我已经用以下的100+结构将百万条记录存储在文件中(真正的数据有更多的列、行和嵌套)
{"id":"2-2-3","key":"value"}{"id":"2-2-3","key":"value"}{"id":"2-2-3","key":"value"}{"id":"2-2-3","key":"value"}{"id":"2-2-3","key":"value"}sqlContext.read.json函数无法解析这一点,因为记录不是在多行上,而是在一条大行上。下面的解决方案解决了这个问题,但却是一个很大的性能杀手。在Apache中处理这个问题的最佳方法是什么?
val rdd = sc.wholeTextFiles("s3://some-bucket/**/*")
val validJSON = rdd.flatMap(_._2.replace("}{", "}\n{").split("\n"))
val df = sqlContext.read.json(validJSON)
df.count()
df.select("id").show()发布于 2018-04-30 16:32:20
这是Antot's answer的一个片段,它应该处理嵌套的JSON
input.toVector
.foldLeft((false, Vector.empty[Char], Vector.empty[String])) {
case ((true, charAccum, strAccum), '{') => (false, Vector('{'), strAccum :+ charAccum.mkString);
case ((_, charAccum, strAccum), '}') => (true, charAccum :+ '}', strAccum);
case ((_, charAccum, strAccum), char) => (false, charAccum :+ char, strAccum)
}
._3它所做的基本上是将数据拆分为一个Vector[Char],并使用foldLeft将输入聚合到子字符串中。诀窍是跟踪有关前一个字符的足够信息,以确定{是否标志着新对象的开始。
我使用这个输入来测试它(基本上是OP的示例输入,其中包含一个嵌套的对象):
val input = """{"id":"2-2-3","key":{ "test": "value"}}{"id":"2-2-3","key":"value"}{"id":"2-2-3","key":"value"}{"id":"2-2-3","key":"value"}{"id":"2-2-3","key":"value"}"""得到了这样的结果,看起来很好:
Vector({"id":"2-2-3","key":{ "test": "value"}},
{"id":"2-2-3","key":"value"},
{"id":"2-2-3","key":"value"},
{"id":"2-2-3","key":"value"})发布于 2018-04-27 17:57:07
原始方法的问题是调用_._2.replace("}{", "}\n{",它从输入字符串中创建另一个巨大的字符串,插入新的行字符,然后再将其分割成一个数组。
通过最小化中间字符串的创建并尽快检索目标字符串,可以进行改进。为此,我们可以使用一些子字符串:
val validJson = rdd.flatMap(rawJson => {
// functions extracted to make it more readable.
def nextObjectStartIndex(fromIndex: Int):Int = rawJson._2.indexOf('{', fromIndex)
def currObjectEndIndex(fromIndex: Int): Int = rawJson._2.indexOf('}', fromIndex)
def extractObject(fromIndex: Int, toIndex: Int): String = rawJson._2.substring(fromIndex, toIndex + 1)
// the resulting strings are put in a local buffer
val buffer = new ListBuffer[String]()
// init the scanning of the input string
var posStartNextObject = nextObjectStartIndex(0)
// main loop terminates when there are no more '{' chars
while (posStartNextObject != -1) {
val posEndObject = currObjectEndIndex(posStartNextObject)
val extractedObject = extractObject(posStartNextObject, posEndObject)
posStartNextObject = nextObjectStartIndex(posEndObject)
buffer += extractedObject
}
buffer
})请注意,只有当输入JSON中的对象不是嵌套的,假设所有大括号都是相同级别的单独对象时,这种方法才能工作。
https://stackoverflow.com/questions/50066307
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