ElementTree parsing in python

Question

I have xml of the form

<root>
  <tag1>   </tag1>
  <tag2>   </tag2>
  <tag3>   </tag3>

  <tag1>   </tag1>
  <tag2>   </tag2>
  <tag3>   </tag3>
</root>

I need to parse the xml in the order

tag1 -> tag2 -> tag3 -> tag1 -> tag2 -> tag3

Currently I'm using

root = tree.getroot()
for data in root.findall('tag1')
    do_operations(data)
for data in root.findall('tag2')
    do_operations(data)

But this approach is giving me and that's obvious

tag1 -> tag1 -> tag2 -> tag2 -> tag3 -> tag3

which is not what I want.

Can you suggest an optimum method in which i can pasrse the XML in the desired manner. tag1 , tag2, tag3 are repeated a lot in the same order as given above.

What module/library are you using??

pradyunsg
– pradyunsg

2013-03-28 09:46:06 +00:00
Commented Mar 28, 2013 at 9:46 — pradyunsg
– pradyunsg, Commented Mar 28, 2013 at 9:46
@Schoolboy ElementTree?

Fredrik Håård
– Fredrik Håård

2013-03-28 09:54:57 +00:00
Commented Mar 28, 2013 at 9:54 — Fredrik Håård
– Fredrik Håård, Commented Mar 28, 2013 at 9:54

DSM · Accepted Answer · 2013-03-28 09:44:46Z

2

IIUC, can't you simply loop over root itself?

>>> for data in root:
...     print data
...     
<Element tag1 at 0x102dea7d0>
<Element tag2 at 0x102dea8c0>
<Element tag3 at 0x102dd6d20>
<Element tag1 at 0x102dea7d0>
<Element tag2 at 0x102dea8c0>
<Element tag3 at 0x102dd6d20>

answered Mar 28, 2013 at 9:44

DSM

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4 Comments

DSM Over a year ago

@Abhishek: those are just the hex versions of the results of id(), as a way to distinguish tags with the same name just from the __repr__.

Abhishek Over a year ago

What if I have them repeating 100 times in the XML. I cannot write them in this fashion

DSM Over a year ago

@Abhishek: I don't understand. I didn't write those, those are the results of print data. Instead of for data in root.findall('tag1'):, just use for data in root:, is the point.

Abhishek Over a year ago

Ok got you now. I thought they have to be hardcoded in program. Got your point.

Fredrik Håård · Accepted Answer · 2013-03-28 09:48:35Z

1

You can iterate over the children instead of using find:

for child in root:
    do operations...

If you do different operations to different tags, you can use child.tag to determine what to do:

for child in root:
    if child.tag == 'tag1':
       do operations
    elif child.tag == 'tag2':
       do other operations
    ...

Or you could put the operations in a dict and avoid the if-elif-else incantation.

answered Mar 28, 2013 at 9:48

Fredrik Håård

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