使用rjson处理JSON [英] Processing JSON using rjson

问题描述

我正在尝试处理JSON格式的一些数据. rjson::fromJSON成功导入数据并将其放入一个笨拙的列表中.

I'm trying to process some data in JSON format. rjson::fromJSON imports the data successfully and places it into a quite unwieldy list.

library(rjson)
y <- fromJSON(file="http://api.lmiforall.org.uk/api/v1/wf/predict/breakdown/region?soc=6145&minYear=2014&maxYear=2020")
str(y)
List of 3
 $ soc                : num 6145
 $ breakdown          : chr "region"
 $ predictedEmployment:List of 7
  ..$ :List of 2
  .. ..$ year     : num 2014
  .. ..$ breakdown:List of 12
  .. .. ..$ :List of 3
  .. .. .. ..$ code      : num 1
  .. .. .. ..$ name      : chr "London"
  .. .. .. ..$ employment: num 74910
  .. .. ..$ :List of 3
  .. .. .. ..$ code      : num 7
  .. .. .. ..$ name      : chr "Yorkshire and the Humber"
  .. .. .. ..$ employment: num 61132
  ...

但是，由于这本质上是表格数据，因此我希望在简洁的data.frame中使用它.经过反复尝试，我得到了结果:

However, as this is essentially tabular data, I would like it in a succinct data.frame. After much trial and error I have the result:

y.p <- do.call(rbind,lapply(y[[3]], function(p) cbind(p$year,do.call(rbind,lapply(p$breakdown, function(q) data.frame(q$name,q$employment,stringsAsFactors=F))))))
head(y.p)
  p$year                   q.name q.employment
1   2014                   London     74909.59
2   2014 Yorkshire and the Humber     61131.62
3   2014     South West (England)     65833.57
4   2014                    Wales     33002.64
5   2014  West Midlands (England)     68695.34
6   2014     South East (England)     98407.36

但是该命令似乎过于复杂和复杂.有更简单的方法吗?

But the command seems overly fiddly and complex. Is there a simpler way of doing this?

推荐答案

我不确定它是否更简单，但是结果是否更完整，我认为更容易阅读.我使用Map的想法是，对于每对夫妇(年份，细目分类)，将细目分类数据汇总到单个表中，然后将其与年份合并.

I am not sure it is simpler, but the result is more complete and I think is easier to read. My idea using Map is, for each couple (year,breakdown), aggregate breakdown data into single table and then combine it with year.

dat <- y[[3]]
res <- Map(function(x,y)data.frame(year=y,
                                   do.call(rbind,lapply(x,as.data.frame))),
        lapply(dat,'[[','breakdown'),
        lapply(dat,'[[','year'))
## transform the list to a big data.frame
do.call(rbind,res)
   year code                     name employment
1  2014    1                   London   74909.59
2  2014    7 Yorkshire and the Humber   61131.62
3  2014    4     South West (England)   65833.57
4  2014   10                    Wales   33002.64
5  2014    5  West Midlands (England)   68695.34
6  2014    2     South East (England)   98407.36

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