我试图迭代地从XML文档中分离和操作节点集,但是我在R中的xml_find_all()函数中得到了一个奇怪的行为,有人能帮助我理解应用于节点集的函数的范围吗?
下面是一个示例:
library( xml2 )
library( dplyr )
doc <- read_xml( "<MEMBERS>
<CUSTOMER>
<ID>178</ID>
<FIRST.NAME>Alvaro</FIRST.NAME>
<LAST.NAME>Juarez</LAST.NAME>
<ADDRESS>123 Park Ave</ADDRESS>
<ZIP>57701</ZIP>
</CUSTOMER>
<CUSTOMER>
<ID>934</ID>
<FIRST.NAME>Janette</FIRST.NAME>
<LAST.NAME>Johnson</LAST.NAME>
<ADDRESS>456 Candy Ln</ADDRESS>
<ZIP>57701</ZIP>
</CUSTOMER>
</MEMBERS>" )
doc %>% xml_find_all( '//*') %>% xml_path()
# [1] "/MEMBERS" "/MEMBERS/CUSTOMER[1]"
# [3] "/MEMBERS/CUSTOMER[1]/ID" "/MEMBERS/CUSTOMER[1]/FIRST.NAME"
# [5] "/MEMBERS/CUSTOMER[1]/LAST.NAME" "/MEMBERS/CUSTOMER[1]/ADDRESS"
# [7] "/MEMBERS/CUSTOMER[1]/ZIP" "/MEMBERS/CUSTOMER[2]"
# [9] "/MEMBERS/CUSTOMER[2]/ID" "/MEMBERS/CUSTOMER[2]/FIRST.NAME"
#[11] "/MEMBERS/CUSTOMER[2]/LAST.NAME" "/MEMBERS/CUSTOMER[2]/ADDRESS"
#[13] "/MEMBERS/CUSTOMER[2]/ZIP"对象customer.01是一个节点集,仅包含来自该客户的数据。
kids <- xml_children( doc )
customer.01 <- kids[[1]]
customer.01
# {xml_node}
# <CUSTOMER>
# [1] <ID>178</ID>
# [2] <FIRST.NAME>Alvaro</FIRST.NAME>
# [3] <LAST.NAME>Juarez</LAST.NAME>
# [4] <ADDRESS>123 Park Ave</ADDRESS>
# [5] <ZIP>57701</ZIP>为什么应用于customer.01节点集的函数也返回customer.02的ID?
xml_find_all( customer.01, "//MEMBERS/CUSTOMER/ID" )
# {xml_nodeset (2)}
# [1] <ID>178</ID>
# [2] <ID>934</ID>如何只返回节点集中的值?
好的,下面的解决方案有一个小问题,同样与xml_find_all()函数的作用域有关。它说,它可以应用于文档、节点或节点集。然而..。
这种情况适用于节点集:
library( xml2 )
url <- "https://s3.amazonaws.com/irs-form-990/201501279349300635_public.xml"
doc <- read_xml( url )
xml_ns_strip( doc )
nd <- xml_find_all( doc, "//LiquidationOfAssetsDetail|//LiquidationDetail" )
nodei <- nd[[1]]
nodei
# {xml_node}
# <LiquidationOfAssetsDetail>
# [1] <AssetsDistriOrExpnssPaidDesc>LAND</AssetsDistriOrExpnssPaidDesc>
# [2] <DistributionDt>2014-11-04</DistributionDt>
# [3] <MethodOfFMVDeterminationTxt>SEE ATTACH</MethodOfFMVDeterminationTxt>
# [4] <EIN>abcdefghi</EIN>
# [5] <BusinessName>\n <BusinessNameLine1Txt>GREENSBURG PUBLIC LIBRARY</BusinessNameLine1Txt>\n</BusinessName>
# [6] <USAddress>\n <AddressLine1Txt>1110 E MAIN ST</AddressLine1Txt>\n <CityNm>GREENSBURG</CityNm>\n <StateAbbreviationCd>IN</StateAb ...
# [7] <IRCSectionTxt>501(C)(3)</IRCSectionTxt>
xml_text( xml_find_all( nodei, "AssetsDistriOrExpnssPaidDesc" ) )
# [1] "LAND"但不是这个:
nodei <- xml_children( nd[[i]] )
nodei
# {xml_nodeset (7)}
# [1] <AssetsDistriOrExpnssPaidDesc>LAND</AssetsDistriOrExpnssPaidDesc>
# [2] <DistributionDt>2014-11-04</DistributionDt>
# [3] <MethodOfFMVDeterminationTxt>SEE ATTACH</MethodOfFMVDeterminationTxt>
# [4] <EIN>abcdefghi</EIN>
# [5] <BusinessName>\n <BusinessNameLine1Txt>GREENSBURG PUBLIC LIBRARY</BusinessNameLine1Txt>\n</BusinessName>
# [6] <USAddress>\n <AddressLine1Txt>1110 E MAIN ST</AddressLine1Txt>\n <CityNm>GREENSBURG</CityNm>\n <StateAbbreviationCd>IN</StateAb ...
# [7] <IRCSectionTxt>501(C)(3)</IRCSectionTxt>
xml_text( xml_find_all( nodei, "AssetsDistriOrExpnssPaidDesc" ) )
# character(0)我猜这是一个将xml_find_all()应用于节点集的所有元素而不是范围问题的问题?
发布于 2017-08-24 18:38:34
目前,您使用的是使用XPath的双正斜杠//的根用户绝对路径搜索,这意味着在文档中查找与包含两个客户ID的路径匹配的所有项。
对于特定节点下的特定子节点,只需在选定节点下使用相对路径:
xml_find_all(customer.01, "ID")
# {xml_nodeset (1)}
# [1] <ID>178</ID>
xml_find_all(customer.01, "FIRST.NAME|LAST.NAME")
# {xml_nodeset (2)}
# [1] <FIRST.NAME>Alvaro</FIRST.NAME>
# [2] <LAST.NAME>Juarez</LAST.NAME>
xml_find_all(customer.01, "*")
# {xml_nodeset (5)}
# [1] <ID>178</ID>
# [2] <FIRST.NAME>Alvaro</FIRST.NAME>
# [3] <LAST.NAME>Juarez</LAST.NAME>
# [4] <ADDRESS>123 Park Ave</ADDRESS>
# [5] <ZIP>57701</ZIP>https://stackoverflow.com/questions/45866491
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