2

我有一个数据框，我希望以特定格式输出到 JSON，下面有一个小示例：

原始数据

dat <- structure(list(unit = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 
  2L), .Label = c("A", "B"), class = "factor"), type = structure(c(1L, 
  1L, 1L, 1L, 2L, 2L, 2L, 2L), .Label = c("X", "Y"), class = "factor"), 
  date = structure(c(1357963687, 1357963869, 1357964048, 1357964230, 
  1357963687, 1357963942, 1357963942, 1357964123), class = c("POSIXct", 
  "POSIXt"), tzone = ""), latitude = c(-21.21, -21.22, -21.23, 
  -21.24, -21.23, -21.23, -21.23, -21.23), longitude = c(116.78, 
  116.77, 116.76, 116.75, 116.74, 116.75, 116.75, 116.76)), .Names = c("unit", 
  "type", "date", "latitude", "longitude"), row.names = c(NA, -8L
  ), class = "data.frame")

我需要的 JSON 格式如下所示：

    [{"unit":"A","type":"X","latitude":[["2013-01-12 12:08:07",-21.21],["2013-01-12 12:11:09",-21.22],["2013-01-12 12:14:08",-21.23],["2013-01-12 12:17:10",-21.24]],
                           "longitude":[["2013-01-12 12:08:07",116.78],["2013-01-12 12:11:09",116.77],["2013-01-12 12:14:08",116.76],["2013-01-12 12:17:10",116.75]]
    },
    {"unit":"B","type":"X", "latitude":[["2013-01-12 12:08:07",-21.23],["2013-01-12 12:12:22",-21.23],["2013-01-12 12:12:22",-21.23],["2013-01-12 12:15:23",-21.23]],
                           "longitude":[["2013-01-12 12:08:07",116.74],["2013-01-12 12:12:22",116.75],["2013-01-12 12:12:22",116.75],["2013-01-12 12:15:23",116.76]]
    }]

到目前为止，我一直无法操纵该RJSONIO::toJSON函数来执行类似的操作，而且我发现文档中的示例并没有太大帮助。

我需要做什么才能获得正确的输出？

注意：type每个unit.

PS：有没有工具可以让这些事情变得简单？也许是拖放的东西？

4

1 回答 1

3

您可以通过首先将数据框转换为列表列表来接近。例如：

> a=list(unit="A",type="X",latitude=c(1,2,3),longitude=c(4,5,6))
> b=list(unit="B",type="Y",latitude=c(11,22,33),longitude=c(43,54,65))
> dlist = list(a,b)
> cat(toJSON(dlist))
[
 {
 "unit": "A",
"type": "X",
"latitude": [      1,      2,      3 ],
"longitude": [      4,      5,      6 ] 
},
{
 "unit": "B",
"type": "Y",
"latitude": [     11,     22,     33 ],
"longitude": [     43,     54,     65 ] 
} 
]

问题实际上是如何将数据框操作为正确的格式。

但是，您的 JSON 输出在向量中具有混合类型 - 字符和数字：["2013-01-12 12:08:07",-21.23]而且我不知道如何从坚持向量是单一类型的 R 中得到它。会["2013-01-12 12:08:07","-21.23"]被接受吗？如果是这样，请继续阅读...

该plyr包有很多用于拆分和操作数据框和列表的代码。例如：

dlply(dat,~unit)

将按unit变量拆分数据框。您可以将一个函数应用于每个部分并返回一个列表。这个功能：

make1 <- function(d){   
   list(
         unit=d$unit[1],
         type=d$type[1],
         latitude=cbind(as.character(d$date),d$latitude),
         longitude=cbind(as.character(d$date),d$longitude))
   }

应该将一个部分转换为正确的列表格式。所以告诉dlply每个部分都这样做，并返回一个列表列表。该列表具有名称，这使得toJSON输出作为命名数组 - 我们需要删除名称以获取 JS 列表。

> L = dlply(dat,~unit,make1)
> names(L)=NULL
> cat(toJSON(L))
 [
 {
 "unit": "A",
"type": "X",
"latitude": [ [ "2013-01-12 04:08:07", "-21.21" ],
[ "2013-01-12 04:11:09", "-21.22" ],
[ "2013-01-12 04:14:08", "-21.23" ],
[ "2013-01-12 04:17:10", "-21.24" ] ],
"longitude": [ [ "2013-01-12 04:08:07", "116.78" ],
[ "2013-01-12 04:11:09", "116.77" ],
[ "2013-01-12 04:14:08", "116.76" ],
[ "2013-01-12 04:17:10", "116.75" ] ] 
},
{
 "unit": "B",
"type": "Y",
"latitude": [ [ "2013-01-12 04:08:07", "-21.23" ],
[ "2013-01-12 04:12:22", "-21.23" ],
[ "2013-01-12 04:12:22", "-21.23" ],
[ "2013-01-12 04:15:23", "-21.23" ] ],
"longitude": [ [ "2013-01-12 04:08:07", "116.74" ],
[ "2013-01-12 04:12:22", "116.75" ],
[ "2013-01-12 04:12:22", "116.75" ],
[ "2013-01-12 04:15:23", "116.76" ] ] 
} 
]

好玩吧？

于 2013-09-12T13:10:39.907 回答