Develop Reference

How do I find text after a given key word?

I am practicing my programming skills (in Python) and I realized that I don't know what to do when I need to find a value that is unknown but introduced by a key word. I am taking the information for this off a website where in the page source it says, '"size":"10","stockKeepingUnitId":"(random number)"'
How can I figure out what that number is.
This is what I have so far --
def stock():
global session
endpoint = '(website)'
reponse = session.get(endpoint)
soup = bs(response.text, "html.parser")
sizes = soup.find('"size":"10","stockKeepingUnitId":')

Off the top of my head there are two ways to do this. Say you have the string mystr = 'some text...content:"67588978"'. The first way is just to search for "content:" in the string and use string slicing to take everything after it:
num = mystr[mystr.index('content:"') + len('content:"'):-1]
Alternatively, as probably a better solution, you could use regular expressions
import re
nums = re.findall(r'.*?content:\"(\d+)\"')
As you haven't provided an example of the dataset you're trying to analyze, there could also be a number of other solutions. If you're trying to parse a JSON or YAML file, there are simple libraries to turn them into python dicts (json is part of the standard library, and PyYaml handles YAML files easily).

Parsing XML Data with Python

actually I am working on a small project and need to parse public available XML data. My goal is to write the data to an mysql database for further processing.
XML Data Link: http://offenedaten.frankfurt.de/dataset/912fe0ab-8976-4837-b591-57dbf163d6e5/resource/48378186-5732-41f3-9823-9d1938f2695e/download/parkdatendyn.xml
XML structure (example):
<parkingAreaStatus>
<parkingAreaOccupancy>0.2533602</parkingAreaOccupancy>
<parkingAreaOccupancyTrend>stable</parkingAreaOccupancyTrend>
<parkingAreaReference targetClass="ParkingArea" id="2[Zeil]"
version="1.0"/>
<parkingAreaStatusTime>2018-02-
04T01:30:00.000+01:00</parkingAreaStatusTime
</parkingAreaStatus>
<parkingAreaStatus>
<parkingAreaOccupancy>0.34625</parkingAreaOccupancy>
<parkingAreaOccupancyTrend>stable</parkingAreaOccupancyTrend>
<parkingAreaReference targetClass="ParkingArea" id="5[Dom / Römer]"
version="1.0"/>
</parkingAreaStatus>
Using the code
import csv
import pymysql
import urllib.request
url = "http://offenedaten.frankfurt.de/dataset/912fe0ab-8976-4837-b591-57dbf163d6e5/resource/48378186-5732-41f3-9823-9d1938f2695e/download/parkdatendyn.xml"
from lxml.objectify import parse
from lxml import etree
from urllib.request import urlopen
locations_root = parse(urlopen(url)).getroot()
locations = list(locations_root.payloadPublication.genericPublicationExtension.parkingFacilityTableStatusPublication.parkingAreaStatus.parkingAreaReference)
print(*locations)
I expected to get a list of all "parkingAreaReference" entries within the XML document. Unfortunately the list is empty.
Playing arround with some code I got the sentiment that only the first block is parsed, I was able to fill the list with the value of "parkingAreaOccupancy" of the "parkingAreaReference" id="2[Zeil]" block by using the code
locations = list(locations_root.payloadPublication.genericPublicationExtension.parkingFacilityTableStatusPublication.parkingAreaStatus.parkingAreaOccupancy)
print(*locations)
-> 0.2533602
which is not the expected outcome
-> 0.2533602
-> 0.34625
MY question is:
What is the best way to get a matrix i can further work with of all blocks incl. the corresponding values stated in the XML document?
Example output:
A = [[ID:2[Zeil],0.2533602,stable,2018-02-
04T01:30:00.000+01:00],[id="5[Dom / Römer],0.34625,stable,2018-02-
04T01:30:00.000+01:00]]
or in general
A = [parkingAreaOccupancy,parkingAreaOccupancyTrend,parkingAreaStatusTime,....],[parkingAreaOccupancy,parkingAreaOccupancyTrend,parkingAreaStatusTime,.....]
After hours of research I hope for some tips from your site
Thank you in advance,
TR

You can just use etree directly and find interesting elements using XPath1 query. One important thing to note is, that your XML has default namespace declared at the root element :
xmlns="http://datex2.eu/schema/2/2_0"
By definition, element where default namespace is declared and all descendant elements without prefix are belong to this default namespace (unless another default namespace found in one of the descendant elements, which is not the case with your XML). This is why we define a prefix d, which references default namespace URI, in the following code, and we use that prefix to find every elements we need to get information from :
root = etree.parse(urlopen(url)).getroot()
ns = { 'd': 'http://datex2.eu/schema/2/2_0' }
parking_area = root.xpath('//d:parkingAreaStatus', namespaces=ns)
for pa in parking_area:
area_ref = pa.find('d:parkingAreaReference', ns)
occupancy = pa.find('d:parkingAreaOccupancy', ns)
trend = pa.find('d:parkingAreaOccupancyTrend', ns)
status_time = pa.find('d:parkingAreaStatusTime', ns)
print area_ref.get('id'), occupancy.text, trend.text, status_time.text
Below is the output of the demo code above. Instead of print, you can store these information in whatever data structure you like :
2[Zeil] 0.22177419 stable 2018-02-04T05:16:00.000+01:00
5[Dom / Römer] 0.28625 stable 2018-02-04T05:16:00.000+01:00
1[Anlagenring] 0.257889 stable 2018-02-04T05:16:00.000+01:00
3[Mainzer Landstraße] 0.20594966 stable 2018-02-04T05:16:00.000+01:00
4[Bahnhofsviertel] 0.31513646 stable 2018-02-04T05:16:00.000+01:00
1) some references on XPath :
XPath 1.0 spec: The most trustworthy reference on XPath 1.0
XPath syntax: Gentler introduction to basic XPath expressions

XML Python Parsing Namespace [duplicate]

I have the following XML which I want to parse using Python's ElementTree:
<rdf:RDF xml:base="http://dbpedia.org/ontology/"
xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
xmlns:owl="http://www.w3.org/2002/07/owl#"
xmlns:xsd="http://www.w3.org/2001/XMLSchema#"
xmlns:rdfs="http://www.w3.org/2000/01/rdf-schema#"
xmlns="http://dbpedia.org/ontology/">
<owl:Class rdf:about="http://dbpedia.org/ontology/BasketballLeague">
<rdfs:label xml:lang="en">basketball league</rdfs:label>
<rdfs:comment xml:lang="en">
a group of sports teams that compete against each other
in Basketball
</rdfs:comment>
</owl:Class>
</rdf:RDF>
I want to find all owl:Class tags and then extract the value of all rdfs:label instances inside them. I am using the following code:
tree = ET.parse("filename")
root = tree.getroot()
root.findall('owl:Class')
Because of the namespace, I am getting the following error.
SyntaxError: prefix 'owl' not found in prefix map
I tried reading the document at http://effbot.org/zone/element-namespaces.htm but I am still not able to get this working since the above XML has multiple nested namespaces.
Kindly let me know how to change the code to find all the owl:Class tags.

You need to give the .find(), findall() and iterfind() methods an explicit namespace dictionary:
namespaces = {'owl': 'http://www.w3.org/2002/07/owl#'} # add more as needed
root.findall('owl:Class', namespaces)
Prefixes are only looked up in the namespaces parameter you pass in. This means you can use any namespace prefix you like; the API splits off the owl: part, looks up the corresponding namespace URL in the namespaces dictionary, then changes the search to look for the XPath expression {http://www.w3.org/2002/07/owl}Class instead. You can use the same syntax yourself too of course:
root.findall('{http://www.w3.org/2002/07/owl#}Class')
Also see the Parsing XML with Namespaces section of the ElementTree documentation.
If you can switch to the lxml library things are better; that library supports the same ElementTree API, but collects namespaces for you in .nsmap attribute on elements and generally has superior namespaces support.

Here's how to do this with lxml without having to hard-code the namespaces or scan the text for them (as Martijn Pieters mentions):
from lxml import etree
tree = etree.parse("filename")
root = tree.getroot()
root.findall('owl:Class', root.nsmap)
UPDATE:
5 years later I'm still running into variations of this issue. lxml helps as I showed above, but not in every case. The commenters may have a valid point regarding this technique when it comes merging documents, but I think most people are having difficulty simply searching documents.
Here's another case and how I handled it:
<?xml version="1.0" ?><Tag1 xmlns="http://www.mynamespace.com/prefix">
<Tag2>content</Tag2></Tag1>
xmlns without a prefix means that unprefixed tags get this default namespace. This means when you search for Tag2, you need to include the namespace to find it. However, lxml creates an nsmap entry with None as the key, and I couldn't find a way to search for it. So, I created a new namespace dictionary like this
namespaces = {}
# response uses a default namespace, and tags don't mention it
# create a new ns map using an identifier of our choice
for k,v in root.nsmap.iteritems():
if not k:
namespaces['myprefix'] = v
e = root.find('myprefix:Tag2', namespaces)

Note: This is an answer useful for Python's ElementTree standard library without using hardcoded namespaces.
To extract namespace's prefixes and URI from XML data you can use ElementTree.iterparse function, parsing only namespace start events (start-ns):
>>> from io import StringIO
>>> from xml.etree import ElementTree
>>> my_schema = u'''<rdf:RDF xml:base="http://dbpedia.org/ontology/"
... xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
... xmlns:owl="http://www.w3.org/2002/07/owl#"
... xmlns:xsd="http://www.w3.org/2001/XMLSchema#"
... xmlns:rdfs="http://www.w3.org/2000/01/rdf-schema#"
... xmlns="http://dbpedia.org/ontology/">
...
... <owl:Class rdf:about="http://dbpedia.org/ontology/BasketballLeague">
... <rdfs:label xml:lang="en">basketball league</rdfs:label>
... <rdfs:comment xml:lang="en">
... a group of sports teams that compete against each other
... in Basketball
... </rdfs:comment>
... </owl:Class>
...
... </rdf:RDF>'''
>>> my_namespaces = dict([
... node for _, node in ElementTree.iterparse(
... StringIO(my_schema), events=['start-ns']
... )
... ])
>>> from pprint import pprint
>>> pprint(my_namespaces)
{'': 'http://dbpedia.org/ontology/',
'owl': 'http://www.w3.org/2002/07/owl#',
'rdf': 'http://www.w3.org/1999/02/22-rdf-syntax-ns#',
'rdfs': 'http://www.w3.org/2000/01/rdf-schema#',
'xsd': 'http://www.w3.org/2001/XMLSchema#'}
Then the dictionary can be passed as argument to the search functions:
root.findall('owl:Class', my_namespaces)

I've been using similar code to this and have found it's always worth reading the documentation... as usual!
findall() will only find elements which are direct children of the current tag. So, not really ALL.
It might be worth your while trying to get your code working with the following, especially if you're dealing with big and complex xml files so that that sub-sub-elements (etc.) are also included.
If you know yourself where elements are in your xml, then I suppose it'll be fine! Just thought this was worth remembering.
root.iter()
ref: https://docs.python.org/3/library/xml.etree.elementtree.html#finding-interesting-elements
"Element.findall() finds only elements with a tag which are direct children of the current element. Element.find() finds the first child with a particular tag, and Element.text accesses the element’s text content. Element.get() accesses the element’s attributes:"

To get the namespace in its namespace format, e.g. {myNameSpace}, you can do the following:
root = tree.getroot()
ns = re.match(r'{.*}', root.tag).group(0)
This way, you can use it later on in your code to find nodes, e.g using string interpolation (Python 3).
link = root.find(f"{ns}link")

This is basically Davide Brunato's answer however I found out that his answer had serious problems the default namespace being the empty string, at least on my python 3.6 installation. The function I distilled from his code and that worked for me is the following:
from io import StringIO
from xml.etree import ElementTree
def get_namespaces(xml_string):
namespaces = dict([
node for _, node in ElementTree.iterparse(
StringIO(xml_string), events=['start-ns']
)
])
namespaces["ns0"] = namespaces[""]
return namespaces
where ns0 is just a placeholder for the empty namespace and you can replace it by any random string you like.
If I then do:
my_namespaces = get_namespaces(my_schema)
root.findall('ns0:SomeTagWithDefaultNamespace', my_namespaces)
It also produces the correct answer for tags using the default namespace as well.

My solution is based on #Martijn Pieters' comment:
register_namespace only influences serialisation, not search.
So the trick here is to use different dictionaries for serialization and for searching.
namespaces = {
'': 'http://www.example.com/default-schema',
'spec': 'http://www.example.com/specialized-schema',
}
Now, register all namespaces for parsing and writing:
for name, value in namespaces.iteritems():
ET.register_namespace(name, value)
For searching (find(), findall(), iterfind()) we need a non-empty prefix. Pass these functions a modified dictionary (here I modify the original dictionary, but this must be made only after the namespaces are registered).
self.namespaces['default'] = self.namespaces['']
Now, the functions from the find() family can be used with the default prefix:
print root.find('default:myelem', namespaces)
but
tree.write(destination)
does not use any prefixes for elements in the default namespace.

Processing large XML file [duplicate]

I have the following XML which I want to parse using Python's ElementTree:
<rdf:RDF xml:base="http://dbpedia.org/ontology/"
xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
xmlns:owl="http://www.w3.org/2002/07/owl#"
xmlns:xsd="http://www.w3.org/2001/XMLSchema#"
xmlns:rdfs="http://www.w3.org/2000/01/rdf-schema#"
xmlns="http://dbpedia.org/ontology/">
<owl:Class rdf:about="http://dbpedia.org/ontology/BasketballLeague">
<rdfs:label xml:lang="en">basketball league</rdfs:label>
<rdfs:comment xml:lang="en">
a group of sports teams that compete against each other
in Basketball
</rdfs:comment>
</owl:Class>
</rdf:RDF>
I want to find all owl:Class tags and then extract the value of all rdfs:label instances inside them. I am using the following code:
tree = ET.parse("filename")
root = tree.getroot()
root.findall('owl:Class')
Because of the namespace, I am getting the following error.
SyntaxError: prefix 'owl' not found in prefix map
I tried reading the document at http://effbot.org/zone/element-namespaces.htm but I am still not able to get this working since the above XML has multiple nested namespaces.
Kindly let me know how to change the code to find all the owl:Class tags.

You need to give the .find(), findall() and iterfind() methods an explicit namespace dictionary:
namespaces = {'owl': 'http://www.w3.org/2002/07/owl#'} # add more as needed
root.findall('owl:Class', namespaces)
Prefixes are only looked up in the namespaces parameter you pass in. This means you can use any namespace prefix you like; the API splits off the owl: part, looks up the corresponding namespace URL in the namespaces dictionary, then changes the search to look for the XPath expression {http://www.w3.org/2002/07/owl}Class instead. You can use the same syntax yourself too of course:
root.findall('{http://www.w3.org/2002/07/owl#}Class')
Also see the Parsing XML with Namespaces section of the ElementTree documentation.
If you can switch to the lxml library things are better; that library supports the same ElementTree API, but collects namespaces for you in .nsmap attribute on elements and generally has superior namespaces support.

Here's how to do this with lxml without having to hard-code the namespaces or scan the text for them (as Martijn Pieters mentions):
from lxml import etree
tree = etree.parse("filename")
root = tree.getroot()
root.findall('owl:Class', root.nsmap)
UPDATE:
5 years later I'm still running into variations of this issue. lxml helps as I showed above, but not in every case. The commenters may have a valid point regarding this technique when it comes merging documents, but I think most people are having difficulty simply searching documents.
Here's another case and how I handled it:
<?xml version="1.0" ?><Tag1 xmlns="http://www.mynamespace.com/prefix">
<Tag2>content</Tag2></Tag1>
xmlns without a prefix means that unprefixed tags get this default namespace. This means when you search for Tag2, you need to include the namespace to find it. However, lxml creates an nsmap entry with None as the key, and I couldn't find a way to search for it. So, I created a new namespace dictionary like this
namespaces = {}
# response uses a default namespace, and tags don't mention it
# create a new ns map using an identifier of our choice
for k,v in root.nsmap.iteritems():
if not k:
namespaces['myprefix'] = v
e = root.find('myprefix:Tag2', namespaces)

Note: This is an answer useful for Python's ElementTree standard library without using hardcoded namespaces.
To extract namespace's prefixes and URI from XML data you can use ElementTree.iterparse function, parsing only namespace start events (start-ns):
>>> from io import StringIO
>>> from xml.etree import ElementTree
>>> my_schema = u'''<rdf:RDF xml:base="http://dbpedia.org/ontology/"
... xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
... xmlns:owl="http://www.w3.org/2002/07/owl#"
... xmlns:xsd="http://www.w3.org/2001/XMLSchema#"
... xmlns:rdfs="http://www.w3.org/2000/01/rdf-schema#"
... xmlns="http://dbpedia.org/ontology/">
...
... <owl:Class rdf:about="http://dbpedia.org/ontology/BasketballLeague">
... <rdfs:label xml:lang="en">basketball league</rdfs:label>
... <rdfs:comment xml:lang="en">
... a group of sports teams that compete against each other
... in Basketball
... </rdfs:comment>
... </owl:Class>
...
... </rdf:RDF>'''
>>> my_namespaces = dict([
... node for _, node in ElementTree.iterparse(
... StringIO(my_schema), events=['start-ns']
... )
... ])
>>> from pprint import pprint
>>> pprint(my_namespaces)
{'': 'http://dbpedia.org/ontology/',
'owl': 'http://www.w3.org/2002/07/owl#',
'rdf': 'http://www.w3.org/1999/02/22-rdf-syntax-ns#',
'rdfs': 'http://www.w3.org/2000/01/rdf-schema#',
'xsd': 'http://www.w3.org/2001/XMLSchema#'}
Then the dictionary can be passed as argument to the search functions:
root.findall('owl:Class', my_namespaces)

I've been using similar code to this and have found it's always worth reading the documentation... as usual!
findall() will only find elements which are direct children of the current tag. So, not really ALL.
It might be worth your while trying to get your code working with the following, especially if you're dealing with big and complex xml files so that that sub-sub-elements (etc.) are also included.
If you know yourself where elements are in your xml, then I suppose it'll be fine! Just thought this was worth remembering.
root.iter()
ref: https://docs.python.org/3/library/xml.etree.elementtree.html#finding-interesting-elements
"Element.findall() finds only elements with a tag which are direct children of the current element. Element.find() finds the first child with a particular tag, and Element.text accesses the element’s text content. Element.get() accesses the element’s attributes:"

To get the namespace in its namespace format, e.g. {myNameSpace}, you can do the following:
root = tree.getroot()
ns = re.match(r'{.*}', root.tag).group(0)
This way, you can use it later on in your code to find nodes, e.g using string interpolation (Python 3).
link = root.find(f"{ns}link")

This is basically Davide Brunato's answer however I found out that his answer had serious problems the default namespace being the empty string, at least on my python 3.6 installation. The function I distilled from his code and that worked for me is the following:
from io import StringIO
from xml.etree import ElementTree
def get_namespaces(xml_string):
namespaces = dict([
node for _, node in ElementTree.iterparse(
StringIO(xml_string), events=['start-ns']
)
])
namespaces["ns0"] = namespaces[""]
return namespaces
where ns0 is just a placeholder for the empty namespace and you can replace it by any random string you like.
If I then do:
my_namespaces = get_namespaces(my_schema)
root.findall('ns0:SomeTagWithDefaultNamespace', my_namespaces)
It also produces the correct answer for tags using the default namespace as well.

My solution is based on #Martijn Pieters' comment:
register_namespace only influences serialisation, not search.
So the trick here is to use different dictionaries for serialization and for searching.
namespaces = {
'': 'http://www.example.com/default-schema',
'spec': 'http://www.example.com/specialized-schema',
}
Now, register all namespaces for parsing and writing:
for name, value in namespaces.iteritems():
ET.register_namespace(name, value)
For searching (find(), findall(), iterfind()) we need a non-empty prefix. Pass these functions a modified dictionary (here I modify the original dictionary, but this must be made only after the namespaces are registered).
self.namespaces['default'] = self.namespaces['']
Now, the functions from the find() family can be used with the default prefix:
print root.find('default:myelem', namespaces)
but
tree.write(destination)
does not use any prefixes for elements in the default namespace.

Extracting XML Attributes

I have an XML file with several thousand records in it in the form of:
<custs>
<record cust_ID="B123456#Y1996" l_name="Jungle" f_name="George" m_name="OfThe" city="Fairbanks" zip="00010" current="1" />
<record cust_ID="Q975697#Z2000" l_name="Freely" f_name="I" m_name="P" city="Yellow River" zip="03010" current="1" />
<record cust_ID="M7803#J2323" l_name="Jungle" f_name="Jim" m_name="" city="Fallen Arches" zip="07008" current="0" />
</custs>
# (I know it's not normalized. This is just sample data)
How can I convert this into a CSV or tab-delimited file? I know I can hard-code it in Python using re.compile() statements, but there has to be something easier, and more portable among diff XML file layouts.
I've found a couple threads here about attribs, (Beautifulsoup unable to extract data using attrs=class, Extracting an attribute value with beautifulsoup) and they have gotten me almost there with:
# Python 3.30
#
from bs4 import BeautifulSoup
import fileinput
Input = open("C:/Python/XML Tut/MinGrp.xml", encoding = "utf-8", errors = "backslashreplace")
OutFile = open('C:/Python/XML Tut/MinGrp_Out.ttxt', 'w', encoding = "utf-8", errors = "backslashreplace")
soup = BeautifulSoup(Input, features="xml")
results = soup.findAll('custs', attrs={})
# output = results [0]#[0]
for each_tag in results:
cust_attrb_value = results[0]
# print (cust_attrb_value)
OutFile.write(cust_attrb_value)
OutFile.close()
What's the next (last?) step?

If this data is formatted correctly -- as in, uses canonical XML -- you should consider lxml rather than BeautifulSoup. With lxml, you read the file, then you can apply DOM logic on it, including XPath queries. With your XPath queries, you can then get the lxml objects that represent each node that you're interested in, extract the data from them that you need, and rewrite them into an arbitrary format of your choosing using something like the csv module..
Specifically, in the lxml documentation, check out these tutorials:
Parsing from Strings and Files
The Element Class: Using XPath to Find Text

I (also) wouldn't use BeautifulSoup for this, and though I like lxml, that's an extra install, and if you don't want to bother, this is simple enough to do with the standard lib ElementTree module.
Something like:
import xml.etree.ElementTree as ET
import sys
tree=ET.parse( 'test.xml' )
root=tree.getroot()
rs=root.getchildren()
keys = rs[0].attrib.keys()
for a in keys: sys.stdout.write(a); sys.stdout.write('\t')
sys.stdout.write('\n')
for r in rs:
assert keys == r.attrib.keys()
for k in keys: sys.stdout.write( r.attrib[k]); sys.stdout.write('\t')
sys.stdout.write('\n')
will, from python-3, produce :
zip m_name current city cust_ID l_name f_name
00010 OfThe 1 Fairbanks B123456#Y1996 Jungle George
03010 P 1 Yellow River Q975697#Z2000 Freely I
07008 0 Fallen Arches M7803#J2323 Jungle Jim
Note that with Python-2.7, the order of the attributes will be different.
If you want them to output in a different specific order, you should sort or
order the list "keys" .
The assert is checking that all rows have the same attributes.
If you actually have missing or different attributes in the elements,
then you'll have to remove that and add some code to deal with the differences
and supply defaults for missing values. ( In your sample data, you have a
null value ( m_name="" ), rather than a missing value. You might want to check
that this case is handled OK by the consumer of this output, or else add some
more special handling for this case.

<product product_id='66656432' name='munch'><category>men</category></product>
In beautiful soup,
product=soup.find("product",attrs={})
then use attribute to access data like product["name"]