CODEWITHSUNDEEP
python
LalitK
LalitK

Reputation: 3

python dictionary values into tuple based on condition

I am trying to get an output from below dict as a tuple mentioned below-

Input: b = {'a':'1','S1':'OptionA','P1':'100','S2':'', 'P2':'','S3':'OptionB','P3':'80'}

Output : [('OptionA', '100'), ('OptionB', '80')]

i have coded for this like below but i want a shorter method , can anyone please suggest -

import re
b = {'a':'1','S1':'OptionA','P1':'100','S2':'', 'P2':'','S3':'OptionB','P3':'80'}

c =[]
for k,v in b.items():
    if k.startswith('S') and v:
        for i,j in b.items():
            if i.startswith('P') and re.search(r'\d+$', k).group() == re.search(r'\d+$', i).group():
                c.append(tuple([v,j]))

print(c)

Upvotes: 0

Views: 140

Answers (2)

randomir
randomir

Reputation: 18697

Maybe with a list comprehension oneliner?

>>> b = {'a':'1','S1':'OptionA','P1':'100','S2':'', 'P2':'','S3':'OptionB','P3':'80'}
>>> [(v, b['P'+k[1:]]) for k,v in b.items() if re.match('^S\d+$',k) and v and 'P'+k[1:] in b]
[('OptionB', '80'), ('OptionA', '100')]

Only non-empty values for matching S<digits> are paired with P<digits>.

Update for the case from comments. If you need to match Stgy1 with Per1, the list comprehension solution starts to lose its charm and becomes a bit unreadable. If you can't simplify your pairing criteria, the for loop is probably a cleaner way to go.

>>> b = {'a':'1','Stgy1':'OptionA','Per1':'100','Stgy2':'', 'Per2':'','Stgy3':'OptionB','Per3':'80'}
>>> [(v, w) for s,v in b.items() for p,w in b.items() if s[0]=='S' and p[0]=='P' and v and w and re.search('\d+$',s).group()==re.search('\d+$',p).group()]
[('OptionB', '80'), ('OptionA', '100')]

Upvotes: 0

chepner
chepner

Reputation: 531808

I would just use exception handling to ignore keys that don't fit your pattern:

c = []
for k, v in b.items():
    if not k.startswith('S') or not v:
        continue
    new_key = v
    try:
        n = int(k[1:])
        new_value = b['P%d' % (n,)]
    except KeyError, ValueError:
        continue
    c.append((new_key, new_value))

Having fewer lines doesn't necessarily improve your code.

Upvotes: 1

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