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Comment convertir un fichier XML en Nice pandas dataframe?

Supposons que j'ai un XML comme celui-ci:

<author type="XXX" language="EN" gender="xx" feature="xx" web="foobar.com">
    <documents count="N">
        <document KEY="e95a9a6c790ecb95e46cf15bee517651" web="www.foo_bar_exmaple.com"><![CDATA[A large text with lots of strings and punctuations symbols [...]
]]>
        </document>
        <document KEY="bc360cfbafc39970587547215162f0db" web="www.foo_bar_exmaple.com"><![CDATA[A large text with lots of strings and punctuations symbols [...]
]]>
        </document>
        <document KEY="19e71144c50a8b9160b3f0955e906fce" web="www.foo_bar_exmaple.com"><![CDATA[A large text with lots of strings and punctuations symbols [...]
]]>
        </document>
        <document KEY="21d4af9021a174f61b884606c74d9e42" web="www.foo_bar_exmaple.com"><![CDATA[A large text with lots of strings and punctuations symbols [...]
]]>
        </document>
        <document KEY="28a45eb2460899763d709ca00ddbb665" web="www.foo_bar_exmaple.com"><![CDATA[A large text with lots of strings and punctuations symbols [...]
]]>
        </document>
        <document KEY="a0c0712a6a351f85d9f5757e9fff8946" web="www.foo_bar_exmaple.com"><![CDATA[A large text with lots of strings and punctuations symbols [...]
]]>
        </document>
        <document KEY="626726ba8d34d15d02b6d043c55fe691" web="www.foo_bar_exmaple.com"><![CDATA[A large text with lots of strings and punctuations symbols [...]
]]>
        </document>
        <document KEY="2cb473e0f102e2e4a40aa3006e412ae4" web="www.foo_bar_exmaple.com"><![CDATA[A large text with lots of strings and punctuations symbols [...] [...]
]]>
        </document>
    </documents>
</author>

Je voudrais lire ce fichier XML et le convertir en un pandas DataFrame:

key                                         type     language    feature            web                         data
e95324a9a6c790ecb95e46cf15bE232ee517651      XXX        EN          xx      www.foo_bar_exmaple.com     A large text with lots of strings and punctuations symbols [...]
e95324a9a6c790ecb95e46cf15bE232ee517651      XXX        EN          xx      www.foo_bar_exmaple.com     A large text with lots of strings and punctuations symbols [...]
19e71144c50a8b9160b3cvdf2324f0955e906fce     XXX        EN          xx      www.foo_bar_exmaple.com     A large text with lots of strings and punctuations symbols [...]
21d4af9021a174f61b8erf284606c74d9e42         XXX        EN          xx      www.foo_bar_exmaple.com     A large text with lots of strings and punctuations symbols [...]
28a45eb2460823499763d70vdf9ca00ddbb665       XXX        EN          xx      www.foo_bar_exmaple.com     A large text with lots of strings and punctuations symbols [...]

C’est ce que j’ai déjà essayé, mais j’obtiens des erreurs et il existe probablement un moyen plus efficace de le faire:

from lxml import objectify
import pandas as pd

path = 'file_path'
xml = objectify.parse(open(path))
root = xml.getroot()
root.getchildren()[0].getchildren()
df = pd.DataFrame(columns=('key','type', 'language', 'feature', 'web', 'data'))

for i in range(0,len(xml)):
    obj = root.getchildren()[i].getchildren()
    row = dict(Zip(['key','type', 'language', 'feature', 'web', 'data'], [obj[0].text, obj[1].text]))
    row_s = pd.Series(row)
    row_s.name = i
    df = df.append(row_s)

Quelqu'un pourrait-il me fournir une meilleure approche de ce problème?

pythonxmlpython-2.7parsingpandas
52
eoriu

Vous pouvez facilement utiliser xml (de la bibliothèque standard Python) pour convertir en pandas.DataFrame. Voici ce que je ferais ( lors de la lecture d'un fichier, remplacez xml_data par le nom de votre fichier ou de l'objet du fichier):

import pandas as pd
import xml.etree.ElementTree as ET
import io

def iter_docs(author):
    author_attr = author.attrib
    for doc in author.iter('document'):
        doc_dict = author_attr.copy()
        doc_dict.update(doc.attrib)
        doc_dict['data'] = doc.text
        yield doc_dict

xml_data = io.StringIO(u'''\
<author type="XXX" language="EN" gender="xx" feature="xx" web="foobar.com">
    <documents count="N">
        <document KEY="e95a9a6c790ecb95e46cf15bee517651" web="www.foo_bar_exmaple.com"><![CDATA[A large text with lots of strings and punctuations symbols [...]
]]>
        </document>
        <document KEY="bc360cfbafc39970587547215162f0db" web="www.foo_bar_exmaple.com"><![CDATA[A large text with lots of strings and punctuations symbols [...]
]]>
        </document>
        <document KEY="19e71144c50a8b9160b3f0955e906fce" web="www.foo_bar_exmaple.com"><![CDATA[A large text with lots of strings and punctuations symbols [...]
]]>
        </document>
        <document KEY="21d4af9021a174f61b884606c74d9e42" web="www.foo_bar_exmaple.com"><![CDATA[A large text with lots of strings and punctuations symbols [...]
]]>
        </document>
        <document KEY="28a45eb2460899763d709ca00ddbb665" web="www.foo_bar_exmaple.com"><![CDATA[A large text with lots of strings and punctuations symbols [...]
]]>
        </document>
        <document KEY="a0c0712a6a351f85d9f5757e9fff8946" web="www.foo_bar_exmaple.com"><![CDATA[A large text with lots of strings and punctuations symbols [...]
]]>
        </document>
        <document KEY="626726ba8d34d15d02b6d043c55fe691" web="www.foo_bar_exmaple.com"><![CDATA[A large text with lots of strings and punctuations symbols [...]
]]>
        </document>
        <document KEY="2cb473e0f102e2e4a40aa3006e412ae4" web="www.foo_bar_exmaple.com"><![CDATA[A large text with lots of strings and punctuations symbols [...] [...]
]]>
        </document>
    </documents>
</author>
''')

etree = ET.parse(xml_data) #create an ElementTree object 
doc_df = pd.DataFrame(list(iter_docs(etree.getroot())))

S'il y a plusieurs auteurs dans votre document d'origine ou si la racine de votre XML n'est pas un author, j'ajouterais le générateur suivant:

def iter_author(etree):
    for author in etree.iter('author'):
        for row in iter_docs(author):
            yield row

et changez doc_df = pd.DataFrame(list(iter_docs(etree.getroot()))) en doc_df = pd.DataFrame(list(iter_author(etree)))

Jetez un coup d'oeil au ElementTreetutoriel fourni dans la bibliothèque xmldocumentation .

40
JaminSore

Voici un autre moyen de convertir un fichier XML en trame de données pandas. Par exemple, j’ai analysé le xml à partir d’une chaîne, mais cette logique est également valable pour la lecture de fichier.

import pandas as pd
import xml.etree.ElementTree as ET

xml_str = '<?xml version="1.0" encoding="utf-8"?>\n<response>\n <head>\n  <code>\n   200\n  </code>\n </head>\n <body>\n  <data id="0" name="All Categories" t="2018052600" tg="1" type="category"/>\n  <data id="13" name="RealEstate.com.au [H]" t="2018052600" tg="1" type="publication"/>\n </body>\n</response>'

etree = ET.fromstring(xml_str)
dfcols = ['id', 'name']
df = pd.DataFrame(columns=dfcols)

for i in etree.iter(tag='data'):
    df = df.append(
        pd.Series([i.get('id'), i.get('name')], index=dfcols),
        ignore_index=True)

df.head()
8
Jai Prakash

Vous pouvez également convertir en créant un dictionnaire d'éléments puis en convertissant directement en un cadre de données:

import xml.etree.ElementTree as ET
import pandas as pd

# Contents of test.xml
# <?xml version="1.0" encoding="utf-8"?> <tags>   <row Id="1" TagName="bayesian" Count="4699" ExcerptPostId="20258" WikiPostId="20257" />   <row Id="2" TagName="prior" Count="598" ExcerptPostId="62158" WikiPostId="62157" />   <row Id="3" TagName="elicitation" Count="10" />   <row Id="5" TagName="open-source" Count="16" /> </tags>

root = ET.parse('test.xml').getroot()

tags = {"tags":[]}
for elem in root:
    tag = {}
    tag["Id"] = elem.attrib['Id']
    tag["TagName"] = elem.attrib['TagName']
    tag["Count"] = elem.attrib['Count']
    tags["tags"]. append(tag)

df_users = pd.DataFrame(tags["tags"])
df_users.head()
2
Naveen Kaushik