Parse Dictionary Values Python

Question

How could I parse the dictionary below, so that it's values only contain ticket numbers?

Current Dictionary:

{'8.8.8.8': 'Open Menu  10A-003272 10A-003328 10A-003652', '8.8.8.9': '10A-003069 10/21/2016', '8.8.8.10': 'Open Menu  10A-003145 10/21/2016'}

Objective Dictionary:

{'8.8.8.8': '10A-003272 10A-003328 10A-003652', '8.8.8.9': '10A-003069', '8.8.8.10': '10A-003145'}

Code used to make dictionary:

with open(esccbList, 'r') as f:
    d = {}
    for line in f:
        d[line.strip()] = next(f, '').strip()

Regex to find ticket numbers:

n = re.search(r'10A-\d{6}',item, re.M|re.I)

@WasiAhmad Yes, I've included the regex code that would find each ticket. — someGuy45
– someGuy45, Commented Dec 13, 2016 at 17:56

Moinuddin Quadri · Accepted Answer · 2016-12-13 17:56:13Z

2

Assuming that your ticket number substring will only contain hyphen -, you may use a dict comprhension to achieve this like:

my_dict = {'8.8.8.8': 'Open Menu  10A-003272 10A-003328 10A-003652', '8.8.8.9': '10A-003069 10/21/2016', '8.8.8.10': 'Open Menu  10A-003145 10/21/2016'}

new = {k: ' '.join(i for i in v.split() if '-' in i) for k, v in my_dict.items()}

Final value hold by new dict will be:

{'8.8.8.9': '10A-003069', 
 '8.8.8.10': '10A-003145', 
 '8.8.8.8': '10A-003272 10A-003328 10A-003652'}

answered Dec 13, 2016 at 17:56

Moinuddin Quadri

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2 Comments

Jean-François Fabre Over a year ago

@someGuy45: now you can also upvote an answer besides accepting it.

Wasi Ahmad Over a year ago

@someGuy45 what is the point of using regular expression then?

Wasi Ahmad · Accepted Answer · 2016-12-13 20:26:59Z

1

I have updated my answer to print the dictionary in desired format.

import re

pattern = re.compile(r'10A-\d{6}')
info = {'8.8.8.8': 'Open Menu  10A-003272 10A-003328 10A-003652', 
        '8.8.8.9': '10A-003069 10/21/2016', 
        '8.8.8.10': 'Open Menu  10A-003145 10/21/2016'}

output = {}
for key, value in info.items():
    tokens = value.split()
    val = ''
    for token in tokens:
        if pattern.match(token):
            val = val + token + ' '
    val = val.strip()
    output[key] = val;

print(output)

It prints:

{'8.8.8.8': '10A-003272 10A-003328 10A-003652', 
 '8.8.8.9': '10A-003069', 
 '8.8.8.10': '10A-003145'}

edited Dec 13, 2016 at 20:26

answered Dec 13, 2016 at 18:00

Wasi Ahmad

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Comments

Shaun · Accepted Answer · 2016-12-13 17:57:58Z

0

d = { k, clean_ticket(v) for k,v in original_dict.items() if is_ticket(v) }

Looks like is_ticket should be something like

def is_ticket(v):
    return "Open Menu" in v

Make a function clean_ticket(v) that strips off the Open Menu

def clean_ticket(v):
    return v.split("Open Menu")[1].strip()

Something like that.

answered Dec 13, 2016 at 17:57

Shaun

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Comments

Erotemic · Accepted Answer · 2016-12-13 17:57:11Z

-2

I assume you have some function

def is_ticket_number(item):
    """ returns True only if item is a ticket number """
    return re.search(r'10A-\d{6}',item, re.M|re.I)

Then all you need to do is

d = {k: v for k, v in d.items() if is_ticket_number(v)}

answered Dec 13, 2016 at 17:57

Erotemic

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