CODEWITHSUNDEEP
pythontry-catch
user3501855
user3501855

Reputation: 131

Condense successive 'try' blocks

I'm trying to take an existing dict with, say, five possible entries and move its entries into a new dict, 'translating' the keys along the way. The keys of the first dict are structured in a way that the move cannot be done procedurally. There is no guarantee that all of the entries will be present in the first dict.

In order to avoid KeyError, each entry can be assigned within its own try block

first_dict = {
   'key_1': foo1,
   'key_two': foo2,
   'key_three': foo3,
   'key_number_4': foo4,
   'key_5': foo5
}

second_dict = {}

try: 
    second_dict['translated_key_1'] = first_dict['key_1']
except KeyError:
    pass

try:
    second_dict['translated_key_2'] = first_dict['key_two']
except KeyError:
    pass

...

try:
    second_dict['translated_key_5'] = first_dict['key_5']
except KeyError:
    pass


   'translated_key_2': first_dict['key_two'],
   'translated_key_3': first_dict['key_three'],
   'translated_key_4': first_dict['key_number_4'],
   'translated_key_5': first_dict['key_5'],
}

Perhaps a better way is to check if the entry exists in the first dict and then assign it.

if 'key_1' in first_dict:
    second_dict['translated_key_1'] = first_dict['key_1']
if 'key_two' in first_dict:
    second_dict['translated_key_2'] = first_dict['key_two']
...

In either case, is there a way to condense this? When the number of elements is large it seems like this would become unnecessarily bulky. Is there a way to iterate through the relationships without creating a try block for each?

Upvotes: 2

Views: 123

Answers (1)

cwallenpoole
cwallenpoole

Reputation: 82008

You could do this with a loop:

# This is a simple mapping which associates the old keys with the new ones
translations = {'translated_key_1': 'key_one',
                'translated_key_2': 'key_two'}

# iterate through the map
for k,v in translations.iteritems():
    # you could also replace with try except
    if v in first_dict:
        second_dict[k] = first_dict[v]

Upvotes: 7

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