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json - 以这种格式将数据帧转换为 JSON

转载 作者:行者123 更新时间:2023-12-04 04:41:58 25 4
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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:

raw data

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")

我将需要的 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 RJSONIO::toJSON函数来做这样的事情,我觉得文档中的例子没有太大帮助。

我需要做什么才能获得正确的输出?

注意:永远只有一个 type每个 unit .

PS:有没有工具可以让这些事情变得简单?也许有些东西拖放?

最佳答案

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帧操作为正确的格式。

但是,您的 JSON 输出在向量中具有混合类型 -character 和 numeric -: ["2013-01-12 12:08:07",-21.23]我不知道如何从 R 中得到它,它坚持将向量作为单一类型。会 ["2013-01-12 12:08:07","-21.23"]可以接受吗?如果是这样,请继续阅读...
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 拆分数据帧变量。您可以对这些部分中的每一个应用一个函数并返回一个列表。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))
}

应该将一节转换为正确的列表格式。所以告诉 dlply对每个部分执行此操作,并返回列表列表。该列表具有名称,这使得 toJSON输出为命名数组 - 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?

关于json - 以这种格式将数据帧转换为 JSON,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/18762229/

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