从Row创建DataFrame会导致"推断架构问题"

Question

当我开始学习PySpark时,我使用了一个列表来创建一个dataframe.现在推断列表中的模式已被弃用,我收到了警告,它建议我使用pyspark.sql.Row.但是,当我尝试创建一个使用时Row,我会推断出架构问题.这是我的代码:

>>> row = Row(name='Severin', age=33)
>>> df = spark.createDataFrame(row)

这会导致以下错误:

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/spark2-client/python/pyspark/sql/session.py", line 526, in createDataFrame
    rdd, schema = self._createFromLocal(map(prepare, data), schema)
  File "/spark2-client/python/pyspark/sql/session.py", line 390, in _createFromLocal
    struct = self._inferSchemaFromList(data)
  File "/spark2-client/python/pyspark/sql/session.py", line 322, in _inferSchemaFromList
    schema = reduce(_merge_type, map(_infer_schema, data))
  File "/spark2-client/python/pyspark/sql/types.py", line 992, in _infer_schema
    raise TypeError("Can not infer schema for type: %s" % type(row))
TypeError: Can not infer schema for type: <type 'int'>

所以我创建了一个架构

>>> schema = StructType([StructField('name', StringType()), 
...                      StructField('age',IntegerType())])
>>> df = spark.createDataFrame(row, schema)

但是,这个错误会被抛出.

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/spark2-client/python/pyspark/sql/session.py", line 526, in createDataFrame
    rdd, schema = self._createFromLocal(map(prepare, data), schema)
  File "/spark2-client/python/pyspark/sql/session.py", line 387, in _createFromLocal
    data = list(data)
  File "/spark2-client/python/pyspark/sql/session.py", line 509, in prepare
    verify_func(obj, schema)
  File "/spark2-client/python/pyspark/sql/types.py", line 1366, in _verify_type
    raise TypeError("StructType can not accept object %r in type %s" % (obj, type(obj)))
TypeError: StructType can not accept object 33 in type <type 'int'>

Answer 1

该createDataFrame函数采用行列表(以及其他选项)加上模式,因此正确的代码将类似于:

from pyspark.sql.types import *
from pyspark.sql import Row

schema = StructType([StructField('name', StringType()), StructField('age',IntegerType())])
rows = [Row(name='Severin', age=33), Row(name='John', age=48)]
df = spark.createDataFrame(rows, schema)

df.printSchema()
df.show()

日期:

root
 |-- name: string (nullable = true)
 |-- age: integer (nullable = true)

+-------+---+
|   name|age|
+-------+---+
|Severin| 33|
|   John| 48|
+-------+---+

在pyspark docs(链接)中,您可以找到有关createDataFrame函数的更多详细信息.