Pyspark:解析一列json字符串

Question

我有一个pyspark数据框,由一列调用json,其中每一行都是一个json的unicode字符串.我想解析每一行并返回一个新的数据帧,其中每一行都是解析的json.

# Sample Data Frame
jstr1 = u'{"header":{"id":12345,"foo":"bar"},"body":{"id":111000,"name":"foobar","sub_json":{"id":54321,"sub_sub_json":{"col1":20,"col2":"somethong"}}}}'
jstr2 = u'{"header":{"id":12346,"foo":"baz"},"body":{"id":111002,"name":"barfoo","sub_json":{"id":23456,"sub_sub_json":{"col1":30,"col2":"something else"}}}}'
jstr3 = u'{"header":{"id":43256,"foo":"foobaz"},"body":{"id":20192,"name":"bazbar","sub_json":{"id":39283,"sub_sub_json":{"col1":50,"col2":"another thing"}}}}'
df = sql_context.createDataFrame([Row(json=jstr1),Row(json=jstr2),Row(json=jstr3)])

我已尝试使用以下方法映射每一行json.loads:

(df
  .select('json')
  .rdd
  .map(lambda x: json.loads(x))
  .toDF()
).show()

但这会返回一个 TypeError: expected string or buffer

我怀疑问题的一部分是,当从a转换为a dataframe时rdd,架构信息会丢失,所以我也尝试手动输入架构信息:

schema = StructType([StructField('json', StringType(), True)])
rdd = (df
  .select('json')
  .rdd
  .map(lambda x: json.loads(x))
)
new_df = sql_context.createDataFrame(rdd, schema)
new_df.show()

但我也是这样TypeError.

看看这个答案,看起来平坦化行flatMap可能在这里很有用,但我也没有成功:

schema = StructType([StructField('json', StringType(), True)])
rdd = (df
  .select('json')
  .rdd
  .flatMap(lambda x: x)
  .flatMap(lambda x: json.loads(x))
  .map(lambda x: x.get('body'))
)
new_df = sql_context.createDataFrame(rdd, schema)
new_df.show()

我收到这个错误:AttributeError: 'unicode' object has no attribute 'get'.

Answer 1

对于Spark 2.1+,您可以使用from_json它来保存数据帧中的其他非json列,如下所示:

from pyspark.sql.functions import from_json, col
json_schema = spark.read.json(df.rdd.map(lambda row: row.json)).schema
df.withColumn('json', from_json(col('json'), json_schema))

您让Spark派生出json字符串列的模式.然后该df.json列不再是StringType,而是正确解码的json结构,即嵌套的StrucType和所有其他列df保持原样.

您可以按如下方式访问json内容:

df.select(col('json.header').alias('header'))

Answer 2

如果您之前将数据帧转换为字符串的RDD,则将带有json字符串的数据帧转换为结构化数据帧实际上非常简单(请参阅:http://spark.apache.org/docs/latest/sql-programming-guide . html #json-datasets)

例如:

>>> new_df = sql_context.read.json(df.rdd.map(lambda r: r.json))
>>> new_df.printSchema()
root
 |-- body: struct (nullable = true)
 |    |-- id: long (nullable = true)
 |    |-- name: string (nullable = true)
 |    |-- sub_json: struct (nullable = true)
 |    |    |-- id: long (nullable = true)
 |    |    |-- sub_sub_json: struct (nullable = true)
 |    |    |    |-- col1: long (nullable = true)
 |    |    |    |-- col2: string (nullable = true)
 |-- header: struct (nullable = true)
 |    |-- foo: string (nullable = true)
 |    |-- id: long (nullable = true)

Answer 3

如果您的JSON格式不是完全/传统格式，则现有答案不起作用。例如，基于RDD的模式推断期望大括号中包含JSON，{}并且在null例如数据如下的情况下，将提供不正确的模式（导致值）：

[
  {
    "a": 1.0,
    "b": 1
  },
  {
    "a": 0.0,
    "b": 2
  }
]

我编写了一个函数来解决此问题，方法是清理JSON，使其驻留在另一个JSON对象中：

def parseJSONCols(df, *cols, sanitize=True):
    """Auto infer the schema of a json column and parse into a struct.

    rdd-based schema inference works if you have well-formatted JSON,
    like ``{"key": "value", ...}``, but breaks if your 'JSON' is just a
    string (``"data"``) or is an array (``[1, 2, 3]``). In those cases you
    can fix everything by wrapping the data in another JSON object
    (``{"key": [1, 2, 3]}``). The ``sanitize`` option (default True)
    automatically performs the wrapping and unwrapping.

    The schema inference is based on this
    `SO Post </sf/answers/3211640211/)/>`_.

    Parameters
    ----------
    df : pyspark dataframe
        Dataframe containing the JSON cols.
    *cols : string(s)
        Names of the columns containing JSON.
    sanitize : boolean
        Flag indicating whether you'd like to sanitize your records
        by wrapping and unwrapping them in another JSON object layer.

    Returns
    -------
    pyspark dataframe
        A dataframe with the decoded columns.
    """
    res = df
    for i in cols:

        # sanitize if requested.
        if sanitize:
            res = (
                res.withColumn(
                    i,
                    psf.concat(psf.lit('{"data": '), i, psf.lit('}'))
                )
            )
        # infer schema and apply it
        schema = spark.read.json(res.rdd.map(lambda x: x[i])).schema
        res = res.withColumn(i, psf.from_json(psf.col(i), schema))

        # unpack the wrapped object if needed
        if sanitize:
            res = res.withColumn(i, psf.col(i).data)
    return res

注意：psf= pyspark.sql.functions。

Answer 4

这是 @nolan-conaway 函数的简洁 (spark SQL) 版本parseJSONCols。

SELECT 
explode(
    from_json(
        concat('{"data":', 
               '[{"a": 1.0,"b": 1},{"a": 0.0,"b": 2}]', 
               '}'), 
        'data array<struct<a:DOUBLE, b:INT>>'
    ).data) as data;

附言。我还添加了爆炸功能：P

您需要了解一些HIVE SQL 类型