是否可以使用pyspark过滤Spark DataFrames以返回列值在列表中的所有行？

Question

如何仅返回列的值在指定列表中的Spark DataFrame的行？

这是我的Python pandas执行此操作的方式:

df_start = df[df['name'].isin(['App Opened', 'App Launched'])].copy()

我看到了这个 SO scala实现并尝试了几种排列,但无法使其工作.

这是使用pyspark执行此操作的失败尝试:

df_start = df_spark.filter(col("name") isin ['App Opened', 'App Launched'])

输出:

Traceback (most recent call last):
  File "/tmp/zeppelin_pyspark-6660042787423349557.py", line 253, in <module>
    code = compile('\n'.join(final_code), '<stdin>', 'exec', ast.PyCF_ONLY_AST, 1)
  File "<stdin>", line 18
    df_start = df_spark.filter(col("name") isin ['App Opened', 'App Launched'])
                                               ^
SyntaxError: invalid syntax

另一种尝试:

df_start = df_spark.filter(col("name").isin(['App Opened', 'App Launched']))

输出:

Traceback (most recent call last):
  File "/tmp/zeppelin_pyspark-6660042787423349557.py", line 267, in <module>
    raise Exception(traceback.format_exc())
Exception: Traceback (most recent call last):
  File "/tmp/zeppelin_pyspark-6660042787423349557.py", line 260, in <module>
    exec(code)
  File "<stdin>", line 18, in <module>
NameError: name 'col' is not defined

Answer 1

正如dmdmdmdmdmd在评论中指出的那样,第二种方法不起作用,因为col需要导入:

from pyspark.sql.functions import col
df_start = df_spark.filter(col("name").isin(['App Opened', 'App Launched']))

这是完成过滤器的另一种方法:

df_start = df_spark.filter(df_spark.name.isin(['App Opened', 'App Launched']))