CODEWITHSUNDEEP
pythonregexstringpython-2.7dictionary
boldbrandywine
boldbrandywine

Reputation: 322

Creating a dictionary from a list of strings

I have a list of strings 

list = ['2(a)', '2(b)', '3', '3(a)', '1d', '5']

where it is intentional that the 1d, 3, and 5 don't involve parentheses.

I would like to create a dictionary which looks like this:

dict = {'2': 'a', '2': 'b', '3': 'a', '1': 'd'}

or

dict = {'2': ['a', 'b'], '3': ['a'], '1': ['d']}.

Essentially, ignore those strings without a letter a-z. I've used regular expressions to extract from the top list the following:

['a', 'b', 'a', 'd'],

but this hasn't helped me much in forming my dictionary easily.

Any help is much appreciated.

Upvotes: 1

Views: 1761

Answers (2)

ospahiu
ospahiu

Reputation: 3525

This is a good time to use setdefault() for a dictionary to define the structure of your dictionary. The first part involves capturing the numbers from the elements using a regex that captures all numbers. That list is then concatenated using join().

We then extract only alphabet characters using either a list comprehension -> [j for j in i if j.isalpha()], or pass as a generator j for j in i if j.isalpha() (generator in our case, joining the elements as a string together once again).

Lastly a check to see that both key and value exist so that we can set our dictionary to be of this format -> { '' : [] , ...}

import re

def to_dict(l):
    d = {}
    for i in l: 
        key = re.findall(r'\d+', i)
        value = ''.join(j for j in i if j.isalpha())
        if key and value:
            d.setdefault(''.join(key), []).append(value)    
    return d

Sample output:

l = ['2(a)', '2(b)', '3', '3(a)', '1c', '5']
print to_dict(l)
>>> {'1': ['c'], '3': ['a'], '2': ['a', 'b']}

Upvotes: 2

TigerhawkT3
TigerhawkT3

Reputation: 49330

Since a dictionary can't contain duplicate keys, use a defaultdict:

import collections
l = ['2(a)', '2(b)', '3', '3(a)', '1c', '5']
d = collections.defaultdict(list)
for item in l:
    num = ''.join(c for c in item if c.isdigit())
    word = ''.join(c for c in item if c.isalpha())
    if word and num:
        d[num].append(word)

Result:

>>> print(d)
defaultdict(<class 'list'>, {'2': ['a', 'b'], '1': ['c'], '3': ['a']})

Upvotes: 4

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