道存在 StringIndexer 但我更愿意避免编码和解码的麻烦,特别是因为我有很多列都有这种行为。

我会寻找如下所示的东西

train = load_from_database()
categorical_cols = ["CategoricalColOfLongs1",
                    "CategoricalColOfLongs2"]
numeric_cols = ["NumericColOfLongs1"]

## This is what I am looking for
## this step detects the min and max value of both columns
## and adds metadata to indicate this as a categorical column
## with (1 + max - min) categories
categorizer = ColumnCategorizer(columns = categorical_cols,
                                autoDetectMinMax = True)
##

vectorizer = VectorAssembler(inputCols = categorical_cols + 
                                         numeric_cols,
                             outputCol = "features")
classifier = DecisionTreeClassifier()
pipeline = Pipeline(stages = [categorizer, vectorizer, classifier])
model = pipeline.fit(train)

I would prefer to avoid the hassle of encoding and decoding,

你无法真正完全避免这一点。分类变量所需的元数据实际上是值和索引之间的映射。不过,无需手动或 执行此操作。假设您有这样的数据框:

import numpy as np
import pandas as pd

df = sqlContext.createDataFrame(pd.DataFrame({
    "x1": np.random.random(1000),
    "x2": np.random.choice(3, 1000),
    "x4": np.random.choice(5, 1000)
}))

您只需要一个汇编器和索引器:

from pyspark.ml.feature import VectorAssembler, VectorIndexer
from pyspark.ml import Pipeline

pipeline = Pipeline(stages=[
    VectorAssembler(inputCols=df.columns, outputCol="features_raw"),
    VectorIndexer(
        inputCol="features_raw", outputCol="features", maxCategories=10)])

transformed = pipeline.fit(df).transform(df)
transformed.schema.fields[-1].metadata

## {'ml_attr': {'attrs': {'nominal': [{'idx': 1,
##      'name': 'x2',
##      'ord': False,
##      'vals': ['0.0', '1.0', '2.0']},
##     {'idx': 2,
##      'name': 'x4',
##      'ord': False,
##      'vals': ['0.0', '1.0', '2.0', '3.0', '4.0']}],
##    'numeric': [{'idx': 0, 'name': 'x1'}]},
##   'num_attrs': 3}}

此示例还显示了您提供哪些类型信息以将向量的给定元素标记为分类变量

{
    'idx': 2,  # Index (position in vector)
    'name': 'x4',  # name
    'ord': False,  # is ordinal?
    # Mapping between value and label
    'vals': ['0.0', '1.0', '2.0', '3.0', '4.0']  
}

因此,如果您想从头开始构建它,您所要做的就是正确的架构:

from pyspark.sql.types import *
from pyspark.mllib.linalg import VectorUDT

# Lets assume we have only a vector
raw = transformed.select("features_raw")

# Dictionary equivalent to transformed.schema.fields[-1].metadata shown abov
meta = ... 
schema = StructType([StructField("features", VectorUDT(), metadata=meta)])

sqlContext.createDataFrame(raw.rdd, schema)

但由于需要序列化、反序列化,效率很低

Spark 2.2 开始,您还可以使用元数据参数:

df.withColumn("features", col("features").alias("features", metadata=meta))

另见

嘿 zero323 我使用了相同的技术来查看元数据并编写了这个 Transformer

def _transform(self, data):
    maxValues = self.getOrDefault(self.maxValues)
    categoricalCols = self.getOrDefault(self.categoricalCols)

    new_schema = types.StructType(data.schema.fields)
    new_data = data
    for (col, maxVal) in zip(categoricalCols, maxValues):
        # I have not decided if I should make a new column or
        # overwrite the original column
        new_col_name = col + "_categorical"

        new_data = new_data.withColumn(new_col_name,
                                       data[col].astype(types.DoubleType()))

        # metadata for a categorical column                                                                                                                                 
        meta = {u'ml_attr' : {u'vals' : [unicode(i) for i in range(maxVal + 1)],
                              u'type' : u'nominal',
                              u'name' : new_col_name}}

        new_schema.add(new_col_name, types.DoubleType(), True, meta)

    return data.sql_ctx.createDataFrame(new_data.rdd, new_schema)