在R中使用多个输入编写JSON

问题描述

我有一个小标题和一个列表，我想将其写入json文件.

I have a tibble and a list which I would like to write to a json file.

# A tibble: 2 x 12
  i     n             c                  x
  <chr> <chr>         <chr>              <chr>
1 NYC   New York City United States      LON,271;BOS,201
2 LON   London        United Kingdom     NYC,270

我想用列表替换'x'列.

I would like to replace the 'x' column with a list.

当我尝试通过'i'列与列表的元素合并时，很多数据被重复...:/

When I try to merge by the 'i' column with the element of the list, a lot of data is duplicated... :/

样本列表:

$NYC
     d   p
1: LON 271
2: BOS 201

$LON
     d   p
1: NYC 270

我想得到的最终结果是这样的:

I would like to end up with something that looks like this:

[
  {
    "i": "NYC",
    "n": "New York City",
    "c": "United States",
    "C": "US",
    "r": "Northern America",
    "F": 66.256,
    "L": -166.063,
    "b": 94.42,
    "s": 0.752,
    "q": 4417,
    "t": "0,0,0,0,0",
    "x": [{
              "d": "LON",
              "p": 271
          },
          {
              "d": "BOS",
              "p": 201
          }]
}
  ...
]

我在想应该有一种方法来写json文件而不合并列表和小标题，或者也许有一种方法可以将它们合并到?

I'm thinking there should be a way to write the json file without merging the list and the tibble, or maybe there is a way to merge them in a ragged way ?

啊.我只是有另一个主意.也许我可以将数据框转换为列表，然后使用Reduce组合列表...

ah. I just had another idea. maybe I can convert my dataframe to a list then use Reduce to combine the lists...

http://www.sharecsv.com/s/2e1dc764430c6fe746d2299f71879c2e/routes-before-split.csv

http://www.sharecsv.com/s/b114e2cc6236bd22b23298035fb7e042/tibble.csv

推荐答案

我们可以执行以下操作:

We may do the following:

tbl
# A tibble: 1 x 13
#       X i     n      c             C     r                 F      L     b     s     q t        x
#   <int> <fct> <fct>  <fct>         <fct> <fct>         <dbl>  <dbl> <dbl> <dbl> <int> <fct>    <fct>
# 1     1 LON   London United Kingd… GB    Northern Eur…  51.5 -0.127  55.4  1.25  2088 0,0,1,3… AAL,15;AAR,15;A…

require(tidyverse)
tbl$x <- map(tbl$x, ~ strsplit(., ";|,")[[1]] %>%
             {data.frame(d = .[c(T, F)], p = as.numeric(.[c(F, T)]))})

后两行是此基本R等效项的简化版本:

The latter two lines are a shortened version of this base R equivalent:

tbl$x <- lapply(tbl$x, function(r) {
  tmp <- strsplit(r, ";|,")[[1]]
  data.frame(d = tmp[seq(1, length(tmp), 2)],
             p = as.numeric(tmp[seq(2, length(tmp), 2)]))
})

我们遍历x列，在可能的情况下将其元素除以;和,，然后使用得出的奇数元素将与d列相对应的事实，得到期望的结果，并且p列中的偶数元素.

We go over the x column, split its elements by ; and , whenever possible, and then use the fact that the resulting odd elements will correspond do the d column in the desired outcome, and the even elements to the p column.

输出:

toJSON(tbl, pretty = TRUE)
[
  {
    "X": 1,
    "i": "LON",
    "n": "London",
    "c": "United Kingdom",
    "C": "GB",
    "r": "Northern Europe",
    "F": 51.508,
    "L": -0.127,
    "b": 55.43,
    "s": 1.25,
    "q": 2088,
    "t": "0,0,1,3,1",
    "x": [
      {
        "d": "AAL",
        "p": 15
      },
      {
        "d": "AAR",
        "p": 15
      },
      {
        "d": "ABZ",
        "p": 48
      }
    ]
  }
]

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