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python – PySpark:TypeError:condition应该是string或Column

作者:互联网

我试图过滤基于如下的RDD:

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

但是得到了以下错误:

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!

解决方法:

DataFrame.filter是DataFrame.where的别名,它需要一个表达为Column的SQL表达式:

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

或等效的SQL字符串:

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:

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

并且不会受益于SQL优化.

标签:python,apache-spark,dataframe,pyspark,apache-spark-sql
来源: https://codeday.me/bug/20191001/1840675.html