apache-spark - 为什么 dropna() 不起作用？

Question

系统:Cloudera Quickstart VM 5.4 上的 Spark 1.3.0 (Anaconda Python dist.)

这是一个 Spark DataFrame:

from pyspark.sql import SQLContext
from pyspark.sql.types import *
sqlContext = SQLContext(sc)

data = sc.parallelize([('Foo',41,'US',3),
                       ('Foo',39,'UK',1),
                       ('Bar',57,'CA',2),
                       ('Bar',72,'CA',3),
                       ('Baz',22,'US',6),
                       (None,75,None,7)])

schema = StructType([StructField('Name', StringType(), True),
                     StructField('Age', IntegerType(), True),
                     StructField('Country', StringType(), True),
                     StructField('Score', IntegerType(), True)])

df = sqlContext.createDataFrame(data,schema)

data.show()

Name Age Country Score
Foo  41  US      3    
Foo  39  UK      1    
Bar  57  CA      2    
Bar  72  CA      3    
Baz  22  US      6    
null 75  null    7 

但是这些都不起作用!

df.dropna()
df.na.drop()

我收到这条消息:

>>> df.show()
Name Age Country Score
Foo  41  US      3    
Foo  39  UK      1    
Bar  57  CA      2    
Bar  72  CA      3    
Baz  22  US      6    
null 75  null    7    
>>> df.dropna().show()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/lib/spark/python/pyspark/sql/dataframe.py", line 580, in __getattr__
    jc = self._jdf.apply(name)
  File "/usr/lib/spark/python/lib/py4j-0.8.2.1-src.zip/py4j/java_gateway.py", line 538, in __call__
  File "/usr/lib/spark/python/lib/py4j-0.8.2.1-src.zip/py4j/protocol.py", line 300, in get_return_value
py4j.protocol.Py4JJavaError: An error occurred while calling o50.apply.
: org.apache.spark.sql.AnalysisException: Cannot resolve column name "dropna" among (Name, Age, Country, Score);
    at org.apache.spark.sql.DataFrame$$anonfun$resolve$1.apply(DataFrame.scala:162)
    at org.apache.spark.sql.DataFrame$$anonfun$resolve$1.apply(DataFrame.scala:162)
    at scala.Option.getOrElse(Option.scala:120)
    at org.apache.spark.sql.DataFrame.resolve(DataFrame.scala:161)
    at org.apache.spark.sql.DataFrame.col(DataFrame.scala:436)
    at org.apache.spark.sql.DataFrame.apply(DataFrame.scala:426)
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:57)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
    at java.lang.reflect.Method.invoke(Method.java:606)
    at py4j.reflection.MethodInvoker.invoke(MethodInvoker.java:231)
    at py4j.reflection.ReflectionEngine.invoke(ReflectionEngine.java:379)
    at py4j.Gateway.invoke(Gateway.java:259)
    at py4j.commands.AbstractCommand.invokeMethod(AbstractCommand.java:133)
    at py4j.commands.CallCommand.execute(CallCommand.java:79)
    at py4j.GatewayConnection.run(GatewayConnection.java:207)
    at java.lang.Thread.run(Thread.java:745)

有没有其他人遇到过这个问题？解决方法是什么？ Pyspark 似乎我正在寻找一个名为“na”的列。任何帮助将不胜感激!

Answer 1

tl;dr 方法 na 和 dropna 仅从 Spark 1.3.1 开始可用。

你犯的几个错误:

1. data = sc.parallelize([....('',75,'', 7 )])，你打算使用'' 来表示None，但是它只是一个String而不是null

2. na 和 dropna 都是 dataFrame 类的方法，因此，您应该使用 df 调用它。

可运行代码:

data = sc.parallelize([('Foo',41,'US',3),
                       ('Foo',39,'UK',1),
                       ('Bar',57,'CA',2),
                       ('Bar',72,'CA',3),
                       ('Baz',22,'US',6),
                       (None, 75, None, 7)])

schema = StructType([StructField('Name', StringType(), True),
                 StructField('Age', IntegerType(), True),
                 StructField('Country', StringType(), True),
                 StructField('Score', IntegerType(), True)])

df = sqlContext.createDataFrame(data,schema)

df.dropna().show()
df.na.drop().show()