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。现在从列表中推断模式已被弃用，我收到一个警告，它建议我改用pyspark.sql.Row。但是，当我尝试使用Row 创建一个时，我遇到了推断架构问题。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)

然后，这个错误被抛出。

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

createDataFrame 函数采用 行列表（以及其他选项）加上架构，因此正确的代码类似于：

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()

输出：

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

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

在 pyspark 文档 (link) 中，您可以找到有关 createDataFrame 函数的更多详细信息。

Answer 2

您需要创建一个 Row 类型的列表并将该列表与架构一起传递给您的 createDataFrame() 方法。示例示例

from pyspark.sql import *
from pyspark.sql.types import *
from pyspark.sql.functions import *
department1 = Row(id='AAAAAAAAAAAAAA', type='XXXXX',cost='2')
department2 = Row(id='AAAAAAAAAAAAAA', type='YYYYY',cost='32')
department3 = Row(id='BBBBBBBBBBBBBB', type='XXXXX',cost='42')
department4 = Row(id='BBBBBBBBBBBBBB', type='YYYYY',cost='142')
department5 = Row(id='BBBBBBBBBBBBBB', type='ZZZZZ',cost='149')
department6 = Row(id='CCCCCCCCCCCCCC', type='XXXXX',cost='15')
department7 = Row(id='CCCCCCCCCCCCCC', type='YYYYY',cost='23')
department8 = Row(id='CCCCCCCCCCCCCC', type='ZZZZZ',cost='10')

schema = StructType([StructField('id', StringType()), StructField('type',StringType()),StructField('cost', StringType())])
rows = [department1,department2,department3,department4,department5,department6,department7,department8 ]
df = spark.createDataFrame(rows, schema)

Answer 3

如果您只是制作 pandas 数据框，则可以将每个 Row 转换为 dict，然后依赖 pandas 的类型推断，如果这足以满足您的需求。这对我有用：

import pandas as pd

sample = output.head(5) #this returns a list of Row objects

df = pd.DataFrame([x.asDict() for x in sample])

Answer 4

我最近遇到了类似的问题，这里的答案帮助我更好地理解了这个问题。

我的代码：

row = Row(name="Alice", age=11)  
spark.createDataFrame(row).show()

导致了一个非常相似的错误：

An error was encountered:  
Can not infer schema for type: <class 'int'>  
Traceback ... 

问题的原因：
createDataFrame 需要一个行数组。因此，如果您只有一行并且不想发明更多，只需将其设为数组：[row]

row = Row(name="Alice", age=11)
spark.createDataFrame([row]).show()