保留字典中的特征定义并将特征返回给客户

Question

我有下面的字典，用于将特征定义保存为字符串。

    features = {
  "journey_email_been_sent_flag": "F.when(F.col('email_14days') > 0,F.lit(1)).otherwise(F.lit(0))",
  "journey_opened_flag": "F.when(F.col('opened_14days') > 0, F.lit(1)).otherwise(F.lit(0))"
}
retrieved_features = {}
non_retrieved_features = {}

或者将其保留为定义本身。

    features = {
  "journey_email_been_sent_flag": F.when(F.col('email_14days') > 0,F.lit(1)).otherwise(F.lit(0)),
  "journey_opened_flag": F.when(F.col('opened_14days') > 0, F.lit(1)).otherwise(F.lit(0))
}

然后下面的代码用于检索功能定义

 def feature_extract(*featurenames):
  for featurename in featurenames:
    if featurename in features:
      print(f"{featurename} : {features[featurename]}")
      retrieved_features[featurename] = features[featurename]
    else:
      print('failure')
      non_retrieved_features[featurename] = "Not Found in the feature defenition"
  return retrieved_features

这就是我调用 function 来检索功能的方式

feature_extract('journey_email_been_sent_flag','journey_opened_flag')

然而，当我试图检索未来时它不起作用，当我将定义保留在字典中时，我收到以下结果

Out[19]: {'journey_email_been_sent_flag': Column<b'CASE WHEN (email_14days > 0) THEN 1 ELSE 0 END'>}

当我在 dataframe 中调用如下特征检索时。

.withColumn('journey_email_been_sent_flag', feature_extract('journey_email_been_sent_flag'))

低于错误

AssertionError: col should be Column

Answer 1

我可以通过这种方式修复它

我将特征定义保留为定义

    features = {
  "journey_email_been_sent_flag": F.when(F.col('email_14days') > 0,F.lit(1)).otherwise(F.lit(0)),
  "journey_opened_flag": F.when(F.col('opened_14days') > 0, F.lit(1)).otherwise(F.lit(0))
}

并使用 F.lit 调用 feature_extract function

F.lit(feature_extract('journey_email_been_sent_flag').get('journey_email_been_sent_flag'))