PySpark:类型错误:条件应该是字符串或列 [英] PySpark: TypeError: condition should be string or Column

问题描述

我正在尝试过滤基于如下的 RDD:

I am trying to filter an RDD based like below:

spark_df = sc.createDataFrame(pandas_df)
spark_df.filter(lambda r: str(r['target']).startswith('good'))
spark_df.take(5)

但出现以下错误:

TypeErrorTraceback (most recent call last)
<ipython-input-8-86cfb363dd8b> in <module>()
      1 spark_df = sc.createDataFrame(pandas_df)
----> 2 spark_df.filter(lambda r: str(r['target']).startswith('good'))
      3 spark_df.take(5)

/usr/local/spark-latest/python/pyspark/sql/dataframe.py in filter(self, condition)
    904             jdf = self._jdf.filter(condition._jc)
    905         else:
--> 906             raise TypeError("condition should be string or Column")
    907         return DataFrame(jdf, self.sql_ctx)
    908 

TypeError: condition should be string or Column

知道我错过了什么吗?谢谢！

Any idea what I missed? Thank you!

推荐答案

DataFrame.filter，它是 DataFrame.where 的别名，需要一个 SQL 表达式表达作为 Column:

DataFrame.filter, which is an alias for DataFrame.where, expects a SQL expression expressed either as a Column:

spark_df.filter(col("target").like("good%"))

或等效的 SQL 字符串:

or equivalent SQL string:

spark_df.filter("target LIKE 'good%'")

我相信你在这里尝试使用 RDD.filter 这是完全不同的方法:

I believe you're trying here to use RDD.filter which is completely different method:

spark_df.rdd.filter(lambda r: r['target'].startswith('good'))

并且不会从 SQL 优化中受益.

and does not benefit from SQL optimizations.

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