我有一个超过100,000个json文件的列表,我想从中获取一个只包含少量变量的data.table。不幸的是文件很复杂。每个json文件的内容如下所示:
样本1
$id
[1] "10.1"
$title
$title$value
[1] "Why this item"
$itemsource
$itemsource$id
[1] "AA"
$date
[1] "1992-01-01"
$itemType
[1] "art"
$creators
list()
样本2
$id
[1] "10.2"
$title
$title$value
[1] "We need this item"
$itemsource
$itemsource$id
[1] "AY"
$date
[1] "1999-01-01"
$itemType
[1] "art"
$creators
type name firstname surname affiliationIds
1 Person Frank W. Cornell. Frank W. Cornell. a1
2 Person David A. Chen. David A. Chen. a1
$affiliations
id name
1 a1 Foreign Affairs Desk, New York Times
我从这组文件中需要的是一个包含创建者名称,项目ID和日期的表格。对于上面的两个示例文件:
id date name firstname lastname creatortype
"10.1" "1992-01-01" NA NA NA NA
"10.2" "1999-01-01" Frank W. Cornell. Frank W. Cornell. Person
"10.2" "1999-01-01" David A. Chen. David A. Chen. Person
到目前为止我做了什么:
library(parallel)
library(data.table)
library(jsonlite)
library(dplyr)
filelist = list.files(pattern="*.json",recursive=TRUE,include.dirs =TRUE)
parsed = mclapply(filelist, function(x) fromJSON(x),mc.cores=24)
data = rbindlist(mclapply(1:length(parsed), function(x) {
a = data.table(item = parsed[[x]]$id, date = list(list(parsed[[x]]$date)), name = list(list(parsed[[x]]$name)), creatortype = list(list(parsed[[x]]$creatortype))) #ignoring the firstname/lastname fields here for convenience
b = data.table(id = a$item, date = unlist(a$date), name=unlist(a$name), creatortype=unlist(a$creatortype))
return(b)
},mc.cores=24))
然而,在最后一步,我收到此错误:
"Error in rbindlist(mclapply(1:length(parsed), function(x){:
Item 1 of list is not a data.frame, data.table or list"
提前感谢您的建议。 相关问题包括: Extract data from list of lists [R] R convert json to list to data.table I want to convert JSON file into data.table in r How can read files from directory using R? Convert R data table column from JSON to data table
答案 0 :(得分:1)
从错误消息中,我想这基本上意味着mclapply()的结果之一是空的,空的意思是NULL或带有0行的data.table,或者只是在并行处理中遇到错误。 / p>
你能做的是:
在mclapply()中添加更多检查,如try-error或检查b和b的类,b是否为空
使用rbindlist时,添加参数fill = T
希望这能解决你的问题。