如何将 XML 解析为 R 数据框 [英] How to parse XML to R data frame

问题描述

我尝试将 XML 解析为 R 数据框，此链接对我帮助很大:

I tried to parse XML to R data frame, this link helped me a lot:

如何创建 R 数据框来自 xml 文件

但我仍然无法弄清楚我的问题:

But still I was not able to figure out my problem:

这是我的代码:

data <- xmlParse("http://forecast.weather.gov/MapClick.php?lat=29.803&lon=-82.411&FcstType=digitalDWML")
xmlToDataFrame(nodes=getNodeSet(data1,"//data"))[c("location","time-layout")]
step1 <- xmlToDataFrame(nodes=getNodeSet(data1,"//location/point"))[c("latitude","longitude")]
step2 <- xmlToDataFrame(nodes=getNodeSet(data1,"//time-layout/start-valid-time"))
step3 <- xmlToDataFrame(nodes=getNodeSet(data1,"//parameters/temperature"))[c("type="hourly"")]

我想要的数据框是这样的:

The data frame I want to have is like this:

latitude  longitude   start-valid-time   hourly_temperature
29.803     -82.411  2013-06-19T15:00:00-04:00    91
29.803     -82.411  2013-06-19T16:00:00-04:00    90

我被困在 xmlToDataFrame()，任何帮助将不胜感激，谢谢.

I'm stuck at the xmlToDataFrame(), Any help would be very much appreciated, thanks.

推荐答案

XML 格式的数据很少以允许 xmlToDataFrame 函数工作的方式组织.您最好提取列表中的所有内容，然后将这些列表绑定到一个数据框中:

Data in XML format are rarely organized in a way that would allow the xmlToDataFrame function to work. You're better off extracting everything in lists and then binding the lists together in a data frame:

require(XML)
data <- xmlParse("http://forecast.weather.gov/MapClick.php?lat=29.803&lon=-82.411&FcstType=digitalDWML")

xml_data <- xmlToList(data)

就您的示例数据而言，获取位置和开始时间非常简单:

In the case of your example data, getting location and start time is fairly straightforward:

location <- as.list(xml_data[["data"]][["location"]][["point"]])

start_time <- unlist(xml_data[["data"]][["time-layout"]][
    names(xml_data[["data"]][["time-layout"]]) == "start-valid-time"])

温度数据有点复杂.首先，您需要到达包含温度列表的节点.然后，您需要提取这两个列表，查看每个列表，然后选择将每小时"作为其值之一的列表.然后您只需要选择该列表，但只保留具有值"标签的值:

Temperature data is a bit more complicated. First you need to get to the node that contains the temperature lists. Then you need extract both the lists, look within each one, and pick the one that has "hourly" as one of its values. Then you need to select only that list but only keep the values that have the "value" label:

temps <- xml_data[["data"]][["parameters"]]
temps <- temps[names(temps) == "temperature"]
temps <- temps[sapply(temps, function(x) any(unlist(x) == "hourly"))]
temps <- unlist(temps[[1]][sapply(temps, names) == "value"])

out <- data.frame(
  as.list(location),
  "start_valid_time" = start_time,
  "hourly_temperature" = temps)

head(out)
  latitude longitude          start_valid_time hourly_temperature
1    29.81    -82.42 2013-06-19T16:00:00-04:00                 91
2    29.81    -82.42 2013-06-19T17:00:00-04:00                 90
3    29.81    -82.42 2013-06-19T18:00:00-04:00                 89
4    29.81    -82.42 2013-06-19T19:00:00-04:00                 85
5    29.81    -82.42 2013-06-19T20:00:00-04:00                 83
6    29.81    -82.42 2013-06-19T21:00:00-04:00                 80

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