如何将XML的一部分转换为数据帧?(适当地)
我试图从ClinicalTrials.gov的XML文件中提取信息。文件的组织方式如下:如何将XML的一部分转换为数据帧?(适当地),xml,r,Xml,R,我试图从ClinicalTrials.gov的XML文件中提取信息。文件的组织方式如下: <clinical_study> ... <brief_title> ... <location> <facility> <name> <address> <city> <state> <zip>
<clinical_study>
...
<brief_title>
...
<location>
<facility>
<name>
<address>
<city>
<state>
<zip>
<country>
</facility>
<status>
<contact>
<last_name>
<phone>
<email>
</contact>
</location>
<location>
...
</location>
...
</clinical_study>
...
...
...
...
我可以在以下代码中使用CRAN的R XML包从XML文件中提取所有位置节点:
library(XML)
clinicalTrialUrl <- "http://clinicaltrials.gov/ct2/show/NCT01480479?resultsxml=true"
xmlDoc <- xmlParse(clinicalTrialUrl, useInternalNode=TRUE)
locations <- xmlToDataFrame(getNodeSet(xmlDoc,"//location"))
库(XML)
clinicalTrialUrl您可以先展平XML
flatten_xml <- function(x) {
if (length(xmlChildren(x)) == 0) structure(list(xmlValue(x)), .Names = xmlName(xmlParent(x)))
else Reduce(append, lapply(xmlChildren(x), flatten_xml))
}
dfs <- lapply(getNodeSet(xmlDoc,"//location"), function(x) data.frame(flatten_xml(x)))
allnames <- unique(c(lapply(dfs, colnames), recursive = TRUE))
df <- do.call(rbind, lapply(dfs, function(df) { df[, setdiff(allnames,colnames(df))] <- NA; df }))
head(df)
# city state zip country status last_name phone email last_name.1
# 1 Birmingham Alabama 35294 United States Recruiting Louis B Nabors, MD 205-934-1813 bnabors@uab.edu Louis B Nabors, MD
# 2 Mobile Alabama 36604 United States Recruiting Melanie Alford, RN 251-445-9649 malford@usouthal.edu Pamela Francisco, CCRP
# 3 Phoenix Arizona 85013 United States Recruiting Lynn Ashby, MD 602-406-6262 LASHBY@CHW.EDU Lynn Ashby, MD
# 4 Tucson Arizona 85724 United States Recruiting Jamie Holt 520-626-6800 jholt1@email.arizona.edu Baldassarre Stea, MD, PhD
# 5 Little Rock Arkansas 72205 United States Recruiting Wilma Brooks, RN 501-686-8530 ALEubanks@uams.edu Amanda Eubanks, APN
# 6 Berkeley California 94704 United States Withdrawn <NA> <NA> <NA> <NA>
flatte\u xml此答案将xml转换为列表,取消列出每个位置节,转换节,将节转换为data.table
,然后使用rbindlist
将所有单个位置合并为一个表。fill=T
参数按名称匹配元素,并用NA
填充缺少的元素值
library(XML); library(data.table)
clinicalTrialUrl <- "http://clinicaltrials.gov/ct2/show/NCT01480479?resultsxml=true"
xmlDoc <- xmlParse(clinicalTrialUrl, useInternalNode=TRUE)
xmlToDT <- function(doc, path) {
rbindlist(
lapply(getNodeSet(doc, path),
function(x) data.table(t(unlist(xmlToList(x))))
), fill=T)
}
locationDT <- xmlToDT(xmlDoc, "//location")
locationDT[1:6]
## facility.name facility.address.city facility.address.state facility.address.zip
## 1: "HYGEIA" Hospital Marousi District of Attica 151 23
## 2: Allina Health, Abbott Northwestern Hospital, John Nasseff Neuroscience Institute Minneapolis Minnesota 55407
## 3: Amrita Institute of Medical Sciences and Research Centre, Kochi Kochi Kerala 682 026
## 4: Anne Arundel Medical Center Annapolis Maryland 21401
## 5: Atlanta Cancer Care Atlanta Georgia 30005
## 6: Austin Health Heidelberg Victoria 3084
## facility.address.country
## 1: Greece
## 2: United States
## 3: India
## 4: United States
## 5: United States
## 6: Australia
库(XML);库(数据表)
clinicalTrialUrl您可以这样做:xpathsaply(xmlDoc,//clinical\u-study/location/facility/name”,xmlValue)
将
的每个成分分别吸出来。但我不知道如何一下子做到。你所做的对我来说非常有效。我的XML文件很简单。谢谢,它成功了。出于某种原因,我的编译器不喜欢函数的语法,所以我不得不将它改为:flatte\uxml是的,我想我们使用的是不同的版本。修正。有机会的时候别忘了接受我的回答。:)