Question

我正在r中做一个项目，正在分析文本。我有

形式的字符串

 "id": 8784, "name": "Daniel", "age":"65", "gender":"M"

和这类事情。我的问题是如何获取数据，以便每个变量都可以成为自己的列，即名称列，第一个条目是Daniel。

Answer 1

最简单的方法是使用sealed trait ILevel case object FooLevel extends ILevel case object CompanionLevel extends ILevel abstract class Foo[T,I](val msg: I) { type Level <: ILevel } object Foo { implicit def provide[T]: Foo[T,String] = new Foo[T,String]("I came from a place you can't touch so subtyping can't help you") { override type Level = FooLevel.type } } class User object User { implicit object userFoo extends Foo[User,Int](42) { type Level = CompanionLevel.type } } type CompanionLevelFoo[T,I] = Foo[T,I] {type Level = CompanionLevel.type } def fooOf[T,I](U: T)(implicit tc2: CompanionLevelFoo[T,I]): Foo[T, I] = tc2 fooOf(new User).msg路径

JSON

如果有多个字符串，我们可以使用

library(jsonlite)
library(tidyverse)
sprintf("{%s}", str1) %>%
   fromJSON %>% 
   as.data.frame
#    id   name age gender
#1 8784 Daniel  65      M

更新

根据OP的评论，如果它还有sprintf("{%s}", str1) %>% map_df(fromJSON)

[{..}]

对于多个元素，

str1 <- '[{"id": 8784, "name": "Daniel", "age":"65", "gender":"M"}]'
fromJSON(str1)
#    id   name age gender
#1 8784 Daniel  65      M

数据

str1 <- c(str1, str1)
str1 %>% 
     map_df(fromJSON)
#    id   name age gender
#1 8784 Daniel  65      M
#2 8784 Daniel  65      M

文本挖掘R将文本分隔成列

1 个答案:

更新

数据