I try to analyze XML-Data in R with dplyr and ggplot2. My code is able to transform the XML data into a data frame. Unfortunately the structure gets lost.
My XML-document have following structure by example:
<?xml version="1.0" encoding="UTF-8"?>
<Budget price="1234" items="1234" year="1990">
<Account name="a" value="123" step="0">
<Account name="1" value="12" step="1"/>
<Account name="1.1" value="12" step="2"/>
<Account name="2" value="12" step="1"/>
<Account name="2.1" value="9" step="2"/>
<Account name="2.2" value="3" step="2"/>
<Account name="3" value="99" step="1"/>
<Account name="3.1" value="78" step="2"/>
<Account name="3.1.1" value="70" step="3"/>
<Account name="3.1.2" value="8" step="3"/>
<Account name="3.2" value="21" step)="2"/>
</Account>
<Account name="b" value="234" step="0">
<Account name="1" value="200" step="1"/>
and so on
At first I save all values:
budget_values = xml_find_all(doc,"//Budget",flatten=FALSE)
Afterwards I select some of the values:
step_ids = purrr::map_chr(budget_values, ~xml_attr(.,"step"))
name_values = purrr::map_chr(budget_values, ~xml_attr(.,"name"))
values = purrr::map_chr(budget_values, ~xml_attr(.,"value"))
Save attributes in a combined list:
values_list <- list((step_ids),(name_values),(values))
And convert it into a data frame:
budget_df <- data.frame(sapply(values_list, c))
That works great. I got an DF like this:
Step-ID | name | vlaue |
---|---|---|
0 | a | 1234 |
1 | 1 | 12 |
2 | 1.1 | 12 |
1 | 2 | 12 |
2 | 2.1 | 9 |
2 | 2.2 | 3 |
1 | 3 | 99 |
0 | b | 234 |
1 | 1 | 200 |
and so on
As you see from the example some names are repeated - usually step 1 and 2; step 3 is usually very unique.
My aim is following dataframe to analyze the data more structured.
Step-ID | name | vlaue |
---|---|---|
0 | a | 1234 |
1 | a1 | 12 |
2 | a1.1 | 12 |
1 | a2 | 12 |
2 | a2.1 | 9 |
2 | a2.2 | 3 |
1 | a3 | 99 |
0 | b | 234 |
1 | b1 | 200 |
and so on
For example: I want the values of all step1. Now I can't tell from which budget it is. With the new name I can see: this value is from budget a, this one from budget b and so on.
I tried following for-loop and stored the result in a new dataframe
df<-for (rows in budget_df) {
if (rows$`Step-ID` == "0") {
saved_name <- rows$name
print(saved_name)
}
else
(rows$`Step-ID` == "1"){
rows$Haushalt saved_name
saved_names<-saved_name rows$name
print(saved_names)
}
else(rows$`Step-ID`=="2"){
rows$Haushalt saved_name
}
else(rows$`Step-ID`=="3"){
rows$name saved_names
}
}
View(df)
And I get following Error:
Error: unexpected '{' in:
" else
(rows$`Step-ID` == "1"){"
My questions is: Is there a better way to analyze the data or rename the values in name?
Thank you very much for your help!
Update:
Thanks again to @jpsmith. I tried following code regarding to his recommondation:
df-budget_df
budget <- ""
df <- for (row in df) {
mutate(
case_when (
df$`Step-ID` == "0" ~ budget <- df$Haushalt,
df$`Step-ID` == "2" ~ mutate(df, sturucture = paste(budget, df$Haushalt)),
df$`Step-ID` == "2" ~ budget <- c(budget, df$Haushalt),
df$`Step-ID` == "3" ~ mutate(df, sturucture = paste(values, df$Haushalt))
)
)
}
Explains logically, what I want to do, but doesn't work. I think, it's because of trying to store the value with <-
? I couldn't find another way at ?case_when
to store values.
Another code (I have overwritten) stores the value of Step-ID
and extended the value of Haushalt
of the same step, instead of: Step-ID 0
to Haushalt
with Step-ID 1
under Step-ID 0
and Step-ID 1
to Haushalt
with Step-ID 2
.
CodePudding user response:
Ok, thanks for all help! I got it with a for-Loop:
n <- 0
df<-mutate(df,Structure=NA)
for (i in 1:nrow(df)) {
if (df[i, 1] == "0") {
first_step <- df[i, 2]
values<-first_step
df$Structure[which(is.na(df$Structure))[1]]<-values
}
else if (df[i, 1] == "1") {
second_step <- df[i, 2]
values <- paste(first_step, second_step)
df$Structure[which(is.na(df$Structure))[1]]<-values
}
else if (df[i, 1] == "2") {
third_step <- df[i, 2]
values <- paste(first_step, second_step, third_step)
df$Structure[which(is.na(df$Structure))[1]]<-values
}
else if (df[i, 1] == "3") {
fourth_step <- df[i, 2]
values <- paste(first_step, second_step, third_step, fourth_step)
df$Structure[which(is.na(df$Structure))[1]]<-values
}
n <- n 1
}
CodePudding user response:
Consider XSLT, the special-purpose language designed to transform XML files, in order to prefix the parent @name
attribute to underlying child @name
attributes. With R's xslt
(complementary package to xml2
) you can run XSLT 1.0 scripts:
XSLT (save as .xsl file, a special .xml file)
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" omit-xml-declaration="no" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:template match="/Budget">
<xsl:copy>
<xsl:apply-templates select="Account"/>
</xsl:copy>
</xsl:template>
<!-- MOVE ATTRIBUTES TO ELEMENTS -->
<xsl:template match="Account">
<xsl:copy>
<stepid><xsl:value-of select="@step"/></stepid>
<name><xsl:value-of select="@name"/></name>
<value><xsl:value-of select="@value"/></value>
</xsl:copy>
<xsl:apply-templates select="*"/>
</xsl:template>
<!-- MOVE ATTRIBUTES TO ELEMENTS AND CONCATENATE PARENT @name ATTRIBUTE -->
<xsl:template match="Account/*">
<xsl:variable name="step">
<xsl:value-of select="../@name"/>
</xsl:variable>
<xsl:copy>
<stepid><xsl:value-of select="@step"/></stepid>
<name><xsl:value-of select="concat($step, @name)"/></name>
<value><xsl:value-of select="@value"/></value>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
R
library(xml2)
library(xslt)
# LOAD XML AND XSLT
doc <- read_xml("inputF.xml")
style <- read_xml("style.xsl", package = "xslt")
# RUN TRANSFORMATION AND SEE OUTPUT
new_xml <- xml_xslt(doc, style)
# RETRIEVE ALL NODES
recs <- xml2::xml_find_all(new_xml, "//Account")
# BIND EACH CHILD TEXT AND NAME
df_list <- lapply(recs, function(r) {
vals <- xml2::xml_children(r)
df <- setNames(
c(xml2::xml_text(vals)),
c(xml2::xml_name(vals))
) |> rbind() |> data.frame()
})
# COMBINE ALL DFS
accounts_df <- do.call(rbind.data.frame, df_list)
Output
accounts_df
# stepid name value
# 1 0 a 123
# 2 1 a1 12
# 3 2 a1.1 12
# 4 1 a2 12
# 5 2 a2.1 9
# 6 2 a2.2 3
# 7 1 a3 99
# 8 2 a3.1 78
# 9 3 a3.1.1 70
# 10 3 a3.1.2 8
# 11 2 a3.2 21
# 12 0 b 234
# 13 1 b1 200