我有一个很大的json文件，其中包含许多有关结果数据的调查。该文件非常复杂，但是我设法使用嵌套循环生成表。解决方案显然是耗时的，而且我相信使用purrrr或dt可以找到更好的方法。不幸的是，我无法共享数据，因此我附上了一个适用于循环的示例：
{
"research": {
"re.id": "id"
},
"surveys": [
{
"id": "10017",
"participant": "31804190",
"answers": [
{
"item": "685022",
"results": [
{
"items": "o",
"value": 2
},
{
"items": "pb",
"value": 3
},
{
"items": "r",
"value": 2
},
{
"items": "s",
"value": 0
},
{
"items": "t",
"value": 0
},
{
"items": "w",
"value": 0
},
{
"items": "z",
"value": 0
},
{
"items": "f",
"value": 2
},
{
"items": "e",
"value": 1
},
{
"items": "l",
"value": 0
}
]
},
{
"item": "90118",
"results": [
{
"items": "o",
"value": 0
},
{
"items": "pb",
"value": 3
},
{
"items": "r",
"value": 1
},
{
"items": "s",
"value": 0
},
{
"items": "t",
"value": 0
},
{
"items": "w",
"value": 0
},
{
"items": "z",
"value": 0
},
{
"items": "f",
"value": 1
},
{
"items": "e",
"value": 1
},
{
"items": "l",
"value": 0
}
]
},
{
"item": "30094",
"results": [
{
"items": "o",
"value": 0
},
{
"items": "pb",
"value": 2
},
{
"items": "r",
"value": 0
},
{
"items": "s",
"value": 1
},
{
"items": "t",
"value": 2
},
{
"items": "w",
"value": 1
},
{
"items": "z",
"value": 0
},
{
"items": "f",
"value": 0
},
{
"items": "e",
"value": -3
},
{
"items": "l",
"value": 3
}]}]}]}
请注意，357个调查中的每个调查都包含13个数据点，最后一个是25个答案的列表，其中包含10个结果。嵌套循环遍历所有调查，不包括未完成的调查，然后另一个循环遍历每个调查的所有答案，而下一个循环遍历每个调查中每个答案的所有结果。而且，结果是以随机顺序给出的，因此需要对其进行排序，这是在嵌套循环的末尾发生的。
以下是该JSON的示例：https://jsonblob.com/2bffde45-d8c0-11e9-9ec2-759b3e404be9
library(rjson)
survey.report <- fromJSON(file = "sample.json")
values.df <- data.frame(matrix(ncol = 13, nrow = 0))
listofitems <- list()
for (a in 1:length(surveytoll[["surveys"]])) {
if (length(surveytoll[["surveys"]][[a]][["answers"]]) == 0) {
listofitems[i] <- 0
i <- i + 1
} else {
for (o in 1:length(surveytoll[["surveys"]][[a]][["answers"]])) {
surveyid <- surveytoll[["surveys"]][[a]][["answers"]][[o]][["item"]]
i <- i + 1
listofitems[i] <- surveyid
for (w in 1:length(surveytoll[["surveys"]][[a]][["answers"]][[o]][["results"]])) {
answers <- surveytoll[["surveys"]][[a]][["answers"]][[o]][["results"]]
}
survey = surveytoll[["surveys"]][[a]][["id"]]
participant = surveytoll[["surveys"]][[a]][["participant"]]
list.items <- list()
list.values <- list()
list.items <- sapply(answers, function(x) x[[1]])
list.values <- sapply(answers, function(x) x[[2]])
lwtt <- t(t(list.values))
lpbtt <- t(t(list.items))
lplw <- cbind.data.frame(lpbtt,lwtt)
lplw$lwtt <- as.character(lplw$lwtt)
lplw$lpbtt <- as.character(lplw$lpbtt)
lplw[order(lplw$lpbtt),]
value <- t(lplw$lwtt)
values.row <- cbind(survey,participant,listofitems[i],value)
values.df <- rbind.data.frame(values.df,values.row)
}
}
}
values.df
循环将生成包含13列的数据框：调查ID，参与者，项目和10个结果。该示例只有3行：
survey participant V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13
1 10017 31804190 685022 2 3 2 0 0 0 0 2 1 0
2 10017 31804190 90118 0 3 1 0 0 0 0 1 1 0
3 10017 31804190 30094 0 2 0 1 2 1 0 0 -3 3
问题在于主json确实很大（10GB），因此循环将需要数周;）。
答案 0 :(得分：1)
因此，您可以利用unnest_wider（在unnest_longer中）中相对较新的tidyr和tidyverse函数-请参阅this vignette on "rectangling: for more information on extracting information from deeply nested structures
library(tidyverse)
json <- jsonlite::fromJSON('sample.json', simplifyVector = FALSE)
surveys <- tibble(survey = json$surveys)
surveys %>%
unnest_wider(survey) %>%
unnest_longer(answers) %>%
unnest_wider(answers) %>%
unnest_longer(results) %>%
unnest_wider(results) %>%
pivot_wider(
names_from = items,
values_from = value
)
# # A tibble: 3 x 13
# id participant item o pb r s t w z f e l
# <chr> <chr> <chr> <int> <int> <int> <int> <int> <int> <int> <int> <int> <int>
# 1 10017 31804190 685022 2 3 2 0 0 0 0 2 1 0
# 2 10017 31804190 90118 0 3 1 0 0 0 0 1 1 0
# 3 10017 31804190 30094 0 2 0 1 2 1 0 0 -3 3