Pyspark将标准列表转换为数据框 [英] Pyspark convert a standard list to data frame

问题描述

情况真的很简单，我需要使用以下代码将python列表转换为数据框

The case is really simple, I need to convert a python list into data frame with following code

from pyspark.sql.types import StructType
from pyspark.sql.types import StructField
from pyspark.sql.types import StringType, IntegerType

schema = StructType([StructField("value", IntegerType(), True)])
my_list = [1, 2, 3, 4]
rdd = sc.parallelize(my_list)
df = sqlContext.createDataFrame(rdd, schema)

df.show()

它失败，并出现以下错误:

it failed with following error:

    raise TypeError("StructType can not accept object %r in type %s" % (obj, type(obj)))
TypeError: StructType can not accept object 1 in type <class 'int'>

推荐答案

该解决方案也是一种使用更少代码，避免序列化为RDD且可能更易于理解的方法:

This solution is also an approach that uses less code, avoids serialization to RDD and is likely easier to understand:

from pyspark.sql.types import IntegerType

# notice the variable name (more below)
mylist = [1, 2, 3, 4]

# notice the parens after the type name
spark.createDataFrame(mylist, IntegerType()).show()

注意:关于命名变量list:术语list是Python内置函数，因此，强烈建议避免使用内置名称作为变量的名称/标签，因为这样最终会覆盖list()函数之类的东西.快速而肮脏的原型制作时，许多人都使用诸如mylist.

NOTE: About naming your variable list: the term list is a Python builtin function and as such, it is strongly recommended that we avoid using builtin names as the name/label for our variables because we end up overwriting things like the list() function. When prototyping something fast and dirty, a number of folks use something like: mylist.

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