loop over list and process into groups

[Sneaky Wombat]

[ {'vlan_or_intf': 'VLAN2021'},
{'vlan_or_intf': 'Interface'},
{'vlan_or_intf': 'Po1'},
{'vlan_or_intf': 'Po306'},
{'vlan_or_intf': 'VLAN2022'},
{'vlan_or_intf': 'Interface'},
{'vlan_or_intf': 'Gi7/33'},
{'vlan_or_intf': 'Po1'},
{'vlan_or_intf': 'Po306'},
{'vlan_or_intf': 'VLAN2051'},
{'vlan_or_intf': 'Interface'},
{'vlan_or_intf': 'Gi9/6'},
{'vlan_or_intf': 'VLAN2052'},
{'vlan_or_intf': 'Interface'},
{'vlan_or_intf': 'Gi9/6'},]

I want it to be converted to:

[{'2021':['Po1','Po306']},{'2022':['Gi7/33','Po1','Po306']},etc etc]

I was going to write a def to loop through and look for certain pre-
compiled regexs, and then put them in a new dictionary and append to a
list, but I'm having trouble thinking of a good way to capture each
dictionary. Each dictionary will have a key that is the vlan and the
value will be a list of interfaces that participate in that vlan.
Each list will be variable, many containing only one interface and
some containing many interfaces.

I thought about using itertools, but i only use that for fixed data.
I don't know of a good way to loop over variably sized data. I was
wondering if anyone had any ideas about a good way to convert this
list or dictionary into the right format that I need. The solution I
come up with will most likely be ugly and error prone, so I thought
i'd ask this python list while I work. Hopefully I learn a better way
to solve this problem.

Thanks!

I also have the data in a list,

[ 'VLAN4065',
'Interface',
'Gi9/6',
'Po2',
'Po3',
'Po306',
'VLAN4068',
'Interface',
'Gi9/6',
'VLAN4069',
'Interface',
'Gi9/6',]

[mk]

Sneaky Wombat wrote:
> I was going to write a def to loop through and look for certain pre-
> compiled regexs, and then put them in a new dictionary and append to a
> list,

regexes are overkill in this case I think.

> [ 'VLAN4065',
> 'Interface',
> 'Gi9/6',
> 'Po2',
> 'Po3',
> 'Po306',
> 'VLAN4068',
> 'Interface',
> 'Gi9/6',
> 'VLAN4069',
> 'Interface',
> 'Gi9/6',]

Why not construct an intermediate dictionary?

elems = [ 'VLAN4065',

'Interface',
'Gi9/6',
'Po2',
'Po3',
'Po306',
'VLAN4068',
'Interface',
'Gi9/6',
'VLAN4069',
'Interface',
'Gi9/6',]

def makeintermdict(elems):
vd = {}
vlan = None
for el in elems:
if el.startswith('VLAN'):
vlan = el.replace('VLAN','')
elif el == 'Interface':
vd[vlan] = []
else:
vd[vlan].append(el)
return vd

def makelist(interm):
finlist = []
for k in interm.keys():
finlist.append({k:interm[k]})
return finlist

if __name__ == "__main__":
intermediate = makeintermdict(elems)
print intermediate
finlist = makelist(intermediate)
print 'final', finlist

{'4068': ['Gi9/6'], '4069': ['Gi9/6'], '4065': ['Gi9/6', 'Po2', 'Po3',
'Po306']}
final [{'4068': ['Gi9/6']}, {'4069': ['Gi9/6']}, {'4065': ['Gi9/6',
'Po2', 'Po3', 'Po306']}]

I hope this is not your homework. :-)

Regards,
mk

[Sneaky Wombat]

Thanks mk,

My approach was a lot more complex than yours, but your's is better.
I like itertools and was using islice to create tuples for (start,end)
string slicing. Too much work though. Thanks!

-j

[lbolla]

===================================

from itertools import groupby

data = \

[ {'vlan_or_intf': 'VLAN2021'},
{'vlan_or_intf': 'Interface'},
{'vlan_or_intf': 'Po1'},
{'vlan_or_intf': 'Po306'},
{'vlan_or_intf': 'VLAN2022'},
{'vlan_or_intf': 'Interface'},
{'vlan_or_intf': 'Gi7/33'},
{'vlan_or_intf': 'Po1'},
{'vlan_or_intf': 'Po306'},
{'vlan_or_intf': 'VLAN2051'},
{'vlan_or_intf': 'Interface'},
{'vlan_or_intf': 'Gi9/6'},
{'vlan_or_intf': 'VLAN2052'},
{'vlan_or_intf': 'Interface'},
{'vlan_or_intf': 'Gi9/6'},]

def clean_up(lst):
return [d.values()[0] for d in data if d.values()[0] != 'Interface']

out = {}
for k, g in groupby(clean_up(data) , key=lambda s:
s.startswith('VLAN')):
if k:
key = list(g)[0].replace('VLAN','')
else:
out[key] = list(g)

print out
===================================

hth,
L.

[nn]

Good use of groupby. Here is what I ended up coming up:

from itertools import groupby
laninfo=[ 'VLAN4065',

'Interface',
'Gi9/6',
'Po2',
'Po3',
'Po306',
'VLAN4068',
'Interface',
'Gi9/6',
'VLAN4069',
'Interface',
'Gi9/6',]

def splitgrp(s, f=[False]):
f[0]^=s.startswith('VLAN')
return f[0]
lanlst=(list(g) for k,g in groupby(laninfo,key=splitgrp))
out={item[0][4:]:item[2:] for item in lanlst}
print(out)

[Paul Rubin]

lbolla <lbo...@gmail.com> writes:
> for k, g in groupby(clean_up(data) , key=lambda s: s.startswith('VLAN')):
> if k:
> key = list(g)[0].replace('VLAN','')

This is the nicest solution, I think. Mine was more cumbersome.

[mk]

Sneaky Wombat wrote:
> [ 'VLAN4065',
> 'Interface',
> 'Gi9/6',
> 'Po2',
> 'Po3',
> 'Po306',
> 'VLAN4068',
> 'Interface',
> 'Gi9/6',
> 'VLAN4069',
> 'Interface',
> 'Gi9/6',]

Hey, I just invented a cute ;-) two-liner using list comprehensions:

# alist = list above

tmp, dk = [], {}
[(x.startswith('VLAN') and (dk.setdefault(x,[]) or tmp.append(x))) or
(not x.startswith('VLAN') and dk[tmp[-1]].append(x)) for x in alist
if x != 'Interface']

No need to use a nuke like itertools to kill a fly. ;-)

Regards,
mk

[nn]

Oh my! You could have at least used some "if else" to make it a little
bit easier on the eyes :-)

[(dk.setdefault(x,[]) or tmp.append(x))
if x.startswith('VLAN')
else dk[tmp[-1]].append(x)

[lbolla]

It looks like Perl ;-)

[mk]

nn wrote:

> Oh my! You could have at least used some "if else" to make it a little
> bit easier on the eyes :-)

That's my entry into """'Obfuscated' "Python" '"''code''"'
'"contest"'""" and I'm proud of it. ;-)

Regards,
mk

[mk]

lbolla wrote:
> It looks like Perl ;-)

A positive proof that you can write perl code in Python. I, for
instance, have had my brain warped by C and tend to write C-like code in
Python. That's only half a joke, sadly. I'm trying to change my habits
but it's hard.

Regards,
mk