以下是原始的json
数据:
json_file <- '{"name":"Doe, John","group":"Red","age":{"v_0":24}}
{"name":"Doe, Jane","group":"Green","age":{"v_0":31}}
{"name":"Smith, Joan","group":"Yellow","age":{"v_0":22}}'
当我要将json_file
转换为数据框时:
library(RJSONIO)
json_file <- fromJSON(json_file)
我收到此错误:
Error: parse error: trailing garbage
:"Red","age":{"v_0":24}} {"name":"Doe, Jane","group":"Gr
(right here) ------^
我知道如果我将原始数据更改为以下数据,一切都会很好:
json_file <- '[{"name":"Doe, John","group":"Red","age":{"v_0":24}},
{"name":"Doe, Jane","group":"Green","age":{"v_0":31}},
{"name":"Smith, Joan","group":"Yellow","age":{"v_0":22}}]'
但实际上我想知道:
1)如何使用[
,,
和]
从原始数据中获取数据帧而不拆分其对象?
2)如果没有办法,如何通过将json
添加到除最后一行之外的每一行的末尾,并添加,
来拆分大型[
文件中的对象]
到文件的第一行和最后一行?
答案 0 :(得分:0)
有多种方法可以执行此操作,而无需编辑文件。
如果要使用data.frame:
library(jsonlite)
# url
zips <- stream_in(url("http://media.mongodb.org/zips.json"))
# file
json_data <- stream_in(file("path/to/file.json"))
或者如果您想要列表:
json_data_as_list <- readLines("path/to/file.json") %>% lapply(fromJSON)
答案 1 :(得分:-1)
您需要那些方括号。将以下内容另存为“ test.json”:
{
"ID":["1","2","3","4","5","6","7","8" ],
"Name":["Rick","Dan","Michelle","Ryan","Gary","Nina","Simon","Guru" ],
"Salary":["623.3","515.2","611","729","843.25","578","632.8","722.5" ],
"StartDate":[ "1/1/2012","9/23/2013","11/15/2014","5/11/2014","3/27/2015","5/21/2013",
"7/30/2013","6/17/2014"],
"Dept":[ "IT","Operations","IT","HR","Finance","IT","Operations","Finance"]
}
现在,加载所需的库并指向您刚刚保存的文件:
# Load the package required to read JSON files.
library("rjson")
# Give the input file name to the function.
result <- fromJSON(file = "C:\\Users\\Excel\\Documents\\test.json")
# Print the result.
print(result)
结果:
print(result)
$ID
[1] "1" "2" "3" "4" "5" "6" "7" "8"
$Name
[1] "Rick" "Dan" "Michelle" "Ryan" "Gary" "Nina" "Simon" "Guru"
$Salary
[1] "623.3" "515.2" "611" "729" "843.25" "578" "632.8" "722.5"
$StartDate
[1] "1/1/2012" "9/23/2013" "11/15/2014" "5/11/2014" "3/27/2015" "5/21/2013" "7/30/2013" "6/17/2014"
$Dept
[1] "IT" "Operations" "IT" "HR" "Finance" "IT" "Operations" "Finance"