简体   繁体   English

PySpark,通过 JSON 文件导入模式

[英]PySpark, importing schema through JSON file

tbschema.json looks like this: tbschema.json看起来像这样:

[{"TICKET":"integer","TRANFERRED":"string","ACCOUNT":"STRING"}]

I load it using following code我使用以下代码加载它

>>> df2 = sqlContext.jsonFile("tbschema.json")
>>> f2.schema
StructType(List(StructField(ACCOUNT,StringType,true),
    StructField(TICKET,StringType,true),StructField(TRANFERRED,StringType,true)))
>>> df2.printSchema()
root
 |-- ACCOUNT: string (nullable = true)
 |-- TICKET: string (nullable = true)
 |-- TRANFERRED: string (nullable = true)
  1. Why does the schema elements gets sorted, when I want the elements in the same order as they appear in the JSON.当我希望元素的顺序与它们在 JSON 中的显示顺序相同时,为什么要对架构元素进行排序。

  2. The data type integer has been converted into StringType after the JSON has been derived, how do I retain the datatype. JSON导出后数据类型integer已经转换为StringType,如何保留数据类型。

Why does the schema elements gets sorted, when i want the elemets in the same order as they appear in the json.为什么模式元素会被排序,当我想要元素的顺序与它们在 json 中出现的顺序相同时。

Because order of fields is not guaranteed.因为不能保证字段的顺序。 While it is not explicitly stated it becomes obvious when you take a look a the examples provided in the JSON reader doctstring.虽然没有明确说明,但当您查看 JSON 阅读器文档字符串中提供的示例时,它变得显而易见。 If you need specific ordering you can provide schema manually:如果您需要特定的排序,您可以手动提供架构:

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

schema = StructType([
    StructField("TICKET", StringType(), True),
    StructField("TRANFERRED", StringType(), True),
    StructField("ACCOUNT", StringType(), True),
])
df2 = sqlContext.read.json("tbschema.json", schema)
df2.printSchema()

root
 |-- TICKET: string (nullable = true)
 |-- TRANFERRED: string (nullable = true)
 |-- ACCOUNT: string (nullable = true)

The data type integer has been converted into StringType after the json has been derived, how do i retain the datatype. json导出后数据类型integer已经转成StringType,如何保留数据类型。

Data type of JSON field TICKET is string hence JSON reader returns string. JSON 字段TICKET数据类型是字符串,因此 JSON 读取器返回字符串。 It is JSON reader not some-kind-of-schema reader.它是 JSON 阅读器,而不是某种模式阅读器。

Generally speaking you should consider some proper format which comes with schema support out-of-the-box, for example Parquet , Avro or Protocol Buffers .一般来说,您应该考虑开箱即用的模式支持附带的一些正确格式,例如ParquetAvroProtocol Buffers But if you really want to play with JSON you can define poor man's "schema" parser like this:但是如果你真的想玩 JSON 你可以像这样定义穷人的“模式”解析器:

from collections import OrderedDict 
import json

with open("./tbschema.json") as fr:
    ds = fr.read()

items = (json
  .JSONDecoder(object_pairs_hook=OrderedDict)
  .decode(ds)[0].items())

mapping = {"string": StringType, "integer": IntegerType, ...}

schema = StructType([
    StructField(k, mapping.get(v.lower())(), True) for (k, v) in items])

Problem with JSON is that there is really no guarantee regarding fields ordering whatsoever, not to mention handling missing fields, inconsistent types and so on. JSON 的问题在于,对于字段的排序确实没有任何保证,更不用说处理缺失的字段、不一致的类型等等。 So using solution as above really depends on how much you trust your data.因此,使用上述解决方案实际上取决于您对数据的信任程度。

Alternatively you can use built-in schema import / export utilities .或者,您可以使用内置模式导入/导出实用程序

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM