python - Sort dictionary by key alphabetically -


this question has answer here:

i'm having trouble sorting dictionary alphabetically keys.

here's code:

colorsizes = {'rust': ['size 8', 'size 10', 'size 12', 'size 14', 'size 16', 'size 18'],                'middle blue': ['size 8', 'size 10', 'size 12', 'size 14', 'size 16', 'size 18'],                'grey': ['size 8', 'size 10', 'size 12', 'size 14', 'size 16', 'size 18'],                'aqua': ['size 8', 'size 10', 'size 12', 'size 14', 'size 16', 'size 18'],                'navy': ['size 8', 'size 10', 'size 12', 'size 14', 'size 16']}  realcolor = {} key in sorted(colorsizes.keys()):     realcolor[key] = colorsizes.get(key)  print(realcolor)

what get:

{'yellow/fuschia':['large', 'extra large'], 'black':['small', 'medium', 'large']}

what wanna get:

{'black':['small', 'medium', 'large'], 'yellow/fuschia':['large', 'extra large']}

thanks!

dictionaries in python versions < 3.6 unordered, sorting , reinserting meaningless.

as fix, either

  1. switch python3.6 (keep in mind caveats), or

  2. use ordereddict

for second option, replace realcolor = {} collections.ordereddict:

from collections import ordereddict     realcolor = ordereddict()

here's example of how ordereddict remembers order of insertion:

dict1 = {} dict1['k'] = 1 dict1['asdfdf'] = 1234  print(dict1) # {'asdfdf': 1234, 'k': 1}  collections import ordereddict dict2 = ordereddict() dict2['k'] = 1 dict2['asdfdf'] = 1234  print(dict2) # ordereddict([('k', 1), ('asdfdf', 1234)])

the __repr__ might different, latter still dictionary , can used accordingly.


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