Read XML with column names as attribute values into pandas DataFrame

Question

I have an XML file like this:

<table chgFlag="i" id="II325">
    <row chgFlag="i" idVal="1">
        <fld id="II325A">
            <datVl chgFlag="i">1</datVl>
        </fld>
        <fld id="II325B">
            <datVl chgFlag="i">2001-12-01</datVl>
        </fld>
        <fld id="II325C">
            <datVl chgFlag="i">2006-04-30</datVl>
        </fld>
        <fld id="II325D">
            <datVl chgFlag="i">01</datVl>
        </fld>
    </row>
    <row chgFlag="i" idVal="2">
        <fld id="II325A">
            <datVl chgFlag="i">2</datVl>
        </fld>
        <fld id="II325B">
            <datVl chgFlag="i">2006-05-01</datVl>
        </fld>
        <fld id="II325C">
            <datVl chgFlag="i">2031-11-30</datVl>
        </fld>
        <fld id="II325D">
            <datVl chgFlag="i">01</datVl>
        </fld>
    </row>
</table>

If I just put it into read_xml I get something like this:

  chgFlag  idVal  fld
0       i      1  NaN
1       i      2  NaN

It take sthe attributes in each row as columns. I don't want that, I want value of id in fld as the column and the text inside datVl as the value.

Something like this.

I manage to get the result I wanted using this code:

    data_dict = xmltodict.parse(ET.tostring(table))

    table_list = []
    if type(data_dict["table"]["row"]) == list:
        for row in data_dict["table"]["row"]:
            row_dict = {}
            for field in row["fld"]:
                row_dict[field["@id"]] = field["datVl"]["#text"]
            table_list.append(row_dict)
        df = pd.DataFrame(table_list)

I was wondering if there is more a general solution, perhaps setting some parameter inside read_xml?

I might need to scale my current solution, that is why I am asking.

score 2 · Answer 1 · answered Jun 17 '22 at 21:34

As mentioned in the comments to this similar question:

read_xml does not parse beyond its immediate descendants.

But you can use the approach mentioned in the above answer: transforming the XML with XSLT to a palatable format. In your case you can use the stylesheet parameter of read_xml with this XSLT

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="xml" omit-xml-declaration="yes"/>    
       
    <xsl:template match="/table">
        <Root><xsl:apply-templates /></Root>
    </xsl:template>

    <xsl:template match="row">
        <Item>
            <xsl:element name="II325A"><xsl:value-of select="fld[@id='II325A']/datVl" /></xsl:element>
            <xsl:element name="II325B"><xsl:value-of select="fld[@id='II325B']/datVl" /></xsl:element>
            <xsl:element name="II325C"><xsl:value-of select="fld[@id='II325C']/datVl" /></xsl:element>
            <xsl:element name="II325D"><xsl:value-of select="fld[@id='II325D']/datVl" /></xsl:element>
       </Item>
    </xsl:template>
    
</xsl:stylesheet>

or, a more general approach for the second template iterating over all of the children of row:

<xsl:template match="row">
    <Item>
        <xsl:for-each select="fld">
            <xsl:element name="{@id}"><xsl:value-of select="datVl" /></xsl:element>
        </xsl:for-each>
   </Item>
</xsl:template>

which creates, in both versions, the following intermediate XML

<Root>
    <Item><II325A>1</II325A><II325B>2001-12-01</II325B><II325C>2006-04-30</II325C><II325D>1</II325D></Item>
    <Item><II325A>2</II325A><II325B>2006-05-01</II325B><II325C>2031-11-30</II325C><II325D>1</II325D></Item>
</Root>

and can be used with the following python code:

dataFolder = './'
df_bulk = pd.read_xml(
    dataFolder+'input.xml', 
    stylesheet='transform.xslt',
    xpath='/Root/Item',    
    )
print(df_bulk.head(10))

The output is:

   II325A      II325B      II325C  II325D
0       1  2001-12-01  2006-04-30       1
1       2  2006-05-01  2031-11-30       1

score 0 · Answer 2 · answered Jun 19 '22 at 00:40

one option is to use the xpath parameter, and pass the specified sections of the xml :

(pd
.read_xml(data, xpath = ".//fld")
.assign(counter = lambda df: df.groupby('id').cumcount())
.pivot('counter', 'id', 'datVl')
.rename_axis(index = None, columns = None)
)
  II325A      II325B      II325C II325D
0      1  2001-12-01  2006-04-30     01
1      2  2006-05-01  2031-11-30     01

Read XML with column names as attribute values into pandas DataFrame

2 Answers2