Change key value in Python 3 dictionary using regex

Question

Given the following dictionary, how can I return new keys that strip away all non-numbers from the keys?

import re 

dict {'abc_12': 5, 'abc_1': 7, 'abc_34': 10}

for k in dict:
    re.sub('[^0-9]', '', k)
    # change value of first key to '12': 5

You can't have duplicate keys. How do you want to handle that issue? — jh314
– jh314, Commented Aug 20, 2014 at 21:04

Daniel Roseman · Accepted Answer · 2014-08-20 21:09:11Z

5

If I understand you correctly, you want to replace the actual dictionary keys with new ones with all non-numeric characters removed.

The easiest way to do that would simply be to create a new dictionary:

new_dict = {re.sub('[^0-9]', '', k): v for k, v in my_dict.items()}

but if for some reason you absolutely need to preserve the existing one, you would need to pop each value and insert a new key with the existing value. You'd need to make sure you're operating on a static copy of the keys, otherwise the iteration would include the new values you've inserted and you'd get into all sorts of trouble.

for k in list(dict.keys()):
    v = my_dict.pop(k)
    new_k = re.sub('[^0-9]', '', k)
    my_dict[new_k] = v

answered Aug 20, 2014 at 21:09

Daniel Roseman

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Comments

dawg · Accepted Answer · 2014-08-20 21:23:51Z

1

You do not need a regex:

>>> di={'abc_12': 5, 'abc_1': 7, 'abc_34': 10}
>>> {''.join(c for c in k if c in '0123456789'):v for k, v in di.items()}
{'1': 7, '12': 5, '34': 10}

answered Aug 20, 2014 at 21:23

dawg

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Comments

Alex Riley · Accepted Answer · 2014-08-20 21:42:06Z

1

I think you're best off creating a new dictionary:

old_dict = {'abc_12': 5, 'abc_1': 7, 'abc_34': 10}

new_dict = {re.sub('[^0-9]', '', key): old_dict[key] for key in old_dict}

edited Aug 20, 2014 at 21:42

answered Aug 20, 2014 at 21:10

Alex Riley

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2 Comments

Stephan Over a year ago

are you sure that you're iterating over keys in that loop?

Alex Riley Over a year ago

Pretty sure - for key in dict iterates over the keys of the dictionary.

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