I don't know list, I not good at list.

Ellerbee, Edward EEllerbee at BBandT.com
Thu Jul 14 15:05:05 CEST 2011

Thank you all for the advice, let me spin this in a different way.

I've built a program that goes to the NANPA website, scrapes area
code/exchange (npa/nxx) digits for a specified area - be it carolina,
alabama, texas, etc - drops it into a file, then massages the data and
prints out the correct format to insert into a voice router. The code is
ugly, but it works great.

The thing I haven't been able to get my script to do is to reduce the
amount of dial-peers. Hence the need to reduce the numbers to the least
common denominator, and put it in the xxxxx[xx] format.If we had a set
number of digits, we could build a dictionary(unless it's possible to do
that dynamically). 

So, a couple assertions:
1. the data will always be 6 digit numbers (in a string format)
2. the data is in a txt file after being put there from my script
3. the data will never be the same (I'm going to use this for site
conversions/new site builds
	e.g. today I might be dealing with 252-, 919- and 704- area
codes, tomorrow might be 304- and 754-
4. I wanted a script to reduce the time taking to build the dial-peers
manually. I'd previously spent 3-4 hours
   on gathering and processing data. The script I have so far pulls data
and massages in about 6 seconds
5. I'm using python 2.7 - it seems like it had more module availability
than 3

And a couple question:
1. Would lists be the best way to handle this data? Would it be better
to process this data from a file?
2. Is there a way to determine the common (first-5 digits) denominator
among a list (or file) of data?
3. and... Could those common numbers be inserted in a dict for

Sorry for the book!


Edward Ellerbee

-----Original Message-----
From: python-list-bounces+eellerbee=bbandt.com at python.org
[mailto:python-list-bounces+eellerbee=bbandt.com at python.org] On Behalf
Sent: Wednesday, July 13, 2011 5:59 PM
To: python-list at python.org
Subject: Re: I don't know list, I not good at list.

 > I've been beating my head against the desk trying to figure out a  >
method to accomplish this:
 > Take a list (this example is 5 items, It could be 150 or more - i.e.
 > it's variable length depending on the city/local calling zones)  >  >
The first 6 digits of phone numbers(NPA/NXX) in a local calling area.
 > I want to concatenate the last digit for insertion into a call  >
routing pattern.
 > I tried this and failed miserably:
 > list1=['252205','252246','252206','252247','252248']
 > for item in list1:
 >         try:
 >                 item1=list1[0]
 >                 item2=list1[1]
 >                 if item1[0:5] == item2[0:5]:
 >                         print item1[0:5] + '[' + item1[5:6] + 
item2[5:6] + ']'
 >                         list1.pop(0)
 >                 else:
 >                         print item1
 >                         list1.pop(0)
 >         except:
 >                 try:
 >                         print item1
 >                         list1.pop(0)
 >                 except:
 >                         pass
 > #-----------------------------------------------------------------
 > My intent is to have the end data come out (from the example list  >
above) in the format of  > 25220[56]  > 25224[678]  >  > I tried putting
together a variable inserted into a regular  > expression, and it
doesn't seem to like:
 > Item1=list1[0]
 > Itemreg = re.compile(Item1[0:5])
 > For stuff in itemreg.list1:
 >         #do something
 > Can somebody throw me a bone, code example or module to read on  >
python.org? I'm a n00b, so I'm still trying to understand functions  >
and classes.
 > I thought the experts on this list might take pity on my pathetic  >
code skillz!
defaultdict comes in handy:

 >>> list1 = ['252205','252246','252206','252247','252248']
 >>> from collections import defaultdict  >>> d = defaultdict(set)  >>>
for item in list1:
	d[item[ : 5]].add(item[5 : ])

 >>> d
defaultdict(<class 'set'>, {'25224': {'8', '7', '6'}, '25220': {'5',
'6'}})  >>> for k, v in d.items():
	print(k + "[" + "".join(v) + "]")


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